What is the effect on the gas’ pressure when compressing a gas to 1/3 of its volume? Explain

Answers

Answer 1

Explanation:

It's pressure become three times larger because according to Boyles Law the pressure of fixed mass of gas is inversely proportional to it's volume provided that temp remains constant. That means a reduction in volume, will result in an increase in pressure and vice versa.


Related Questions

John is gliding along on his scooter at a speed of 2.8 m/s when
suddenly he collides with Sam who is walking around the corner. If
John is brought to rest in 0.15 s, what is his acceleration?

Answers

Answer: His acceleration is -18.66m/s^2

Explanation:

Ok, the initial speed is 2.8m/s. (we can define the initial direction as the positive direction).

And he wants to stop, so he must accelerate in the opposite direction as the initial movement, then we would have:

a(t) = -A.

So we have a constant, and negative acceleration.

Now, if we want to find the velocity we must integrate over time, and we will get:

v(t) = -A*t + V0

where V0 is the initial velocity, we know that it is 2.8m/s, and t is the time in seconds.

Then the velocity is:

v(t) = -A*t + 2.8m/s.

Now we know that John is brought to rest in 0.15 seconds after he starts slowing down, this means that at t = 0.15 seconds, his velocity is equal to zero.

v(0.15s) = 0m/s = -A*0.15s + 2.8m/s

2.8m/s = A*0.15s

2.8m/s/0.15s = 18.66m/s^2 = A.

So his acceleration is -A, then we have that:

His acceleration is -18.66m/s^2

when the temperature of matter decrease , the particles do what

Answers

Answer:

When the temperature decreases the particals start to slow down.

A falling object satisfies the initial value problem dv dt = 9.8 − v 5 , v(0) = 0 where v is the velocity in meters per second. (a) Find the time that must elapse for the object to reach 95% of its limiting velocity. (Round your answer to two decimal places.) s (b) How far does the object fall in the time found in part (a)? (Round your answer to two decimal places.) m Additional Materials

Answers

Answer:

a.  t [tex]\simeq[/tex] 14.98 sec

b.   x = 501.27 m

Explanation:

From the given information:

[tex]\dfrac{dv}{dt}=9.8-(\dfrac{v}{5 })[/tex]  and   [tex]v(0)=0[/tex]

[tex]\dfrac{dv}{dt}=\dfrac{49-v}{5 }[/tex]

[tex]\dfrac{dv}{49-v}=\dfrac{dt}{5 }[/tex]

Taking  Integral of  both sides

[tex]- ln(49-v) = \dfrac{t}{5} + C[/tex]  

at t=0 we have v=0

This implies that

[tex]- ln(49-0) = \dfrac{0}{5} + C[/tex]

[tex]C= - ln(49)[/tex]

Thus:

[tex]\dfrac{t}{5} - In (49) = - In (49 -v) \\ \\ In(49) - \dfrac{t}{5} = In (49-v)[/tex]

[tex]49-v = e^{(-\frac{t}{5} +ln(49))}\\ \\ v = 49 - 49e^{(-\dfrac{t}{5})}[/tex]

The limiting velocity when the time is infinite is :

95% of 49 = 46.55

[tex]0.05= e^{(-\dfrac{t}{5})}[/tex]

[tex]\dfrac{t}{5}= In(\dfrac{1}{0.05})[/tex]

[tex]\dfrac{t}{5}=2.9957[/tex]

t = 5 × 2.9957

t [tex]\simeq[/tex] 14.98 sec

b.) [tex]v = 49 - 49e^{(-\dfrac{t}{5})}[/tex]

[tex]v = \dfrac{dx}{dt}=49 - 49e^{(-\dfrac{t}{5})}[/tex]

[tex]dx=(49 - 49e^{(-\frac{t}{5})}) \ dt[/tex]

Taking integral of both sides.

[tex]x = 49t + 245 e^{(\frac{-t}{5})} +C[/tex]

 at time t = 0 , distance x traveled = 0

C= - 245

Therefore

[tex]x = 49t + 245 e^{(\frac{-t}{5})} -245[/tex]

replacing the value of t = 14.98

[tex]x = 49(14.98) + 245 e^{(\frac{-14.98}{5})} -245[/tex]

x = 501.27 m

An ionized oxygen molecule (O2+) at point A has charge +e and moves at 1.24 ✕ 103 m/s in the positive x-direction. A constant electric force in the negative x-direction slows the molecule to a stop at point B, a distance of 0.766 mm past A on the x-axis. Calculate the x-component of the electric field and the potential difference between points A and B. (The mass of an oxygen molecule is 5.31 ✕ 10−26 kg and the fundamental charge is e = 1.60 ✕ 10−19 C.)

