I NEED HELP ASAP!!!
23. Formulating Hypotheses Suppose you want
to know if objects with different masses fall to
the ground at different rates. State a hypothesis
about falling objects.
24. Designing Experiments Explain how you can
test the hypothesis in Question 23. What will be
the manipulated variable in your experiment?
What will be the responding variable?

I already have a hypothesis for 23 I just need to know how I will test my hypothesis.

Also I don’t know if Im right but would the manipulated variable be the Weight??

Answers

Answer 1

A plausible hypothesis in this case is; "different masses of objects fall to the ground at different rates".

What is a hypothesis?

A hypothesis is a plausible explanation for an observation. A hypothesis is always formulated prior to an experiment. The experiment would now confirm or disprove the hypothesis that has been put forward.

Recall that an experiment must have a dependent variable and an independent variable. The independent variable is the variable that we have to continue to manipulate while the dependent variable is the variable that dependent variable changes with the independent variable.

A plausible hypothesis in this case is; "different masses of objects fall to the ground at different rates". The independent variable would now be mass of the object while the dependent variable would be the rate at which the object falls.

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Related Questions

A speeding car passes a highway patrol checkpoint, then decelerates at a constant rate. After 5 s, the car is 225 m from the checkpoint, and its speed is then 30 m/s. What was the car’s velocity when it passed the checkpoint?.

Answers

The car’s velocity when it passed the checkpoint was of: 60 m/s

The formula and procedure we will use to solve this exercise is:

vi = [(2 *x)/ t] - vf

Where:

x = distancet = timevi = initial velocityvf = final velocity

Information about the problem:

x = 225mt = 5 svf= 30 m/svi = ?

Applying the initial velocity formula we have:

vi = [(2 *x)/ t] - vf

vi = [(2 * 225m)/5 s ] - 30 m/s

vi = [(450 m)/5 s ] - 30 m/s

vi = 90 m/s - 30 m/s

vi = 60 m/s

What is velocity?

It is a physical quantity that indicates the displacement of a mobile per unit of time, it is expressed in units of distance per time, for example (miles/h, km/h).

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____?_______ is the muscular partition which between chest and abdominal cavity.

Answers

Answer: The diaphragm

Explanation:

The diaphragm is a thin dome-shaped muscle which separates the thoracic cavity (lungs and heart) from the abdominal cavity (intestines, stomach, liver, etc.)

Is my 1-8 right? And if not what is the right answer and show how you got the answer. Also hat would be the answer for 9 and 10 and why because I can’t seem to figure them out.

Answers

The answers for question numbers 3 and 4 is correct. For question number 9, the average speed is 2.5 mi /hr and for question number 10 the speed is 5 yards / s.

We know that,

v = d / t

where,

v = Speed

d = Distance

t = Time

1 ) v = 360 / 6 = 60 km / h

2 ) v = 120 / 3 = 40 mi / h

3 ) v = 18 / 6 = 3 m / s

4 ) v = 1000 / 20 = 50 m / min

5 ) t = 6pm - 5 pm = 1 hr

    v = 2.5 / 1 = 2.5 mi / hr

6 ) v = 1.5 / 0.33 = 4.5 mi / hr ( Since 20 min = 20 / 60 = 0.33 hr )

   d = 4.5 * 1 = 4.5 mi

7 ) d = 20 * 1 = 20 mi ( Since 60 min = 1 hr )

8 ) d = 60 * 2 = 120 mi

9 ) Distance per lap = 0.5 mi

     Total laps = 10

     Total distance = 10 * 0.5 = 5 mi

     v = 5 / 2 = 2.5 mi / hr

10 ) v = 100 / 20 = 5 yards / s

Therefore, the answer for:

v = 60 km / hv = 40 mi / hv = 3 m / sv = 50 m / minv =2.5 mi / hrd = 4.5 mid = 20 mid = 120 miv = 2.5 mi / hrv = 5 yards / s

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A balloon clings to a wall after it is negatively charged by rubbing.(a) Does that occur because the wall is positively charged?

Answers

A balloon clings to a wall after it is negatively charged by rubbing because the wall is positively charged.

Is opposite charge attract each other?

