Answer:
Explanation:
Speed of sound in air at room temperature = 346 m /s
Let speed of sound in metal given = v m /s
Time take by sound wave travelling through air = 4 / 346 = .01156 s
Time taken by sound wave travelling through metal = 4 / v s
Given ,
.01156 - 4 / v = 11 x 10⁻³ s = .011 s ( time taken by sound travelling in air is more )
4 /v = .01156 - .011 = .00056
v = 4 / .00056 = 7142.85 m /s
In traveling a distance of 2.3 km between points A and D, a car is driven at 83 km/h from A to B for t seconds and 41 km/h from C to D also for t seconds. If the brakes are applied for 4.4 seconds between B and C to give the car a uniform deceleration, calculate t and the distance s between A and B.
Answer:
- time t taken for car to travel is 64.57 s
- distance travelled between A and B is 1.4887 km
Explanation:
Given the data in the question;
[tex]U_{BC}[/tex] = 83 km/h = ( 83×1000 / 60×60) = 23.0555 m/s
[tex]U_{CD}[/tex] = 41 km/h = ( 41×1000 / 60×60) = 11.3888 m/s
now, we calculate the acceleration;
a = ( [tex]U_{BC}[/tex] - [tex]U_{CD}[/tex] ) / t
we substitute
a = ( 23.0555 - 11.3888 ) / 4.4
a = 11.6667 / 4.4
a = 2.6515 m/s²
Now equation for displacement from BC
[tex]S_{BC}[/tex] = [tex]U_{BC}[/tex]t + 1/2.at²
we substitute
[tex]S_{BC}[/tex] = 23.0555×4.4 + 1/2×a×(4.4)²
we substitute -2.6515m/s² for a
[tex]S_{BC}[/tex] = 23.0555×4.4 + 1/2×(-2.6515)×(4.4)²
= 101.4442 - 25.6665
[tex]S_{BC}[/tex] = 75.7792 m
Now, for total distance covered = 2.3km = ( 2.3×1000) = 2300m
so
[tex]S_{AB}[/tex] + [tex]S_{BC}[/tex] + [tex]S_{CD}[/tex] = 2300 m
we substitute substitute
[tex]S_{AB}[/tex] + 75.7792 m + [tex]S_{CD}[/tex] = 2300 m
[tex]S_{AB}[/tex] + [tex]S_{CD}[/tex] = 2300 - 75.7792
[tex]S_{AB}[/tex] + [tex]S_{CD}[/tex] = 2224.2208 m
so we substitute 23.0555t for [tex]S_{AB}[/tex] and 11.3888t for [tex]S_{CD}[/tex]
23.0555t + 11.3888t = 2224.2208
34.4443t = 2224.2208
t = 2224.2208 / 34.4443
t = 64.57 s
Therefore, time t taken for car to travel is 64.57 s
Distance Between A to B
[tex]S_{AB}[/tex] = t × [tex]U_{AB}[/tex]
we substitute
[tex]S_{AB}[/tex] = 64.57 s × 23.0555
[tex]S_{AB}[/tex] = 1488.69 m
[tex]S_{AB}[/tex] = 1.4887 km
Therefore, distance travelled between A and B is 1.4887 km
The standard test to determine the maximum lateral acceleration of a car is to drive it around a 200-ft diameter circle painted on a level asphalt surface. The driver slowly increases the vehicle speed until he is no longer able to keep both wheel pairs straddling the line. If the maximum speed is 35 mi/hr for a 3000-lb car, compute the magnitude F of the total friction force exerted by the pavement on the car tires.
Answer:
the magnitude F of the total friction force is 2456.7 lb
Explanation:
Given the data in the question;
maximum speed = 35 mi/hr = ( 35×5280 / 60×60) = 51.3333 ft/s
diameter = 200ft
radius = 200/2 = 100 ft
First we calculate the normal component of the acceleration;
[tex]a_{n}[/tex] = v² / p
where v is the velocity of the car( 51.3333 ft/s)
p is the radius of the curvature( 100 ft)
so we substitute
[tex]a_{n}[/tex] = (51.3333 ft/s)² / 100ft
[tex]a_{n}[/tex] = (2635.1076 ft²/s²) / 100ft
[tex]a_{n}[/tex] = 26.35 ft/s²
we convert Feet Per Second Squared (ft/s²) to Standard Gravity (g)
1 ft/s² = 0.0310809502 g
[tex]a_{n}[/tex] = 26.35 ft/s² × 0.0310809502 g
[tex]a_{n}[/tex] = 0.8189g
Now consider the dynamic equilibrium of forces in the Normal Direction;
∑[tex]F_{n}[/tex] = m[tex]a_{n}[/tex]
F = m[tex]a_{n}[/tex]
we know that mass of the car is 3000-lb = 3000lb([tex]\frac{1}{g}slug[/tex]/1 lb)
so
we substitute
F = 3000lb([tex]\frac{1}{g}slug[/tex]/1 lb) × 0.8189g
F = 2456.7 lb
Therefore; the magnitude F of the total friction force is 2456.7 lb
I love buying physics toys. I recently broke out my new electromagnetic field meter and started playing with it. After turning it on, I noticed the device kept showing an electric field value of 200 N/C towards the ground, without being near anything obvious (e.g., an electronic device) that would be producing the electric field. I even took a long walk to check if the reading was somehow localized to my house, but I got the same result. How might you explain the reading (assuming the device is working properly)
Answer:
ionized particles from the sun.