Answers

Answer:

[tex]\mathbf{E =3.33 \times 10^2 \ N/C}[/tex]

[tex]\mathbf{ V_A - V_B = 0.2551 \ Volts}[/tex]

Explanation:

Given that:

The charge on the ionized oxygen molecule = +e

The speed of the ionized oxygen molecule with this charge is 1.24 × 10³ m/s

distance travelled by the particle before rest is d = 0.766 m

According to the third equation of motion.

[tex]v^2 = u^2 +2as[/tex]

[tex]v^2 = u^2 +2(\dfrac{-eE}{m}) s[/tex]

[tex]0^2= u^2 +2(\dfrac{-eE}{m}) s[/tex]

[tex]E = \dfrac{mu^2}{2e* \ s}[/tex]

[tex]E = \dfrac{5.31 *10^{-26}* (1.24*10^3)^2}{2*1.6*10^{-19}*0.766*10^{-3}}[/tex]

[tex]\mathbf{E =3.33 \times 10^2 \ N/C}[/tex]

Thus, the electric field shows to be in the negative x-direction.

The potential difference between point A and B now is:

[tex]\Delta V = E.d \\ \\ V_A - V_B = 3.33 \times 10^2 \times 0.766 \times 10^{-3}[/tex]

[tex]\mathbf{ V_A - V_B = 0.2551 \ Volts}[/tex]

Which situation is most likely to lead to improving a scientific theory?
A. Analyzing the history of how a theory was developed
O B. Gathering scientists for a brainstorming session
c. Increasing public awareness of the need for scientific research
O D. Performing experiments that have never been done before
SUBMIT

Answers

Performing experiments that have never been done before is most likely to lead to improving a scientific theory. The correct option is D.

What is scientific theory?

A theory is a well-thought-out elaboration for natural-world observational data that has been built using the scientific process and incorporates many facts and hypotheses.

A theory not only helps to explain established information; it also allows scientists to predict what they will see if the hypothesis is correct.

Scientific theories can be tested. Emerging data should be consistent with an existing theory.

To improve a scientific theory, conduct novel experiments on it. This will provide greater clarity on the theory.

It may reveal the limitations of this theory, as well as aspects of the theory that can be improved.

Thus, the correct option is D.

For more details regarding scientific theory, visit:

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what is the net force on a backpack with a mass of 12.0 kg and an acceleration of 0.5 m/s^2?

Answers

Answer:

The answer is

6.0 N

Explanation:

The net force acting on an object when given the mass and acceleration can be found by using the formula

Force = mass × acceleration

From the question

mass of backpack = 12 kg

acceleration = 0.5 m/s²

So the force is

Force = 12 × 0.5

We have the final answer as

6.0 N

Hope this helps you

4. What is the instantaneous acceleration at t= 10 s?

Answers

Answer:

I am fairly certain the answer is 2m/s^2

Explanation:

The morning after a massive snowstorm, Michaela gets into her car to drive to work. The storm caused her windows to freeze, so she first needs to defrost the car. While the engine is running, she checks the thermometer. It shows the air inside of her car has a temperature of 0 °C. Does this mean the air inside of her car has no kinetic energy? Explain your answer.

Answers

Answer:

Hope it helps

A Brainliest please

In a circus act, an acrobat rebounds upward from the surface of a trampoline at the exact moment that another acrobat, perched 9.0 m above him, releases a ball from rest. While still in flight, the acrobat catches the ball just as it reaches him.Required:If he left the trampoline with a speed of 5.6 m/s, how long is he in the air before he catches the ball?