Yes, opposite charges attract each other. When a positive charge and a negative charge interact with each other, their forces act from the direction of positive to the direction of negative charge. As a result opposite charges attract each other while on the other hand, similar charges repel each other because their forces move in the opposite direction so they repel each other.

So we can conclude that a balloon clings to a wall after it is negatively charged by rubbing because the wall is positively charged.

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Help pls Important!!!!

Answers

No,plate motion would not take place on that planet.

a water tank in the shape of a hemispherical bowl of radius 5 m5 m is filled with water to a depth of 2 m.2 m. how much work is required to pump all the water over the top of the tank? (the density of water is 1000 kg/m31000⁢ kg/m3 ). (use symbolic notation and fractions where needed.) W

Answers

The work required to pump all the water over the top of the tank of radius 5 m and depth 2 m is 5.12 x 10^6 J.

The work required to pump the water over the top of the tank is equal to the gravitational potential energy of the water inside the tank.

The potential energy is given by the following equation,

U= mgh

Here, m is the mass of water present inside the tank.

We know that, Density( Mass/Volume

Hence, m= Density x volume

So, the equation becomes

U= ρ x V x g x h

Where V is the volume of hemispherical tank= 2/3π x r^3

Putting the given values in the above equation,

U= 1000 x 2/3 x 3.14 x 53 x 9.8 x 2 = 5.12 x 10^6 J

Hence, the work required to pump all the water over the top of a tank of radius 5 m and a depth of 2 m is 5.12 x 10^6 J.

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the displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s − 2 sin t 1 3 cos t, where t is measured in seconds.

Answers

The average velocity or displacement of a particle for the first time interval is Δs / Δt = 6 cm/s.

Solution:

As we know that displacement is calculated in centimeters and the unit of time is second.

The average velocity for the first interval [1,2] is given

Δs / Δt = s (t2) - s (t) / t2 - t1

Δs / Δt = 2sin2  π  + 3cos 2 π -  ( 2sin π + 3cos π ) / 2 - 1

Δs / Δt = 2(0) + 3(1) - 2(0) - 3 (-1) / 1

Δs / Δt = 6 cm/s

Thus the average velocity or displacement of a particle for the first time interval is Δs / Δt = 6 cm/s

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The complete question is:

The displacement of a particle moving back and forth along a line is given by the following equation s(t) = 2sin π t + 3cos π t. Estimate the instantaneous velocity of the particle when t = 1

A cannonball is shot horizontally off a high castle wall at 47.4 m/s. What is the magnitude of the cannonball's velocity after 1.23 s? (Ignore direction)

Answers

Answer:

48.9 m/s

Explanation:

I am ignoring air resistance in this:

horizontal velocity stays constant, so we need to find vertical velocity

v_0y = 0 m/s, t = 1.23s, a = 9.81 m/s^2 <- (g)

v_y = v_0y + at = 0 + 9.81 * 1.23 = 12.066 m/s

the magnitude v = sqrt(v_x ^ 2 + v_y ^2) = sqrt(47.4^2 + 12.066^2) = 48.9m/s

Answer:

V = 49.2 m/s

Explanation:

Given:

Vₓ  = 47.4 m/s

t = 1.23 s

___________

V - ?

Vy =  Voy + g·t = 0 + 9.8·1.23  ≈ 12,1 m/s

V = √( (Vx)² + (Vy)² ) = √ ( 47,7² + 12,1²) = 49.2 m/s

If a circuit has a power of 50 W with a current of 4.5 A, what is the resistance in the circuit?

Answers

Answer:

[tex]{ \rm{power, \: p = current \times p.d}} \\ { \rm{50 = 4.5 \times (current \times resistance)}} \\ { \rm{50 = 4.5 \times (4.5 \times r)}} \\ { \rm{resistance = \frac{50}{ {4.5}^{2} } }} \\ \\ { \rm{resistance = 2.5 \: ohms}}[/tex]

2. A farmer moves along the boundary of a square field side 10m in 40 sec. What will be the magnitude of displacement of the farmer at the end of 2 min 20 seconds from his initial position.