* interactions in radiation belts.
* the friction of the planet in the solar wind
q = +9 10⁵ C
Explanation:
Due to being made up of matter, the planet Earth has a series of positive and negative charges, in general these charges should be balanced and the net charge of the planet should be zero, but there are several phenomena that introduce unbalanced charges, for example:
* ionized particles from the sun.
* interactions in radiation belts.
* the friction of the planet in the solar wind
This creates that the planet has a net electrical load
We can roughly calculate the charge of the planet
E = k q / r²
q = E r² / k
let's calculate
q = 200 (6.37 10⁶)²/9 10⁹
q = +9 10⁵ C
A toy car accelerates uniformly from rest at a constant rate. The car travels 1.0 meters in 1.0 seconds. The acceleration of the car is ________ meters per second squared.
Answer:
a=1m/s^2
Explanation:
1÷1÷1=1m/s^2
A football of mass 2.5kg is lifted up to the top of a cliff that is 180m high. How much
potential energy does the football gain?
The potential energy of the football with mass 2.5 kg which is lifted up to the top of a cliff is 4410 Joules.
What is Potential energy?Potential energy is the stored energy which depends upon the relative position of the various parts of a system of objects. Potential energy is the product of mass of the object, acceleration due to gravity, and the height. The SI unit of potential energy is Joule (J).
PE = m × g × h
PE = Potential energy,
m = mass of the object,
g = acceleration due to gravity,
h = height
PE = 2.5 × 9.8 × 180
PE = 4410 Joules
Therefore, the potential energy of the football is 4410 Joules.
Learn more about Potential energy here:
https://brainly.com/question/24284560
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A pendulum has a period of 5.14s and a length of 0.25m. What is the acceleration
due to gravity? *
Answer:
Acceleration due to gravity, g = 2.68m/s²
Explanation:
Given the following data;
Period = 5.14s
Length = 0.25m
To find acceleration due to gravity, g;
[tex] Period, T = 2 \pi \sqrt {lg} [/tex]
Substituting into the equation, we have;
[tex] 5.14 = 2*3.142 \sqrt {0.25g} [/tex]
[tex] 5.14 = 6.284 \sqrt {0.25g} [/tex]
[tex] \frac {5.14}{6.284} = \sqrt {0.25g} [/tex]
[tex] 0.8180 = \sqrt {0.25g} [/tex]
Taking the square of both sides
[tex] 0.8180^{2} = 0.25g [/tex]
[tex] 0.6691 = 0.25*g[/tex]
[tex] g = \frac {0.6691}{0.25} [/tex]
Acceleration due to gravity, g = 2.68m/s²
A lunar eclipse occurs when the Moon passes through Earth's
A baseball player hits a baseball. The mass of the ball is 0.15 kg. The ball accelerates at a rate of 60 m/s 2 . What is the net force on the ball to the nearest newton?
Answer:
Please find attached pdf
Explanation:
If an ocean wave has a wavelength of 2 m and a frequency of 1 wave/s, then its speed is m/s Enter the answer Check it CRATCHPAD Improve this questic 트
Answer:
2m/s
Explanation:
v=f×wavelength
v=2×1
=2m/s
The head of a rattlesnake can accelerate at 50 m/s2 in striking a victim. If a car could do as well, how long would it take to reach a speed of 100 km/h from rest
Answer:
the time for the car to reach the final velocity is 0.56 s.
Explanation:
Given;
acceleration of the car, a = 50 m/s²
final velocity of the car, v = 100 km/h = 27.778 m/s
the initial velocity of the car, u = 0
The time for the car to reach the final velocity is calculated as;
v = u + at
27.778 = 0 + 50t
27.778 = 50t
t = 27.778 / 50
t = 0.56 s
Therefore, the time for the car to reach the final velocity is 0.56 s.
