Suppose a 100-gram mass, when attached to a spring, causes the spring to stretch 8 cm.
What would you expect a 25-gram mass to do when attached to the same spring?
(1 point)
O Stretch the spring 4 cm.
O Stretch the spring 8 cm.
O Stretch the spring 32 cm.
O Stretch the spring 2 cm.

Answers

Answer 1

Answer:Apparently the further displacement should be of other 3 cm; but, let's consider the development of the spring on one vertical plain, and fixed on a wall by the upper end.

Now the spring doesn't look circular; but as a stick fixed on the wall. Before placing the first 100 grams weight, the stick is not horizontal; but its overhanging end is lower by the measured length of the spring at rest. The length line of the spring with its horizontal projection line forms an angle.

The first 100 grams load is decomposed in function of the new angle: one component in line with the stick, the other one perpendicularly respect to it. The same happens with 200 grams.

The flection momentum is given by the perpendicular component of the load with respect to the stick line in overload, and the measure of the stick itself. Such component is inversely proportional to the other component, and decreases in relation to higher load while the other component (that doesn't produce flexion) encreases.

I assume that I can't calculate the new displacement because is not given the height of the spring at rest, the length of the development of the spring. Accordingly, I cannot calculate all various angles of the stick in relation with gravity.

Explanation:

Answer 2

When a 25 g mass is attached to the same spring, the spring is stretched by 2 cm.

What is Hook's law?

Hooke's law states that the applied force [F] equals a constant k times the displacement or change in length [x].

Mathematically - F = - Kx

Given is a situation in which a stretch of 8 cm is caused by a 100g mass.

From this, we can write -

Mass [m] = 100 g = 0.1 Kg

Displacement [x] = 8 cm = 0.08 m

According to the question -

The spring constant [K] can be calculated using Hook's law as -

F = -Kx

mg = -Kx

K = - mg/x

K = - (0.1 x 9.8) / 0.08

K = - 12.25 N/m

|K| = 12.25 N/m

Now, when mass [M] = 25g = 0.025 Kg, using Hook's law -

|F| = |- kx| = kx

Mg = 12.25x

x = Mg/12.25

x = (0.025 x 9.8)/12.25

x = 0.02

x = 2 cm

Therefore, when a 25 g mass is attached to the same spring, the spring is stretched by 2 cm.

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Related Questions

Two kg of refrigerant 134a undergoes a polytropic process in a piston–cylinder assembly from an initial state of saturated vapor at 2 bar to a final state of 12 bar, 80°c. Determine the work for the process, in kj.

Answers

The work done by polytropic process is -69.88 kJ.

The work for the polytropic process is determined using  

P . v^n = constant.

where P is pressure, V is specific volume

As we know that

W = m∫Pdv

W = m(P2v2 - P1v1) / (1 - n)

In order to evaluate this expression, we need to determine the specific volumes and the polytropic exponent, n.

State 1: v1 = vg1 = 0.0993 m3/kg

State 2: at 12 bar, 80oC; v2 = 0.02051 m3/kg

Hence by using polytropic exponent

P1 / P2 = (v2 / v1)^n

n = ln(2/12) / ln(0.02051/0.0993)

n = 1.136

Find the work done

W = m(P2v2 - P1v1) / (1 - n)

W = 2kg(12bar . 0.02051m³/kg - 2 . 0.0993) / (1 - 1.136) x (10⁵ N/m²/1 bar) x (1 kJ / 10³ Nm)

W = -69.88 kJ

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Use differentials to estimate the amount of metal in a closed cylindrical can that is 10 cm high and 4 cm in diameter if the metal in the top and bottom is 0. 1 cm thick and the metal in the sides is 0. 05 cm thick.

Answers

The amount of metal in a closed cylindrical can that is 10 cm high and 4 cm in diameter if the metal on the top and the bottom is 0.1 cm thick and the metal on the sides is 0.05 cm thick is 8.8 cm.

The formula for calculating the volume of a cylinder is given below.

V = πr^2 h

Get the differential of the volume as shown:

dV = V/ h dh + V / r dr

V/ h = πr^2

V/ h  = 2  πr h

Now, the differential becomes

dV =  πr^2dh +  2πrh dr

Given the following parameters i.e. diameter and height

dh = 0.1 + 0.1 = 0.2 cm

dr = 0.05 cm

h = 10 cm

d = 4 cm  

r = 2cm

Substituting the values in the above equation, we get

dV = 3.14(2)^2(0.2)  + 2(3.14)(2)(10)(0.05)

dV = 2.512 + 6.28

dV = 8.792 cm

dV = 8.8 cm

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a baseball is thrown straight into the air with an initial velocity of 85 mph. How far will the ball travel if it takes 5.4 seconds to land?