Answers

Answer:

The farmer is moving at 10 m / 40 s = .4 m/s

2 min 20 sec = 140 sec

The farmer will have moved 140 * .4 = 56 m

To cover the x-axis twice he will have moved x = 10

56 - 50 = 6 m   and he will have moved a distance of 6 on the y-axis

S = (10^2 + 6^2)^1/2 = 11.7 m from the origin

a spring whose equilibrium length is 7 m7 m is compressed to a length of 12 m12 m when a force of 14 n14 n is applied. find the work done by the spring force while it is compressed to a length of 6 m.6 m. (give an exact answer. use symbolic notation and fractions where needed.) W

Answers

The work done by the spring force while it is compressed to the length of 6m is 0.58 J.

What do you mean by work done?

The definition of work done includes both the forces applied to the body and the total displacement of the body. When we apply force "F" to a block, the body moves with some acceleration, or, additionally, its speed increases or decreases depending on the direction of the force. The system's kinetic energy changes as speed increases or decreases. Since we are aware that energy cannot be created or destroyed, it must be changed into another form. This perspective refers to it as completed work. When negative energy is finished, the energy declines, and when positive energy is finished, the energy rises.

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the universe originated from the explosion and expansion of all matter and energy. this statement is an example of .

Answers

The Big Bang Theory is an example of the universe originated from the explosion and expansion of all matter and energy

The Big Bang Theory is the way that astronomers explain how the universe began as a tiny, dense, fireball that exploded 13.8 billion years ago. It includes Albert Einstein's general theory of relativity along with standard theories of fundamental particles.

The theory was born of the statement that other galaxies are moving away from our own at countless speed in all directions, as if they had all been forced by an ancient explosive force.

What is the solar system?

The solar system is the gravitationally bound system of the sun, the objects that orbit it or travel around the sun. It consists of an average star (sun), the planets (Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto) and their moons, dwarf planets and countless asteroids, comets, and other small icy objects.

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How quickly would a 60kg object accelerate if the person applied a 500N force?

Answers

Answer:

8.33 m/s^2

Explanation:

The equation for force is Force = mass * acceleration. The force is 500 N, and the mass is 60 kg, so substituting those into the equation you get that 500 = 60 * acceleration. Divide 500 by 60 to get the acceleration, which is 8.33 m/s^2.

Hope this helps! Let me know if you have any confusion!

Oliver repeats his run five times.
Give two reasons why.

Answers

Answer:

•So he stays healthy.

•He has energy and time to do it.

A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the magnitude of the electric field(a) 0 cm

Answers

The magnitude of the electric field is 0 N/C.

We choose a Gaussian sphere of radius r > R concentric with the solid sphere.

Apply Gauss's law from Equation (1):

[tex]\Phi_{E} = \oint E\cdot dA = \frac{q_{in}}{\in_{0} }[/tex]

Since the electric field vector E is parallel to the area vector dA of the curved part the dot product can be the simple product EdA:

                                 [tex]\oint E\cdot dA = \frac{q_{in}}{\in_{0} }[/tex]

since the electric field is constant we can put E out of the integral:

                                [tex]E \oint dA = \frac{q_{in}}{\in_{0} }[/tex]

                                    [tex]EA = \frac{q_{in}}{\in_{0} }[/tex]

Substitute for A from Equation (2) and for [tex]q_{in}[/tex] from Equation (3)

recognizing that the enclosed charge by the Gaussian sphere is less than Q:

                              [tex]E(4\pi r^{2}) = \frac{pVgaussian}{\in_{0} }[/tex]

Substitute for V gaussian (volume of the Gaussian surface) from Equation (4):

                                          [tex]E(4\pi r^{2}) = \frac{\rho (\frac{4}{3} \pie r^{2})} {\in_{0} }[/tex]

                                           [tex]E = \frac{\rho r}{3\in_{0} }[/tex]

Where p is the charge per unit volume of the solid sphere which is equal to the total charge Q of the sphere divided by the volume V of the sphere :

                             [tex]E = \frac{(\frac{Q}{V} )r}{3\in_{0} }[/tex]

                                [tex]E = \frac{Qr}{(\frac{4}{3} \pi R^{3})3\in_{0} }[/tex]

                                [tex]E = \frac{Qr}{4\pi \in_{0}R^{3} }[/tex]

Substitute numerical values:

                                                  [tex]E = \frac{(26\times10^{-6})(0)}{4\pi(8.8542 \times10^{-12})(0.4)^{3}}[/tex]

                                 [tex]E = 0 N/C[/tex]

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A circular coil of five turns and a diameter of 30.0 cm is oriented in a vertical plane with its axis perpendicular to the horizontal component of the Earth's magnetic field. A horizontal compass placed at the coil's center is made to deflect 45.0° from magnetic north by a current of 0.600 A in the coil.(a) What is the horizontal component of the Earth's magnetic field?