6.
ribbon
AA
SON
120 N
Two teams of students are competing in a tug-o-war contest, as shown in the
picture above. How does the ribbon move?
Answer:
The ribbon will move to the right.
Explanation:
To know the the correct answer to the question, we shall determine the net force and direction. This can be obtained as follow:
Force to the right (Fᵣ) = 120 N
Force to the left (Fₗ) = 80 N
Net force (Fₙ) =?
Fₙ = Fᵣ – Fₗ
Fₙ = 120 – 80
Fₙ = 40 N to the right.
From the calculation made above, the net force is 40 N to the right. Thus, the ribbon will move to the right.
Suppose a car is traveling at 22.8 m/s, and the driver sees a traffic light turn red. After 0.404 s has elapsed (the reaction time), the driver applies the brakes, and the car decelerates at 9.00 m/s2. What is the stopping distance of the car, as measured from the point where the driver first notices the red light?
Answer:
38.09 m
Explanation:
We'll begin by calculating the distance travelled by the car during the reaction time. This can be obtained as follow:
Reaction time (tᵣ) = 0.404 s
Initial velocity (u) = 22.8 m/s,
Distance travelled during the reaction time (sᵣ) =?
sᵣ = utᵣ
sᵣ = 22.8 × 0.404
sᵣ = 9.21 m
Next, we shall determine the distance travelled by the car when the brake was applied. This can be obtained as follow:
Initial velocity (u) = 22.8 m/s
Acceleration (a) = –9 m/s² (since the car is decelerating)
Final velocity (v) = 0 m/s
Distance travelled when the brake was applied (s₆) =?
v² = u² + 2as₆
0² = 22.8² + (2 × –9 × s₆)
0 = 519.84 – 18s₆
Collect like terms
0 – 519.84 = –18s₆
–519.84 = –18s₆
Divide both side by –18
s₆ = –519.84 / –18
s₆ = 28.88 m
Finally, we shall determine the stopping distance of the car, as measured from the point where the driver first notices the red light. This can be obtained as follow:
Distance travelled during the reaction time (sᵣ) = 9.21 m
Distance travelled when the brake was applied (s₆) = 28.88 m
Stopping distance =?
Stopping distance = sᵣ + s₆
Stopping distance = 9.21 + 28.88
Stopping distance = 38.09 m
What is the displacement of the particle in the time interval 7 seconds to 8 seconds?
Answer:
it 1.5 meters
Explanation:
if u could put the option number it will be cool and hope it help and if it doesnt am really sorry ;)
In the 1986 Olympic Games, Abdon Pamich of Italy won the 50 km walk, in
4h, 11 min and 11.2 s. Find his average speed in m/s.
Answer: 4
Explanation: because my pet monkey said it was
In July 2015, Oregon State University, the National Oceanic and Atmospheric Administration, and the Coast Guard cooperated to send a hydrophone into Challenger Deep, the deepest part of the Mariana Trench. The titanium shelled recording device withstood the pressure 10,994 meters (nearly 7 miles!) under the ocean's surface. The hydrophone recorded 23 days of audio from the deepest part of the ocean floor. If the spherical hydrophone has a radius of 10 cm, what is the total force exerted on the titanium shell by the ocean water
Answer:
Explanation:
Pressure due to water column as deep as 10994 meters can be given by the following expression
Pressure = h d g , where h is depth of water , d is density of water and g is acceleration due to gravity .
Pressure = 10994 x 10³ x 9.8
= 10.77 x 10⁷ N / m²
Pressure will act on curved surface of the spherical shell , the effective surface area will be π R² where R is radius of the surface .
Effective surface = 3.14 x 0.1²
= .0314 m²
Total force = pressure due to water column x effective surface
= 10.77 x 10⁷ x .0314 N.
= 33.82 x 10⁵ N .
Which of the following is an example of Newton's Third Law?* O A stack of pennies will not move unless you flick them over. O Falling off of a skateboard after you run into a curb A ball hits the ground and the ground pushes up on it with the same force
Answer:
A ball hits the ground and the ground pushes up on it
Explanation:
Newton's third law basically states that for every action, there's a reaction.
a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.
Hope this Helps!!! :)
5. An astronaut has a mass of 65kg where the gravitational field strength is 10N/kg
a. Calculate the weight of the astronaut on earth
[3]
Answer: a) weight on Earth = mass of the object and gravity n the Earth. = 65*10 = 650 kg.