A) 316.0 m
B) 179.3 m
C) 432.5 m
D) 62.71 m ​

Answers

D. The distance traveled by the baseball during the given time is 62.32 m.

Distance traveled by the baseball

The distance traveled by the baseball is calculated as follows;

s = vt - ¹/₂gt²

where;

s is the distance traveled by the baseballv is the initial velocity of the baseballg is acceleration due to gravityt is the time of motion of the baseball

The given parameters;

v, initial velocity of the baseball = 85 mph = 38 m/s

g is acceleration due to gravity = 9.8 m/s²

t is the time of motion, = 5.4 seconds

Substitute the given parameters and solve for the distance travelled by the baseball as shown below.

s = (38 m/s)(5.4 s) - ¹/₂(9.8 m/s²)(5.4 s)²

s = 205.2 - 142.88

s = 62.32 m

Thus, the distance traveled by the baseball during the given time is 62.32 m.

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A man fire a bullet with a velocity of 500 m/s velocity upward.( use gravity 10m/s2.
calculate
A the maximum could the bullet rise
B THE VELOCITY OF THE BULLET AT t=10s ,20s and 60s
C. the time require to move up
D. the time of flight
E the height it rise at t=20s
F. the height it rise when first velocity equal to 40 m/s

Answers

The maximum height reached by the bullet is 12,500 m.

The velocity of the of the bullet at 10s is 600 m/s.

The velocity of the of the bullet at 20s is 700 m/s.

The velocity of the of the bullet at 60s is 1,700 m/s.

The time required for the object to move to maximum height is 50.5 seconds.

The height reacted by the bullet at time, t = 20 seconds is 12,000 m.

The height risen by the bullet when the first velocity is 40 m/s is 80 m.

What is the maximum height reached by the bullet?

The maximum height reached by the bullet is calculated as follows;

v² = u² - 2gh

v² = 2gh

h = v² /2g

h = (500²)/(2 x 10)

h = 12,500 m

Velocity of the of the bullet at various time interval

v = u + at

when t = 10 s

v = 500 m/s  +  10(10)

v = 600 m/s

when the time, t = 20 seconds

v = 500 + 20(10)

v = 700 m/s

when the time, t = 60 s

v = 500 + 20(60)

v = 1,700 m/s

Time required for the object to move to maximum height

t = √(2h/g)

t = √(2 x 12,500 / 9.8)

t = 50.5 seconds

Height reacted by the bullet at time, t = 20 seconds

h = ut + ¹/₂gt²

h = (500 x 20)  +  ¹/₂(10)(20²)

h = 12,000 m

Height risen by the bullet when the first velocity is 40 m/s

h = v² /2g

h = (40²)/(2 x 10)

h = 80 m

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Abby runs at a constant speed and travels 10 meters in 4 seconds. if abby runs for 8 seconds at this constant speed, how far will he travel?

Answers

If Abby runs for 8 seconds at this constant speed, Abby will travel 20 meters at this constant speed.

Calculation

The Speed = 10/4 = 2.5 m/s

Distance = speed x time d=st

In 8 seconds, d=2.5 x 8 = 20 meters

What is the definition of movement speed?

Velocity is the pace and direction of an object's movement, whereas speed is the time rate at which an object is traveling along a path. In other words, velocity is regarded as a vector, whereas speed is a scalar value.

Use the speed formula, s = d/t, which states that speed is equal to distance divided by time, to solve for speed or rate.

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a chain lying on the ground is 10 m long and its mass is 90 kg. how much work (in j) is required to raise one end of the chain to a height of 6 m? (use 9.8 m/s2 for g.)

Answers

Amount of work done required to raise one end of the chain is 158.6 J

Work is said to be done when a  force act on a body and it displaced from its position.

It's unit is joule.

W = F × S

where

W = work done

F = force applied

S = displacement

Force acting on the chain with respect to the distance is given by

F = mgx / 10

m is mass

m = 90 kg

g is gravity (9.8 m/s2 for g.)

90× 9.8× x / 10

F = 88.2x Newton

To calculate the amount of work done.