Answers

The horizontal component of the magnetic field is 12.6 μT.

The magnetic influence on moving electric currents, electric charges, and magnetic materials is described by a magnetic field, which is a vector field. When a charge moves through a magnetic field, a force that is perpendicular to both its own velocity and the magnetic field operates on it.

The horizontal component of the Earth's magnetic field is perpendicular to the axis of a circular coil with five turns and a diameter of D = 30.0 cm that is vertically orientated.

A coil current of I = 0.600 A causes a horizontal compass to deflect 45.0° from magnetic north when it is positioned in the coil's center.

Let B be the magnetic field and R be the radius of the circular coil.

Then the horizontal component of the Earth's magnetic field is given as:

[tex]B(h) = B(coil) = \frac{\mu_{0} NI}{2R}[/tex]

[tex]B(h) =\frac{4 \pi \times 10^{-7}\times(5)\times(0.6)}{0.3}[/tex]

B(h) = 12.6 μT

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The tensile stress in a thick copper bar is 99.5 % of its elastic breaking point of 13.0× 10¹⁰ N/m² . If a 500-Hz sound wave is transmitted through the material, (b) What is the maximum speed of the elements of copper at this moment?

Answers

The speed of the wave in the rod is

[tex]The speed of the wave in the rod is $v = \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{20\times 10^{10}\ \mathrm{N/m^2}}{7.86\times 10^3\ \mathrm{kg/m^3}}} = 5044\ \mathrm{m/s}$[/tex]

5044m/s

What is sound wave?

A sound wave is the pattern of disruption brought on by the movement of energy moving through a medium as it propagates away from the source of the sound (such as air, water, or any other liquid or solid substance). Pressure waves are produced when an item vibrates, such as a ringing phone, and these waves are known as sound waves. The surrounding medium's particles are disturbed by the pressure wave, and those particles disrupt the particles next to them, and so on. Like ocean waves, the disturbance's pattern causes outward movement in all directions. Usually in all directions and with decreased intensity as it gets further away from the source, the wave transmits the sound energy through the medium.

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Me.Henderson Porsche accelerates from 0 to 60 mi/hr in 4 seconds. It’s acceleration is?

Answers

The acceleration of Mr. Henderson Porsche is 54545.45 mi/hr².

Velocity is the directional speed of an object, body, or particle in motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time.

We know that acceleration of a body is Change in Velocity per unit time.

[tex]a=\frac{v-u}{t}[/tex]

Here ,

v is the Final Velocity,

u is the Initial Velocity and;

t is the time taken.

In the given question ,

The Porsche accelerated from 0 to 60 miles per hour in four seconds.

u = 0 mi/hr

v = 60 mi/hr

t = 4 sec

[tex]t=\frac{4}{3600}{~}hr\\[/tex]

[tex]t=\frac{1}{900} {~}hr[/tex]

t  = 0.0011 hr

Then,

[tex]a=\frac{v-u}{t}[/tex]

[tex]a=\frac{60-0}{0.0011}[/tex]

[tex]a=54545.45 {~}mi/hr^{2}[/tex]

Hence, the acceleration of the Porsche is 54545.45 mi/hr².

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forces of 10 pounds and 14 pounds act on each other with an angle of 50°. the magnitude of the resultant force

Answers

The resultant force is 10.77 pounds.

We need to know about vector to solve this problem. Force is included to vector because it has magnitude and also direction. The resultant force can be determined using cosine law. It can be written as

F² = F1² + F2² - 2F1 . F2 cos(θ⁰)

From the question above, we know that:

F1 = 10 pounds

F2 = 14 pounds

θ = 50°

By substituting the parameter, we get

F² = F1² + F2² - 2F1 . F2 cos(θ⁰)

F² = 10² + 14² - 2. 10 . 14 cos(50⁰)

F² = 100 + 196 - 280 . 0.64

F² = 100 + 196 - 280 . 0.64

F² = 116.02

F = 10.77 pounds

Hence, the resultant force is 10.77 pounds.