Explanation:
An astronaut has a mass of 65 kg on Earth where the gravitational field strength is 10 N kg A calculate the astronaut's weight on Earth
hope this helps :)
Answer:
650
Explanation:
use the equation
weight = gm
Your favorite golfer taps the golf ball with just enough force that it rolls into the ninth hole is an example of what law of motion???
Answer:
mass and on the net force acting on it. ... Tap again to see term ... Newton's second law of motion states that an object's acceleration depends on its ... You hit a ping-pong ball & a tennis ball with a tennis rack
6th grade science I mark as brainliest !
Answer:
first is Atoms
4) is True
The design of interior spaces is relatively unimportant to good
architecture?
6. The petrol in a petrol can weighs 2000g. The density of petrol is 0.8g/cm3.
What is the volume of the petrol in the can in a) cm3 b)litres (1000cm3=1 litre)
Pls help :((
Answer:
a. 2500 cm³.
b. 2.5 litres.
Explanation:
Given the following data:
Density = 0.8g/cm³
Mass = 2000g
To find the volume of the petrol;
Density can be defined as mass all over the volume of an object.
Simply stated, density is mass per unit volume of an object.
Mathematically, density is given by the equation;
[tex]Density = \frac{mass}{volume}[/tex]
Making volume the subject of formula, we have;
[tex]Volume = \frac{mass}{density}[/tex]
Substituting into the equation, we have:
[tex]Volume = \frac{2000}{0.8}[/tex]
Volume = 2500 cm³
a. The volume of the petrol in the can in cubic centimeters (cm³) is 2000.
b. The volume of the petrol in the can in litres;
1000 cm³ = 1 litre
2500 cm³ = x litres
Cross-multiplying, we have;
1000x = 2500
x = 2500/1000
x = 2.5 litres.
Therefore, the volume of the petrol in the can in litres is 2.5.
In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is launched vertically into the air at point T. In case B, a 1 kg block slides without friction down an identically shaped ramp and is also launched vertically at point T. Select the statement that best describes which object will go higher after launch, and why
Answer:
the block reaches higher than the sphere
\frac{y_{sphere}} {y_block} = 5/7
Explanation:
We are going to solve this interesting problem
A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp
Let's use the concept of conservation of energy
starting point. At the top of the ramp
Em₀ = U = m g y₁
final point. At the exit of the ramp
Em_f = K + U = ½ m v² + ½ I w² + m g y₂
notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂
energy is conserved
Em₀ = Em_f
m g y₁ = ½ m v² + ½ I w² + m g y₂
angular and linear velocity are related
v = w r
the moment of inertia of a sphere is
I = [tex]\frac{2}{5}[/tex] m r²
we substitute
m g (y₁ - y₂) = ½ m v² + ½ ([tex]\frac{2}{5}[/tex] m r²) ([tex]\frac{v}{r}[/tex])²
m g h = ½ m v² (1 + [tex]\frac{2}{5}[/tex])
where h is the difference in height between the two sides of the ramp
h = y₂ -y₁
mg h = [tex]\frac{7}{5}[/tex] ([tex]\frac{1}{2}[/tex] m v²)
v = √5/7 √2gh
This is the exit velocity of the vertical movement of the sphere
v_sphere = 0.845 √2gh
B) is the same case, but for a box without friction
starting point
Em₀ = U = mg y₁
final point
Em_f = K + U = ½ m v² + m g y₂
Em₀ = Em_f
mg y₁ = ½ m v² + m g y₂
m g (y₁ -y₂) = ½ m v²
v = √2gh
this is the speed of the box
v_box = √2gh
to know which body reaches higher in the air we can use the kinematic relations
v² = v₀² - 2 g y
at the highest point v = 0
y = vo₀²/ 2g
for the sphere
y_sphere = 5/7 2gh / 2g
y_esfera = 5/7 h
for the block
y_block = 2gh / 2g
y_block = h
therefore the block reaches higher than the sphere
[tex]\frac{y_{sphere}} {y_bolck} = 5/7[/tex]
A fan is set on a desk next to a stack of paper. The fan is turned on and then turned
on HIGH SPEED. Which of the following would best apply Newton's First Law to this
example.
The papers accelerated due to the force of the fan.
The papers are acted upon by an unbalanced force from the fan.
O The papers exerted an equal force on the air blown by the fan.
O The papers that were the heaviest were blown the closest.
Answer:
The second option - the papers are acted upon by an unbalanced force from the fan.
Explanation:
In order for the eye to see an object _____ from the object myst be reflected to your eye.
Light or particle ?
Answer: light from the object
Explanation:
When light is reflected off an object like a lamp or a door is travels in a straight line but in a new direction so if the light enters our eyes we will see the object because our eyes can detect light
Starting from rest, a wheel with constant angular acceleration turns through an angle of 25 rad in a time t. What will its angular velocity be after 3t?