[tex]W = \int\limits^f_i{F(x)} \, dx[/tex]

[tex]W = \int\limits^6_0 {F(x)} \, dx[/tex]

[tex]W = 88.2\int\limits^6_0 {x} \, dx[/tex]

[tex]W = 88.2 | \frac{x^2}{2}|^{6} _{0}[/tex]

W = 88.3( 6² - 0²)/2

W = 88.2 (36- 0) /2

W = 88.2 × 18

W = 158.6 J

Amount of work done required to raise one end of the chain is 158.6 J

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a train startes from rest and accelerate uniformly until it attains a velocity of 8m/s in 10 seconds. calculate the acceleration of the train​

Answers

The rate at which an object's velocity with respect to time changes is referred to as acceleration in mechanics. They are vector quantities, accelerations. The direction of the net force applied on the item determines the object's acceleration.

A train starts from rest and accelerate uniformly until it attains a velocity of 8m/s in 10 seconds. calculate the acceleration of the train​.

a = 8/10

a = 0.8m/s2

The equation a = v/t denotes acceleration (a), which is the change in velocity (v) over the change in time (t). This enables you to calculate the change in velocity in meters per second squared (m/s2). Since acceleration is a vector quantity, both its magnitude and its direction are included.

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An ocean liner is cruising at 10 meters/second and is about to approach a stationary ferryboat. A parcel is released from the ocean liner from a height of 5.5 meters above the ferryboat’s deck. Calculate the distance at which the ferryman should position the boat from the point horizontally below the point of release so that the parcel lands inside the boat.

Answers

Answer:

If a stream channel is 5 meters wide and 3 meters deep, and the velocity of the water moving through the channel is 10 meters/second, what is the stream's discharge (include units)? ... and deposition (point bars) occurs on the inside

Explanation:

You are driving down the road at a constant speed. Another car going a bit faster catches up with you and passes you. Choose the correct position graph for both vehicles on the same set of axes, and notes the point on the graph where the other vehicle passes you.

Answers

Answer: Option 4 (last graph)

Explanation:

1. Since you're driving at a constant speed, the line for your car will be linear. This rules out the first option.

2. This is a position time graph (x-t) graph. This means that to be moving forwards, there should be a positive slope. Option 3 can be eliminated as your car has a slope of 0. A slope of 0 on a position time graph shows an object at rest, not an object in motion.

3. The 2nd option can be ruled out as it represents that you are going faster than the other car. The steeper the slope = the faster the speed.

4. This ultimately leaves you with the last option as your final answer!

Which timeline best shows the history development of cell theory

Answers

Answer:

you did not show a timeline luckly i did the the quiz it is the second timeline

Explanation:

Answer:

b

Explanation:

I did the test

A car acccelerates at 4m/s/s. Assuming the car starts from rest, how much time does it need to accelerate to a speed of 40 m/s?

Answers

A car is travelling with a acceleration of 4 m/s². Assuming the car starts from rest, The time need to accelerate to the speed of 40 m/s is (t)=10 s.

What is time?

Time is a scalar quantity. its shows the period when the process happened. It can be measured in hour, minute, seconds etc.

How can we calculate the time?

To calculate the time need to accelerate to the wanted speed we are using the formula, v=u+at

Here we are given,

u= the initial velocity of the car = 0m/s [Note: the car starts from rest]

a= the  acceleration of the car = 4 m/s².

v= the final velocity of the car = 40 m/s.

we have to calculate the time need to accelerate to the wanted speed =t

Now, we put the values in the above equation,

v=u+at

Or, t= v/a [Note: u=0]

Or, t= 40/4

Or, t=10 m/s.

From the above calculation, we can say that the time need to accelerate to the wanted speed is (t)=10 s

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How far did car A travel in the time interval t = 0 to t = 6?

Answers

Answer:  Problem 03.043 - Velocity-time graph The figure shows a plot of vx(for a car traveling in a straight line. Vx (m/s) 20 | 15 20 7(s) What is aav.x between t = 9.00 s and t = 110 s? 7 m/s2 What is Vav,x for the time interval from t= 9.00 s to r= 110 s? What is Vav.x for the interval t = 0 to t= 15.0 s? m/s What is the increase in the car's speed between 10.0 s and 15.0 s? m/s How far does the car travel from time t = 100 s to t = 170 s?

The answer is 03.043 but i dont know maybe my brain is incorrect but I got 03.043. :| hope it helped if its correct.