Answers

Answer:

1.6 secs

Explanation:

In a circus act, an acrobat upwards from the surface of a trampoline

At that same moment another acrobat perched 9.0m above him

A ball is released from rest

While still in motion the acrobat catches the ball

He left the ball with a trampoline of 5.6m/s

Since the ball is falling downwards from a distance then acceleration will be negative

a= -9.8

s= d

s= 1/2at^2

= 1/2 × (-9.8)t^2

= 0.5× (-9.8)t^2

d = -4.9t^2

Therefore the time the acrobat stays in the air before catching the ball can be calculated as follows

9 - 4.9t^2= 5.6t + 1/2(-9.8)t^2

9 - 4.9t^2= 5.6t + (-4.9)t^2

9 - 4.9t^2= 5.6t - 4.9t^2

9= 5.6t

t= 9/5.6

t= 1.6 secs

The slope of the x-t curve at any point represents:_______a. Displacementb. Velocityc. Accerlationd. None of these

Answers

Answer:

Velocity

Explanation:

x-t curve means position vs time graph in which position (x) is in x-axis and time is in y-axis.

Slope of a graph = [tex]\dfrac{\Delta y}{\Delta t}[/tex].

We know that, velocity = displacement/time

or we can say that slope = displacement/time = velocity

Hence, the correct option is (b) "velocity".

An owl is carrying a mouse to the chicks in its nest. It is 4.00 m west and 12.0 m above the center of the 30 cm diameter nest and is flying east at 2.50 m/s at an angle 32° below the horizontal when it accidentally drops the mouse. Will it fall into the nest? Find out by solving for the horizontal position of the mouse (measured from the point of release) when it has fallen the 12.0 m.

Answers

Answer:

4.61 m away

Explanation:

No, the owl will not fall into the nest.

The vertical component of the velocity, v of mouse after travelling vertical distance of 12 m is given by

v² - (4 * sin 32)² = 2 * 9.81 * 12

v² - (4 * 0.53)² = 235.44

v² - 4.494 = 235.44

v² = 235.44 + 4.494

v² = 239.934

v = 15.49 m/s

The time taken to travel the said 12 m downwards =

(15.49 - 4 * sin 32) / 9.81

(15.49 - 4 * 0.53) / 9.81

(15.49 - 2.12) / 9.81

13.37 / 9.81 = 1.36 s

Horizontal distance traveled in this time = 4 * (cos 32) * 1.36

= 4 * 0.848 * 1.36

= 4.61 m

Therefore, the horizontal position of the mouse is 4.61 m away from the point of release.

Your office has a 0.025 m^3 cylindrical container of drinking water. The radius of the container is about 13 cm. Required:a. When the container is full, what is the pressure that the water exerts on the sides of the container at the bottom of the container above the atmospheric pressure? b. When the container is full, what is the pressure that the water exerts on the sides of the container halfway down from the top above the atmospheric pressure?

Answers

Answer:

a) 105935.7 Pa

b) 103630.35 Pa

Explanation:

The volume of the container = 0.025 m^3

The radius of the container = 13 cm = 0.13 m

We have to find the height of the tank

From the equation for finding the volume of the cylinder,

V = [tex]\pi r^2h[/tex]

where

V is the volume of the cylinder

h is the height of the cylinder

substituting values, we have

0.025 = 3.142 x [tex]0.13^2[/tex] x h

0.025 = 0.0531h

h = 0.025/0.0531 = 0.47 m

Pressure at the bottom of the tank P = ρgh

where

ρ is the density of water = 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

h is the depth of water which is equal to the height of the tank

substituting values, we have

P = 1000 x 9.81 x 0.47 = 4610.7 Pa

atmospheric pressure = 101325 Pa

therefore, the pressure in the tank bottom above atmospheric pressure = 101325 Pa + 4610.7 Pa = 105935.7 Pa

b) For half way down the container, depth of water will be = 0.47/2 = 0.235 m

pressure P = 1000 x 9.81 x 0.235 = 2305.35 Pa

This pressure above atmospheric pressure = 101325 Pa + 2305.35 Pa = 103630.35 Pa

(a) The pressure exerted by water on the sides of the container at the bottom is 4606 Pa.

(b) The pressure exerted by water on the sides of the container halfway down from the top is of 2303 Pa.

Given data:

The volume of cylindrical container is, [tex]V = 0.025 \;\rm m^{3}[/tex].

The radius of container is, r = 13 cm = 0.13 m.

(a)

When the container is full, the pressure that the water exerts on the sides of the container is given as,

[tex]P = \rho gh[/tex]

Here, [tex]\rho[/tex]  is the density of water, g is the gravitational acceleration and h is the height of container and its value is obtained as,

[tex]V = \pi r^{2}h\\\\0.025= \pi \times (0.13)^{2} \times h\\\\h = 0.47 \;\rm m[/tex]

Then pressure is,

[tex]P = 1000 \times 9.8 \times 0.47\\\\P=4606 \;\rm Pa[/tex]

Thus, the pressure exerted by water on the sides of the container at the bottom is 4606 Pa.