Answer:
θ = 225 rad
Explanation:
given data
angle = 25 rad
to find out
angular velocity after 3t?
solution
let angular acceleration α in t
θ = ω × t + 0.5 × α × t² ........................1
here ω = 0 (initial velocity )
so put this value here
25 = 0 + 0.5 × α × t² ..........................2
α = 25 ÷ (0.5 t²)
α = 50 ÷ t² .........................3
now here we take in 3t
θ = ω × 3t + 0.5 × α × (3t)²
for ω = 0
θ = 0 + 0.5 × α × 9t²
now put value in eq 2
so
θ = (0.5) × (50 ÷ t²) × (3t)²
θ = 25 × 9
θ = 225 rad
A 7300 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.20 m/s2 and feels no appreciable air resistance. When it has reached a height of 575 m , its engines suddenly fail so that the only force acting on it is now gravity. Part A What is the maximum height this rocket will reach above the launch pad
Answer:
Explanation:
We shall first calculate the velocity at height h = 575 m .
acceleration a = 2.2 m /s²
v² = u² + 2 a s
u is initial velocity , v is final velocity , s is height achieved
v² = 0 + 2 x 2.2 x 575
v = 50.3 m /s
After 575 m , rocket moves under free fall so g will act on it downwards
If it travels further by height H
from the relation
v² = u² - 2 g H
v = 0 , u = 50.3 m /s
H = ?
0 = 50.3² - 2 x 9.8 H
H = 129.08 m
Total height attained by rocket
= 575 + 129.08
= 704.08 m .
A laser positioned on a ship is used to communicate with a small two man research submarine resting on the bottom of a lake. The laser is positioned 12 m above the surface of the water, and it strikes the water 20 m from the side of the ship. The water is 76 m deep and has an index of refraction of 1.33. How far is the submarine from the side of the ship
Answer:
84.1 m
Explanation:
Given :
The distance from the ship to submarine :
20 + y
Using Pythagoras :
Tan θ = opposite / Adjacent
Tan θ = 20 / 12
12 tan θ = 20
θ = tan^-1(20/12)
20
θ = 59.036°
The angle phi;
n1sinθ1 = n2sin θ
Sin 59.036 = 1.33 * sin phi
Sin phi = sinsin(59.04) ÷1.33
0.8574907 = 1.33 * sin phi
Sin phi = 0.8574907 / 1.33
Sin phi = 0.6447298
phi = sin(0.6447298
Phi = 40.15°
From Pythagoras :
y = 76tan40.15°
y = 76 * 0.8435707
y = 64.11
20 + y
20 + 64.11 = 84.11
If the car has a mass of 0.2 kg, the ratio of height to width of the ramp is 12/75, the initial displacement is 2.25 m, and the change in momentum is 0.58 kg*m/s, how far will it coast back up the ramp before changing directions
Answer:
l = 0.548 m
Explanation:
For this exercise we compensate by finding the speed of the car
p = m v
v = p / m
v = 0.58 / 0.2
v = 2.9 m / s
this is how fast you get to the ramp, let's use conservation of energy
starting point. Lowest point
Em₀ = K = ½ m v²
final point. Point where it stops on the ramp
[tex]Em_{f}[/tex] = U = m g h
mechanical energy is conserved
Em₀ = Em_{f}
½ m v² = m g h
h = [tex]\frac{m v^2}{2 g}[/tex]
let's calculate
h = [tex]\frac{0.2 \ 2.9^2}{2 \ 9.8}[/tex]
h = 0.0858 m
to find the distance that e travels on the ramp let's use trigonometry, we look for the angle
tan θ = y / x
tan θ = 12/75 = 0.16
θ = tan⁻¹ 0.16
θ = 9º
therefore
sin 9 = h / l
l = h / sin 9
l = 0.0858 / sin 9
l = 0.548 m
The number of complete wavelengths that pass a point in a given time is referred to as...
A. Wavelength
B. Frequency
C. Amplitude
D. Reflection
Pluto has a radius of 1.15 x 10^6 m, and its acceleration due to gravity is 0.61 | m/s^2. What is Pluto's mass?
so we will use Newton's gravitational law :
gravitational acceleration = G*m/r^2
G is the gravitational constant = G = 6.673×10^-11 N m^2 kg^-2
after substitution :
6.673×10^-11 * m / (1.15 x 10^6)^2 = 0.61
= 5.04575*10^-23 * m = 0.61
dividing over 5.04575*10^-23 :
m = 1.20894*10^22 kg
pls give me brainliest