Explanation: Well, I havent done math in a while bc im not in school anymore so i forgot all that stuff :|

steam is accelerated in a steady-flow nozzle from a low velocity to a velocity of 210 m/s at a rate of 3.2 kg/s. if the temperature and pressure of the steam at the nozzle exit are 200 ºc and 600 kpa, what is the exit area of the nozzle?

Answers

The exit area of the nozzle is 0.000861 m² which is calculated using ideal gas equation.

What is an ideal gas?

Velocity = 280 m/s

Rate = 2.5 kg/s

Pressure = 2 MPa

Temperature = 400°C

Using equation of Ideal gas

PV= RT

V= RT/P

V= 0.0965 [tex]m^{3}[/tex]/ kg

Using equation of continuity

(dm/dt) = [tex]A_{1}[/tex]v/V

[tex]A_{1}[/tex]= 0.000861 [tex]m^{2}[/tex].

A theoretical gas known as an ideal gas is made up of several randomly moving point particles with no interparticle interactions. Because it abides by the ideal gas law, a condensed equation of state, and is amenable to statistical mechanics analysis, the ideal gas concept is helpful.

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One way to control avalanches is to send explosive charges to key areas on mountain slopes to trigger small avalanches before larger ones can build up. Norway, for instance, uses solar- powered launchers that fire pre-timed charges. Your launcher fires charges at an angle of 70 degrees from the horizontal and a speed of 200 m/s. If you fire a charge and it travels a horizontal distance of 300 m away from you, how high up the slope will it strike?.

Answers

By using uniform motion, the high up the slope will it strike is 730.62 m.

We need to know about uniform motion to solve this problem. The uniform motion is an object's motion under acceleration. It should follow the rule

vt = vo + a . t

vt² = vo² + 2a . s

s = vo . t + 1/2 . a . t²

where vt is final velocity, vo is initial velocity, a is acceleration, t is time and s is displacement.

From the question above, we know that

θ = 70⁰

vi = 200 m/s

x = 300 m

Find the initial velocity in x and y axis

vx = vi cosθ = 68.40 m/s

vy = vi sinθ = 187.94 m/s

Find the time taken to reach 300 m

vx = x / t

68.4 = 300 / t

t = 4.39 second

Find the time taken to reach maximum height

vt = vo + a . t

0 = vy - g . t

vy = gt

187.94 = 9.8 . t

t = 19.18 second

The motion of projectile is raising to maximum height, hence

s = vo . t + 1/2 . a . t²

h = vy . t - 1/2 . g . t²

h = 187.94 . 4.39 - 1/2 . 9.8 . 4.39²

h = 730.62 m

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________ is a measure of your ability to work with and adapt to people from other backgrounds.

Answers

Answer:

Cultural intelligence

Explanation:

Cultural intelligence

A wave as represented by the equation Y = 2 sin π (0.5x - 200t) Where all distances are measured in cm and in Second. For this wave, calculate a) frequency b)wave length C )Speed d) Amplitude e)period f)wave number g)angular velocity ​

Answers

By calculating, a) frequency = 100 Hz b)wavelength = 4 cm  C )Speed = 400 cm/s d) Amplitude =  2 cm e)period = 0.01 s f)wave number  = 0.5π  g)angular velocity = 200π. ​

Given in question

Y = 2 sin π (0.5x - 200t)

Y = 2 sin (0.5πx - 200πt)

Comparing it with the standard form of a wave equation:

Y = Asin (kx - ωt) where

A= amplitude

K = wave number = 2π/λ where λ = wavelength

ω = Angular velocity = 2πf where f = frequency

We get

a) 2πf = 200π

    f = 100 Hz

b) 2π/λ =0.5π

   λ = 2π/0.5π

   λ = 4 cm

c) Speed v = ω/k

v = 200π/0.5π

v = 400 cm/s

d) A= amplitude = 2 cm

e)  period = 1/f = 0.01 s

f) K = wave number = 0.5π

g) ω = Angular velocity = 200π

Therefore, by comparing  a) frequency = 100 Hz b)wavelength = 4 cm  C )Speed = 400 cm/s d) Amplitude =  2 cm e)period = 0.01 s f)wave number  = 0.5π  g)angular velocity = 200π. ​

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Niagra falls is 170 feet tall. How much gravitational potential energy is lost when a pound of water falls from
the top of the falls to the bottom of the falls? Answer in BTU.