(b)

For the ides of the container halfway down from the top, the height is,

h' = h/2

h' = 0.47/2 = 0.235 m

Then the pressure is obtained as,

[tex]P' = \rho \times g \times h'\\\\P' = 1000 \times 9.8 \times 0.235\\\\P'=2303 \;\rm Pa[/tex]

Thus, the pressure exerted by water on the sides of the container halfway down from the top is of 2303 Pa.

Learn more about the atmospheric pressure here:

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measurement conversions [metric to metric] 568 cm = m​

Answers

Answer:

5.68 meters

Explanation:

hope this helps!

Answer:

5.68

Explanation:

to convert cm to m you  move the decimal point 2x to the left

A two-liter bottle is one-fourth full of water and three-quarters full of air. The air in the bottle has a gage pressure of 340 kPa. The bottle is turned upsidedown and the cap is released so that the water is rapidly forced out of the bottle. If the air in the bottle undergoes an adiabatic pressure change, what is the pressure in the bottle when the bottle is five-sixths full of air

Answers

Answer:

The value is     [tex] P_G  =  2.925 *10^{5} \  Pa[/tex]

Explanation:

From the question we are told that

   The  volume of the bottle is  [tex]v  =  2 \  L  =  2 *  10^{-3} \  m^3[/tex]

   The gauge pressure of the air is [tex]P_g  =  340 \  kPa  =  340 *10 ^{3}  \  Pa[/tex]

   

Generally the volume of air before the bottle is turned upside down is  

      [tex]V_a  =  \frac{3}{4}  * V[/tex]

       [tex]V_a  =  \frac{3}{4}  *  2 *10^{-3}[/tex]

       [tex]V_a  =  0.0015 \  m^3 }[/tex]

Generally the volume air when the bottle is turned upside-down is

      [tex]V_u  =  \frac{5}{6}  *  2 *10^{-3}[/tex]

       [tex]V_u  =  0.00167 \  m^3 [/tex]

From the the mathematical relation of adiabatic process we have that

     [tex]P_g *  V_a^r  =  P_G *  V_u^r[/tex]

Here r is a constant with  a value  r =  1.4

So

      [tex] 340 *10 ^{3}  *  0.0015^{1.4}  =  P_G *  0.00167^{1.4}[/tex]

        [tex] P_G  =  2.925 *10^{5} \  Pa[/tex]

Part A
Suppose that you want to construct a line with slope m=3 that passes through the point (2,1). You would begin by setting up the equation
y=3x+b.
If you plug in the coordinates for any point on that line, the two sides of the equation will be equal. Once you've done this, you can solve for b. What is the value of b?
Express your answer as an integer.
Part B
Suppose that you want to find the equation for a line that passes through the two points (0,3) and (4,9). What is the slope of this line?
Express your answer numerically.

Answers

Answer:

a) b = -5

b) slope = 3/2

Explanation:

a) The equation of a line is given as y = mx + b, where m is the slope of the line and b is the intercept on the y axis.

Given that y = 3x + b and it passes through the point (2, 1). Hence when x = 2, y = 1. Therefore, substituting for x and y:

1 = 3(2) + b

1 = 6 + b

b = 1 - 6

b = -5

b) The equation of a line passing through two points ([tex]x_1,y_1[/tex]) and [tex]x_2,y_2[/tex] is given by:

[tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]

The equation of the line passing through the two points (0,3) and (4,9) is:

[tex]y-3=\frac{9-3}{4-0}(x-0)\\ \\y-3=\frac{3}{2}x\\ \\y = \frac{3}{2}x+3[/tex]

Comparing y = (3/2)x + 3 with y = mx + b, the slope (m) is 3/2

The volume of water and an egg in a graduated beaker is 200mL. After the egg is removed the volume of the water is found to be 125mL . What is the volume of the egg in cm ^3

Answers

Answer:

75 cm³

Explanation:

The volume of the egg is equal to the volume of the egg and water minus the volume of the water.

V = 200 mL − 125 mL

V = 75 mL

V = 75 cm³

Answer:

75 cm

Explanation:

The equation would be 200ml -125 ml thus finding the volume of the egg.