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A football player kicks a football in a field goal attempt. When the football reaches its maximum height, what is the relationship between the direction of the velocity and acceleration vectors? assume air resistance is negligible.

Answers

The dot product of velocity vector and acceleration vector at maximum height is equal to zero.

[tex]$\overrightarrow{v}.\overrightarrow{a}= 0[/tex]

We have a football player who kicked a football in order to do a goal.

We have to determine the relationship between the direction of the velocity and acceleration vectors when the football reaches its maximum height.

What is Projectile Motion ?

Projectile motion is the two - dimensional motion of an object thrown or projected into the air such that it moves under the influence of acceleration of gravity.

According to the question -

Assume that the velocity with which the projectile was fired is u m/s.

Therefore, it can be resolved into two components as -

[tex]\overrightarrow{u} =[/tex] u(x) + u(y) = u cosθ [tex]a_{x}[/tex] + u sinθ [tex]a_{y}[/tex].

Now, when the projectile reaches the maximum height, it will no longer cover any vertical height but it will keep moving in the horizontal direction. Therefore - the vertical component of the velocity will become zero. Therefore -

[tex]\overrightarrow{u} =[/tex] u(x) + u(y) = u cosθ [tex]a_{x}[/tex] + 0 = u cosθ [tex]a_{x}[/tex]

Now, refer to the figure attached for reference -

At the maximum height, the velocity vector is in the horizontal direction and the vector for the acceleration due to gravity is vertically downwards.

Therefore - the acceleration vector and velocity vector will have an angle of 90° between them. Assume -

Velocity at maximum height = v = u cosθ [tex]a_{x}[/tex]

and acceleration = a = g = 9.8 m/[tex]s^{2}[/tex]

Therefore -

The dot product of velocity vector and acceleration vector is equal to zero.

[tex]\overrightarrow{v}.\overrightarrow{a}= 0[/tex]

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Suppose a point charge is located at the center of a spherical surface. The electric field at the surface of the sphere and the total flux through the sphere are determined. Now the radius of the sphere is halved. What happens to the flux through the sphere and the magnitude of the electric field at the surface of the sphere? (a) The flux and field both increase.(b) The flux and field both decrease.(c) The flux increases, and the field decreases. (d) The flux decreases, and the field increases.(e) The flux remains the same, and the field increases. (f) The flux decreases, and the field remains the same.

Answers

Answer:

gogle it

Explanation:

i cant help

Four toy cars are moving at a constant speed until they experience an unbalanced force of 12 n each. Which toy car would have an acceleration of 3 m/s2 ?.

Answers

The car's acceleration would be for a mass of 4 kg is 3 m/s²

Given the information below:

12 Newtons of force.

Increasing speed Equals

By using Newton's Second Law of Motion, we can determine the mass of the car:

According to Newton's second law of motion, force is equal to the rate at which momentum changes. Force is defined as mass times acceleration for a constant mass.

force= mass* acceleration

This formula provides Newton's Second Law of Motion mathematically;

12= mass*3

mass=12/ 3

mass=4 kilogram

By changing the parameters in the formula, we obtain;

= 4 kilos of mass.

As a result, the vehicle with a mass of 4 kg would accelerate at 3 m/s².

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The cosmic rays of highest energy are protons that have kinetic energy on the order of 10¹³ MeV (a) As measured in the proton's frame, what time interval would a proton of this energy require to travel across the Milky Way galaxy, which has a proper diameter ≅10⁵ly ?

Answers

The relativistic time it takes the proton to travel the milky way is 9.877 × 10⁻⁷ s

How to find the time it takes the proton to travel the milky way?