Answers

The amount of gravitational potential energy lost when a pound of water falls from the top of the falls to the bottom is 233.55 Joules.

What is Gravitational Potential Energy?

The gravitational potential energy of a body is due to virtue of its position. A body of mass 'm' at height 'h' is given by -

P.E. = mgh

Given in the is Niagara falls which 170 feet tall. A pound of water falls from the top of the falls to the bottom of the fall. From this we can write -

Height of Niagra falls = 170 feet tall = 51.81 m

Mass of water = 1 pound = 0.46kg

Assume that the potential energy of the pound of water at 170 feet tall is (P[1]) and that at the bottom is (P[2]).

Now, at the top of falls, the potential energy of 1 pound of water = P[1] =

m x g x h = 0.46 x 9.8 x 51.81 = 233.55 J

At the bottom of the falls -

P[2] = 0

The amount of gravitational potential energy lost =  P[1]  -  P[2] =

233.55 - 0 = 233.55 joules.

Therefore, the amount of gravitational potential energy lost when a pound of water falls from the top of the falls to the bottom is 233.55 Joules.

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Show your work on the following problems.
1.
An airplane accelerates down a run-way at 3.20 m/s? for 32.8 s until is finally lifts off the ground
Determine the distance traveled before take-off.
2.A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. De
acceleration of the car and the distance traveled.
3.
A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the
acceleration of the car and the distance traveled.
4.
A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of
the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a
uniform acceleration).

Answers

The distance travelled by the airplane before takeoff is 1721 m.

2. The acceleration of the race car is 11.2 m/s², and Distance travelled by the car is 79.9 m.

3. The acceleration of the bullet is 1.62 * 10⁵ m/s².

What is the distance travelled by the airplane before takeoff?

The distance travelled by the airplane before takeoff is determined using the formula:

s = ut + ¹/₂at²

u = 0; a = 3.20 m/s², t = 32.8 s

s = 0 * 32.0 +  ¹/₂ * 3.2 * 32.8²

s = 1721 m

2. The acceleration of the race car is calculated as follows:

a = (v - u)/t

v = 46.1 m/s, u = 18.5 m/s, t = 2.47 s

a = (46.1 - 18.5)/2.47

a = 11.2 m/s²

Distance travelled, s = 18.5 * 2.47 + ¹/₂ * 11.2 * 2.47²

s = 79.9 m

3. The acceleration of the bullet is calculated as follows:

a = (v² - u²)/2s

where v = 0 m/s, u = 521 m/s, s = 0.840 m

a = (0² - 521²)/2 * 0.840

a = 161572.0238

a = 1.62 * 10⁵ m/s²

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A truck driver has a shipment of apples to deliver to a destination 500 miles away. The trip usually takes him 9.0 hours. Today he finds himself daydreaming and realizes 110 miles into his trip that that he is running 20 minutes later than his usual pace at this point.
At what speed must he drive for the remainder of the trip to complete the trip in the usual amount of time?

Answers

The speed at which the driver must drive for the remaining time to complete the trip in the usual time is 61.25 miles per hour.

What are the total distance the driver needs to cover and the approximate time?

The total distance the driver should cover is 500 miles and this should be completed in 9 hours.

What is the regular speed?

Based on the information provided, the regular speed is:

500 miles/ 9 hours = 55.55 miles per hour

However, we know today he has covered 110 miles and he is 20 minutes late. This means, there are 390 miles to cover and the remaining time is 6.7 hours.

500  miles= 9 hours or 540 minutes110 miles =  x118.8 minutes + 20 extra minutes = 138.8 minutes = 2.3 hours9 hours - 2.3 hours = 6.7 hours

Based on this, let's calculate the new speed:

390 miles/ 6.7 = 61.25 miles per hour

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a wrench 0.3 meters long lies along the positive yy-axis, and grips a bolt at the origin. a force is applied in the direction of ⟨0,−3,1⟩⟨0,−3,1⟩ at the end of the wrench. find the magnitude of the force in newtons needed to supply 100 newton-meters of torque to the bolt.

Answers

The magnitude of the force in newtons needed to supply 100 newton-meters of torque to the bolt is   359N.

What is magnitude of force?

The amount that encapsulates the force's power is known as its magnitude. Consider the following scenario: the force is 10 N in the east. The direction is indicated by "towards east," while the force is indicated by "10." Magnitude may be thought of as simply the "value" or "amount" of any physical quantity.