What is the force applied by the ground called? A. applied force B. friction C. gravity

Answers

Answer:

gravity i think hope this helps

Explanation:

A constant electric field with magnitude 1.50 ✕ 103 N/C is pointing in the positive x-direction. An electron is fired from x = −0.0200 m in the same direction as the electric field. The electron's speed has fallen by half when it reaches x = 0.190 m, a change in potential energy of 5.04 ✕ 10−17 J. The electron continues to x = −0.210 m within the constant electric field. If there's a change in potential energy of −9.60 ✕ 10−17 J as it goes from x = 0.190 m to x = −0.210 m, find the electron's speed (in m/s) at x = −0.210 m.

Answers

Answer:

The speed of electron is [tex]1.5\times10^{7}\ m/s[/tex]

Explanation:

Given that,

Electric field [tex]E=1.50\times10^{3}\ N/C[/tex]

Distance = -0.0200

The electron's speed has fallen by half when it reaches x = 0.190 m.

Potential energy [tex]P.E=5.04\times10^{-17}\ J[/tex]

Change in potential energy [tex]\Delta P.E=-9.60\times10^{-17}\ J[/tex] as it goes x = 0.190 m to x = -0.210 m

We need to calculate the work done by the electric field

Using formula of work done

[tex]W=-eE\Delta x[/tex]

Put the value into the formula

[tex]W=-1.6\times10^{-19}\times1.50\times10^{3}\times(0.190-(-0.0200))[/tex]

[tex]W=-5.04\times10^{-17}\ J[/tex]

We need to calculate the initial velocity

Using change in kinetic energy,

[tex]\Delta K.E = \dfrac{1}{2}m(\dfrac{v}{2})^2+\dfrac{1}{2}mv^2[/tex]

[tex]\Delta K.E=\dfrac{-3mv^2}{8}[/tex]

Now, using work energy theorem

[tex]\Delta K.E=W[/tex]

[tex]\Delta K.E=\Delta U[/tex]

So, [tex]\Delta U=W[/tex]

Put the value in the equation

[tex]\dfrac{-3mv^2}{8}=-5.04\times10^{-17}[/tex]

[tex]v^2=\dfrac{8\times(5.04\times10^{-17})}{3m}[/tex]

Put the value of m

[tex]v=\sqrt{\dfrac{8\times(5.04\times10^{-17})}{3\times9.1\times10^{-31}}}[/tex]

[tex]v=1.21\times10^{7}\ m/s[/tex]

We need to calculate the change in potential energy

Using given potential energy

[tex]\Delta U=-9.60\times10^{-17}-(-5.04\times10^{-17})[/tex]

[tex]\Delta U=-4.56\times10^{-17}\ J[/tex]

We need to calculate the speed of electron

Using change in energy

[tex]\Delta U=-W=-\Delta K.E[/tex]

[tex]\Delta K.E=\Delta U[/tex]

[tex]\dfrac{1}{2}m(v_{f}^2-v_{i}^2)=4.56\times10^{-17}[/tex]

Put the value into the formula

[tex]v_{f}=\sqrt{\dfrac{2\times4.56\times10^{-17}}{9.1\times10^{-31}}+(1.21\times10^{7})^2}[/tex]

[tex]v_{f}=1.5\times10^{7}\ m/s[/tex]

Hence, The speed of electron is [tex]1.5\times10^{7}\ m/s[/tex]

A tennis ball is hit into the air and moves along an arc. (a) Neglecting air resistance, where along the arc is the speed of the ball a minimum? (Select all that apply.) at the initial position of motion at the highest point in the trajectory at the final position of motion

Answers

Answer:

at the highest point in the trajectory

Explanation:

When the tennis ball is hit, it moves in the air along a curve or an arc. This path is the parabola curve. Such a motion in the two dimension is known as projectile motion. It is constant accelerated motion in the downward direction.

The velocity of the ball is minimum at the highest point of the motion. When we hit the ball, the ball moves up to certain eight and then it gradually fall back to the earth surface along a curve.

The horizontal velocity of the ball is always the same along the curve. Only the vertical velocity varies. As the ball reaches the top of the curve or the maximum height, its vertical velocity becomes zero.

So, speed of the tennis ball is minimum at the highest point of the path.

Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 6.00 m away from the slits.A. Which laser has its first maximum closer to the central maximum?B. What is the distance delta ymax-min between the first maxima (on the same side of the central maximum) of the two patterns?C. What is the distance Delta ymax-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?

Answers

Answer:

Now

Using

ym = m x λ x L/d

therefore for each of first Max we have m =1

And laser 1 had

y = (d/20) x (6.0m)/d

y = 6.0m /20 = 0.3

Then laser 2 will be

y = 6.0m/15 = 0.4

y ( laser1 ) < y ( laser2 )

So the first maxima of laser 1 will be closer to the Central maxima

So

(0.4m -0 .3 m) = 0.1m

( C)

Now for laser 1 we say

y= 26.0 m / 20 = 0.6 m

Laser 2

We have

ym=(m+1/2) x λ x L/d

So

Because there is no central minimum the first minimum is at m = 0

We can way 3rd minimum is at m = 2

So

y = (2.5) x 6.0 / 15 = 1m

So

Δy= 1m - 0.6m = 0.4 m

What physical property of Earth gives rise to the seasons?

Answers

Answer:

Earth's tilted axis causes the seasons. Throughout the year, different parts of Earth receive the Sun's most direct rays. So, when the North Pole tilts toward the Sun, it's summer in the Northern Hemisphere. And when the South Pole tilts toward the Sun, it's winter in the Northern Hemisphere.

Explanation:

Which formula is used to find an object's acceleration? ​

Answers

a=v-v0 divided by t equals delta(the triangle) V divided by delta t

Answer:

The answer is C.

Explanation:

A car traveling at 50 km/h hits a bridge abutment. A passenger in the car moves forward a distance of 61 cm (with respect to the road) while being brought to rest by an inflated air bag. What magnitude of force (assumed constant) acts on the passenger's upper torso, which has a mass of 44 kg?

Answers

Answer:

6957.04N

Explanation:

Using

vf2=vi2+2ad

But vf = 0 .

So convert 50km/hr to m/s, and you need to convert 61 cmto m

(50km/hr)*(1hr/3600s)*(1000m/km) = 13.9m/s

61cm * (1m/100cm) = .61m

So n

0 = (13.9m/s)^2 + 2a(.61m)

a = 158.11m/s^2

So

using F = ma

F = 44kg(158.11m/s^2) = 6957.04N

Answer:

6.95 kN

Explanation:

Given that

Speed of the car, u = 50 kmph

Distance moved by the passenger, s = 61 cm = 0.061 m

Mass of the passenger, m = 44 kg

Converting the initial velocity from km/h to m/s, we have

50 kmph = 50 * 1000/3600

50 kmph = 13.89 m/s

Using one of the equations of motion,

v² = u² + 2as, where v = 0,making a the subject of formula

a = -u²/2s

a = -(13.89²) / 2 * 0.61

a = -192.93 / 1.22

a = -158 m/s²

The force acting on the passenger's leg, F = m.a, so

-F = 44 * -158

F = 6952 N, or 6.95 kN

A tennis ball is shot vertically upward inside a tower with an initial speed of
20.0 m/s. Neglect air resistance. Approximately how long does it take the tennis ball to reach its maximum height?
0.50 s
2.04 s
4.08 s
6.08 s
9.80 s

Determine the velocity of the ball 3.0 s after it is thrown
9.40 m/s, downward
9.40 m/s, upward
29.4 m/s, downward
38.8 m/s, upward
38.8 m/s, downward

Answers

The ball's height at time t is

y = (20.0 m/s) t - 1/2 g t²

where g is the acceleration due to gravity, with magnitude 9.80 m/s².

Also, recall that

v² - u² = 2 ay

where u is the initial velocity, v is the final velocity, a is the acceleration, and ∆y is the change in height. Let Y be the maximum height. At this height, v = 0, so

- (20.0 m/s)² = 2 (-g) Y

==>  Y ≈ 20.408 m

Plug this into the first equation and solve for t :

Y = (20.0 m/s) t - 1/2 (9.80 m/s²) t²

==>  t ≈ 2.04 s

The ball's velocity at time t is

v = 20.0 m/s - g t

After t = 3.0 s, its velocity will be

v = 20.0 m/s - (9.80 m/s²) (3.0 s)

v = -9.40 m/s

or 9.40 m/s in the downward direction.