Since the question has to do with relativistic motion, we need to find the relativistic time it takes the proton to travel the milky way. First, we use the equation for relativistic kinetic energy, we have

K = mc(γ - 1) where

K = kinetic energy of proton = 10¹³ MeV = 10¹³ × 1.602 × 10⁻¹³ J = 1.602 J, m = rest mass of proton = 1.6726 × 10⁻²⁷ kg, c = speed of light = 3 × 10⁸ m/s and γ = 1/[√(1 - (v/c)²] where v = speed of proton

So, making γ subject of the formula, we have

γ  = K/mc + 1

Substituting the values of the variables into the equation, we have

γ  = K/mc + 1

γ  = 1.602J/(1.6726 × 10⁻²⁷ kg × 3 × 10⁸ m/s) + 1

γ  = 1.602J/(5.0178 × 10⁻¹⁹ J) + 1

γ  = 0.3193 × 10¹⁹ + 1

γ  = 3.193 × 10¹⁸

The distance moved by the proton

Now since the proper diameter of the milky way d = 10⁵ ly, since the length moved by the proton is going to be contracted, the actual distance moved by the proton is

d' = d/γ

= 10⁵ ly/3.193 × 10¹⁸

= 0.3132 × 10⁻⁻¹³ ly

= 0.3132 × 10⁻⁻¹³ × 9.46 × 10¹⁵ m

= 2.963 × 10² m

The velocity of the proton

Next we need to find the velocity of the proton from

γ = 1/[√(1 - (v/c)²]

3.193 × 10¹⁸ = 1/[√(1 - (v/c)²]

√(1 - (v/c)² = 1/3.193 × 10¹⁸

√(1 - (v/c)² = 0.3132 × 10¹⁸

1 - (v/c)² = (0.3132 × 10⁻¹⁸)²

1 - (v/c)² = 0.0981 × 10⁻³⁶

(v/c)² = 1 - 0.0981 × 10⁻³⁶

(v/c)² = 1 - 0.0000000000000000000000000000000000981

(v/c)² = 0.9999999999999999999999999999999999019

(v/c)² ≅ 0.99999

(v/c) = √0.99999

v/c = 0.999994999987

v = 0.999994999987c

The time it takes the proton to travel the milky way galaxy

So, the time it takes the proton to travel the milky way is t = d/v

= 2.963 × 10² m/0.999994999987c

= 2.963 × 10² m/2.99998499996 × 10⁸ m/s

= 0.9877 × 10⁻⁶ s

= 9.877 × 10⁻⁷ s

So, the time it takes the proton to travel the milky way is 9.877 × 10⁻⁷ s

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What happens when you push or pull a water tube

Answers

The thing that happens when you push or pull a water tube is that the force makes the tube to speed up, slow down, or be in one place.

It is very hard when pulling  water and if a person need to consistently pull air out of the tube before water makes it to the pump, it is very  difficult and some pumps may or cannot do it.

What is the pulling about?

In regards to pushing, when a person is pushing above the Center of Gravity, one is pushing the said object into the ground and thus is increasing the rate of friction.

If a person is pulling you are said to be pulling above the Center of Gravity and lowering friction.

So, The thing that happens when you push or pull a water tube is that the force makes the tube to speed up, slow down, or be in one place.

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if , determine the angles θ and ϕ so that the resultant force is directed along the positive x axis and has a magnitude of .

Answers

The angle θ is 36.3° and the angle ϕ is 26.4°.

A vector quantity is a quantity having both magnitude and direction.

To add vector quantities and get their resultants, we need to apply the parallelogram law of vector addition

Let A and B are two vectors and C is their resultant, then according to the parallelogram law

C^2 = A^2 + B^2 + 2ABcosθ, where  θ is the angle between the two vectors

On applying the parallelogram law in the given diagram,

(40)^2 = 30^2 + 60^2 - 2(30)(60)cosθ

Θ = 36.3°

And

30^2 = 40^2 + 60^2 - 2(40)(60)cosϕ

Φ = 26.4°

Thus, the angle θ is 36.3° and the angle ϕ is 26.4°.

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when the syringe volume is suddenly reduced (by roughly a factor of 2), the pressure changes by more than a factor of 2. why does the pressure spike to a value more than twice its initial value?

Answers

When the syringe is compressed the pressure spike to a value more than twice its initial value because the temperature of the gas increases.

What is pressure?

Pressure is the force applied perpendicularly to an object's surface in relation to the area across which it is applied (symbol: p or P). Gauge pressure is the pressure when compared to the ambient pressure.

Various units are used to express pressure. The pound-force per square inch (psi), which is equal to one newton per square metre (N/m2) in the SI, is the standard unit of pressure in the imperial and U.S. customary systems. Some of these results derive from dividing a unit of force by a unit of area. The atmosphere, which is equal to this pressure, and the torr, which is defined as 1760 of this, are two other ways to describe pressure.