The force applied that applies torque is:

Fz = 100 N-m / .3m = 333.33 N

Find the angle:

theta = tan^-1 ( 5/2) = 68.19 degrees

Total force:

F= 333.33 / sin(68.19) = 359N

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Karen is asked to look at two spots of light and judge which one gets brighter over time. Both spots start out at 100 units of brightness. She notices that the right spot is brighter when it is 110 units. If she is presented with two new spots of light both starting at 300 units, how much brighter will one have to get before she will be able to reliably detect the difference?.

Answers

One of the light spots will have to get 330 units brighter than the other for her to be able to tell the difference.

In the case of the first bulb, she noticed the difference when the right was 0.10 % brighter than the other one.

100  * 0.10 = 10

100 + 10 = 110

If we apply the same sense in the second case, one of the spots will have to be 0.10% brighter than the other.

300 * 0.10=30

300+30 = 330

In order to tell the difference, the other spot should be at 330 units of brightness.

The unit of brightness is Lumen. It is use to calculate the total amount of brightness emitted from a source over a unit of time.

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The road is 6 kilometers long. How
long is the road in meters?

Answers

Answer:

6000 meters

Explanation:

Because 6kilometers is 6000 maters

If Tim walks to the mailbox 600 m south and then to his
neighbor's house 300 m north and the entire trip takes him 5
minutes, then what is his average velocity?
a 2 m/s
b 3 m/s
C 60 m/s
d 1 m/s

What’s the answer ???

Answers

The average velocity of Tim is 1 m/s.

What is average velocity?

Average velocity is the ratio of total displacement to total time.

The S.I unit of velocity is m/s.

To calculate the average velocity, we use the formula below.

Formula:

v = (D-d)/t.........equation 1

Where:

v = Average velocity

From the question

Given:

D = 600 md = 300 m t = 5 munites = 300 s

Substitute these values into equation 1

v = (600-300)/300v = 300/300v = 1 m/s

Hence, the average velocity of Tim is 1 m/s.

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Venus has the smallest orbital eccentricity of any of the planets. What is the eccentricity of Venus?(1 point)
Responses

1.0300


0.8040


0.0068


0.0000

Answers

The eccentricity of Venus is = 0.0068. That is option C.

What is orbital eccentricity?

The orbital eccentricity is defined as the measurement that is used to show how an orbit deviates from the shape of a circle.

The Venus is one of the planets of the universe that is the second closest to the sun.

The Venus is one of the planets that has a small orbital eccentricity being 0.006772 which is approximately 0.0068. This means that it is closest to being circular in shape.

The planets with the largest orbital eccentricity is the planet Mercury with orbital eccentricity of 0.2056.

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During which of the following intervals is no unbalanced force acting on the car?

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The friction force that is exerted on the car's rotating wheels causes an imbalanced force to be applied to the vehicle, which then accelerates. The center of the circle the car is turning is the target of both the unbalanced force and the acceleration. Your body, however, continues to move and is in motion.

An object's motion may vary if an imbalanced force is applied to it. The small car will alter speed, direction, or even halt if a heavier, faster toy car moves into its path or along its side. The small toy car is being subjected to a stronger force, which is why its motion has changed.

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QUESTION 2: Switch on your light source. Make sure it is shining onto a wall Hold your largest cardboard square between the light source and the wall. . • 2a. What do you observe? . Now do the same with the second cardboard square. 2b. What do you observe? ▸ Now do the same with the third cardboard square. 2c. What do you observe? 2d. What differences do you observe between the shadows of the three sh QUESTION 33 Switch on your light source. Make sure it is shining onto a wali. Hold your largest plastic shopping bag square between the light sourc​

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When a light source falls on an object if it is a opaque object it will form shadow.

And the light falls on the largest cardboard there will be a big shadow formed in the wall. Now when we take a second cardboard the shadow will form in accordance with that cardboard that if the cardboard is more it will form small shadow

And when we do with the same third cardboard or small cardboard the shadow will be according to the size of the cupboard now when we keep a plastic shopping bag in between the light source and the wall if the plastic bag is transparent the light will pass through the plastic bag and if the plastic bag is opaque the light will not pass through the bag and a shadow will be not be formed

Here we can conclude that if a light is passing through a pack object the light cannot pass through the object and it will format shadow but when the light is passing through a transparent object that light will pass through the object and there will be no shadow formed

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A rock is dropped from a cliff and takes 6 seconds to hit the ground. At what time is the speed of the rock the highest??