Which is larger, the Sun's pull on Earth or Earth's pull on the Sun?
a. The Sun's pull on Earth is twice as large as Earth's pull on the much larger Sun.
b. They pull on each other equally.
c. The Sun's pull on Earth is larger.
d. Earth's pull on the Sun is larger.
e. There is no pull or force between Earth and the Sun.

Answers

B it’s the most logical answer

At the time when the sun pulls on earth should be large so it should be pulled on each other equally.

Gravitational force:

Here the gravitational force that lies between two masses should be proportional with respect to the product of mass that should be divided via the square of the distance that lies between them. The acceleration should be measured when the force should be divided by the mass. Here the two forces should be equal and opposite

Therefore, the option b is correct.

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A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853 m/s2. What was the car’s velocity at point “A”?

Answers

Recall that

[tex]{v_f}^2-{v_i}^2=2a\Delta x[/tex]

where [tex]v_i[/tex] and [tex]v_f[/tex] are the initial and final velocities, respecitvely; [tex]a[/tex] is the acceleration; and [tex]\Delta x[/tex] is the change in position.

So we have

[tex]\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)[/tex]

[tex]\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}[/tex]

(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)

A train 4.00 3 102 m long is moving on a straight track with a speed of 82.4 km/h. The engineer applies the brakes at a crossing, and later the last car passes the crossing with a speed of 16.4 km/h. Assuming constant acceleration, determine how long the train blocked the crossing. Disregard the width of the crossing.

Answers

Answer:

The  value  is  [tex]t  =29.2  \  s [/tex]

Explanation:

From the question we are told that

Generally the average velocity of the train is mathematically represented as

          [tex]v  =  \frac{u +  v}{2}[/tex]

substituting  82.4 km/h for  u and   16.4 km/h. for  v

       [tex]v  =  \frac{82.4 + 16.4}{2}[/tex]

          [tex]v  =  49.4 \  km/h[/tex]

Generally the time taken is mathematically represented as

     [tex]t  =  \frac{ L}{v}[/tex]

substituting   49.4 \  km/h for  v and  [tex]4.00 * 10^2 \  m  =  0.400 \  km[/tex]

         [tex]t  =  \frac{ 0.400}{49.4}[/tex]

          [tex]t  = 0.00809 \  h [/tex]

converting to seconds

         [tex]t  = 0.00809 * 3600  [/tex]

          [tex]t  =29.2  \  s [/tex]

     

An open freight car rolls friction-free along a horizontal track in a pouring rain that falls vertically. As water accumalates in the car, its speed

Answers

Answer:

decrease

Explanation:

weight

As water accumulates in the car, its speed decrease according to Newton's second law of motion

What is newton's  second law of motion?

Newton's second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

according to newton's second law

force = mass * acceleration

as water accumulate in the car , mass will increase

since mass and acceleration are inversely proportional to each other

hence , acceleration will decrease and speed will decrease

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A group of engineers is preparing a satellite to land by moving it 10% closer to
Earth in each rotation. Which statement is correct about the rotational inertia of
the satellite? (1 point)
O Rotational inertia does not change because it is conserved.
Rotational inertia increases proportional to the decrease in the
radius of rotation.
Rotational inertia first decreases and then increases as the
satellite is ready to land.
O
Rotational inertia decreases proportional to the decrease in the
radius of rotation.

Answers

Answer:

Rotational inertia decreases proportional to the decrease in the radius of rotation.

Explanation:

The rotational inertia decreases proportional to the decrease in the

radius of rotation.

Rotational inertia is directly proportional to the angular momentum of the object and inversely proportional to its angular velocity.

[tex]I = \frac{L}{\omega}[/tex]

where;

I is the rotational inertiaL is the angular momentumω is the angular speed

The angular momentum is given as;

[tex]L = mvr[/tex]

where;

m is the mass of the objectv is the velocity of the objectr is the radius of the object

The new rotational inertia equation becomes;

[tex]I = \frac{mvr}{\omega}[/tex]

From this equation, we can observe that the rotational inertia is directly proportional to the radius of the object.

A 10% closer to the Earth, means a decrease in the radius of the satellite by 10%.

Thus, a decrease in the rotational inertia as well.

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1.
Which of the follow
following
cannot be zero?
A. Distance B. Displacement
" Velocity D. Acceleration
C.​

Answers

A.Distance , because distance cannot be zero
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