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In an ultrahigh vacuum system (with typical pressures lower than 10⁻⁷ pascal), the pressure is measured to be 1.00 ×10⁻¹° torr (where 1 torr = 133 Pa). Assuming the temperature is 300 K , find the number of molecules in a volume of 1.00m³ .

Answers

The number of molecules in the given volume is 5.33 x 10⁻¹² moles.

What is the number of molecules of the gas?

The number of molecules of the gas is calculated as follows;

PV = nRT

n = PV/RT

where;

P is the pressure of the gasV is the volume of the gasT is temperature of the gasn is number of molecules of the gas

1 torr = 133 Pa

1 x 10⁻¹⁰ torr =  ?

= 1.33 x 10⁻⁸ Pa

n = (1.33 x 10⁻⁸ x 1) / (8.314 x 300)

n = 5.33 x 10⁻¹² moles

Thus, the number of molecules in the given volume is 5.33 x 10⁻¹² moles.

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the density of osmium (the densest metal) is . if a -kg rectangular block of osmium has two dimensions of cm, calculate the third dimension of the block.

Answers

The block's third dimension is 2.77 cm.

The mass of the osmium block is 1 kg.

The dimensions of the rectangular osmium block are 4 cm × 4 cm.

Density is defined as the mass of the object by the volume of the object.

density = mass / volume

Now, the volume of a rectangle is:

Volume = length × breadth × height

And now let's assume that the block's height is y.

Then,

Volume = length × breadth × height

Volume = 4 cm × 4 cm × y cm

Osmium now has a density of ρ = 22.57 g/cm³.

Therefore, the density of the rectangular osmium will be:

density = mass / volume

[tex]\rho = \frac{1{~}kg}{4{~}cm {~}\times {~}4{~}cm {~}\times {~} y}[/tex]

[tex]22.57 {~}g/cm^{3}=\frac{1000{~}g}{4{~}cm \times 4{~}cm \times y}[/tex]

[tex]16{{~}cm^{2} \times y{~} =\frac{1000{~}g}{22.57{~} g/cm^{3}}[/tex]

16y cm² = 44.31 cm³

Dividing each side of the equation by 16 cm²,

( 16y cm² ÷ 16 cm²) = ( 44.31 cm³ ÷ 16 cm² )

y = 2.77 cm

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The complete question is mentioned below:

The density of osmium is 22.57 g/cm³. If a 1.00kg rectangular block of osmium has two dimensions of 4.00 cm x 4.00 cm, calculate the third dimension of the block.

how many times higher could an astronaut jump on the moon than on earth if his takeoff speed is the same in both locations (gravitational acceleration on the moon is about 1/6 of size 12{g} {} on earth)?

Answers

An astronaut could 6 times higher on the moon than on earth if his takeoff speed is the same in both locations.

What was it like for astronauts on the lunar surface?

From lunar circle, space explorers guided cameras through the window of their space apparatus toward catch photographs of the moon's surface. The nearest look we've had at the moon came from the sendoff of NASA's Apollo program during the 1960s. Somewhere in the range of 1967 and 1972, a progression of missions handled the primary men on the moon. At the point when the following space explorer to arrive at the moon strolls on the lunar surface in 2024, she'll confront radiation levels multiple times higher than on Earth. While Apollo mission space travelers. The prompt surface was extremely fine, as best we could see peering down from 15 feet. Off somewhere out there was an exceptionally clear skyline, perhaps with a rock.

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which si unit for legnth is greater than a meter a. decimeter b. centimeter c. kilometer d. milimeter

Answers

Option(c) is Kilometer is correct. Kilometer unit for length is greater than a meter.

What do you mean by SI units?

Worldwide adoption of the International System of Units (SI) as a system of measurement units is widespread. For convenience, this updated version of the metric system is based on the number 10. The metric prefixes or the SI prefixes are a collection of prefixes that have been established. According on the prefixes, the unit is either a multiple or a fraction of base ten. It enables the zeros of very small or very high numbers to be reduced, for example, from 7,500,000 Joules to 1 nanometer and from 0.000000001 meter to 7.5 Megajoules, respectively. Additionally, some SI prefixes have a group of symbols before the unit symbol.

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