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Answer:

Time to reach ground = t = 6 sec

Explanation:

Acceleration due to gravity = g = 9.8 m/s²

Initial Velocity of the Rock = Vi = 0 m/s (Because, the rock will be at rest, initially)

When a rock is dropped and takes 6 seconds to touch the ground, its maximum speed will be [tex]6^{th}[/tex] second.

The force of gravity causes a rock to accelerate when it is dropped from a cliff. On Earth, the acceleration caused by gravity is roughly 9.8 meters per second squared (m/s2). This indicates that the rock's speed increases by 9.8 m/s per second as it falls.

According to the question, it takes the rock 6 seconds to hit the ground. It takes the rock exactly this long to fall from the cliff to the ground.

One must comprehend that the rock's speed rises steadily as it falls in order to determine the moment at which it is moving at its fastest. The rock's initial velocity (Vi) is 0 m/s because it begins at rest. The acceleration brought on by gravity causes it to move faster as it drops.

The rock will reach its highest speed just before it touches the ground because its velocity is constantly rising. As a result, the rock is moving at its fastest at the moment of impact (t = 6 sec).

Thus, the maximum speed will be at T= 6 seconds.

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compare copper sulfate with the elements that form copper sulfate

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Copper sulfate is a salt that is made up of the copper ion and the sulfate ion further, the sulfate ion is the combination of the sulfur and the oxygen atom.

What is a Chemical compound?

The chemical compound is a combination of two or more  either similar or dissimilar chemical elements

As given in the problem we have to compare copper sulfate with the elements that form copper sulfate,

The salt copper sulfate is composed of the copper ion and the sulfate ion. In addition, the sulfate ion is a mixture of sulfur and oxygen atoms.

The chemical formula for the copper sulfate is CuSO₄ which is made up of a Copper (Cu) atom, Sulphur (S)  atom, and Oxygen (O) atom.

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A parallel-plate capacitor has a capacitance of 10 mf and is charged with a 20-v power supply. The power supply is then removed and a dielectric material of dielectric constant 4. 0 is used to fill the space between the plates. What is the voltage now across the capacitor?.

Answers

The capacitor's energy becomes 0.5 J when the dielectrics have been inserted between its plates.

How does capacitance work?

The quantity of energy held in the form of an electric charge in an electric device is referred to as capacitance.

Data provided

C = 10 mF = 0.01 F is the parallel-plate capacitor's capacitance.

V' = 20 V is the power supply's potential.

The material has a dielectric constant of 4.0.

First, let's determine the charge that the capacitor has stored:

[tex]C^{\prime}=\epsilon \times C[/tex]

Solving as,

[tex]\begin{aligned}&C^{\prime}=4.0 \times 0.01 \\&C^{\prime}=0.04 \mathrm{~F}\end{aligned}[/tex]

Now, the expression for the energy stored in the capacitor is,

[tex]U^{\prime}=\frac{q^2}{2 C^{\prime}}[/tex]

Hence

[tex]\begin{aligned}U^{\prime} &=\frac{0.2^2}{2 \times 0.04} \\U^{\prime} &=0.5 \mathrm{~J}\end{aligned}[/tex]

Thus, we may say that the capacitor's energy reserves are 0. 5J

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Bob and carol are both observing the moon at exactly the same time. However, bob is in miami and carol is in los angeles. Bob observes the moon in the first quarter phase. In what phase will carol observe the moon to be?.

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Bob and Carol are both observing the moon at exactly the same time. However, bob is in Miami and carol is in Los Angeles. Bob observes the moon in the first quarter phase. Carl will also observe the moon in the first quarter phase.

The moon cycle consists of four primary and secondary phases.

The four primary phases are New Moon, First Quarter Phase , Full Moon, and the Last Quarter.

During the new moon and full moon periods, there are definite chances of solar and lunar eclipses.

The secondary phases of the moon are waxing crescent, waxing gibbous, waning gibbous, and waning crescent.

In the southern hemisphere and northern hemisphere, people observe the same phases of the moon.

As Miami and Los Angeles are in the southern hemisphere i.e. south of the equator, therefore, people living in that hemisphere will observe the moon in the same phases.  Thus Bob and Carol will also observe the moon in the first quarter phase from different cities.

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