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Stels [109]
3 years ago
15

When scientists record the volume of a gas, why do they also record the temperature and the pressure

Physics
1 answer:
alekssr [168]3 years ago
6 0
They record the temp and Pressure as well because gas will change its volume based on the temperature of the gas and the pressure applied
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Two rollerbladers face each other and stand at rest on a flat parking lot. tracey has a mass of 32 kg, and jonas has a mass of 4
galina1969 [7]
Momentum is a product of mass and the velocity of a body. The initial momentum is always equal to the final momentum during collisions between two bodies.
Therefore; M1U1 +M2U2 = M1V1+ M2V2, where m1 is the mass of tracey and m2 is the mass of jonas, while u is the initial velocity and v is the final velocity.
 (32 ×0)+ (45×0) = (32 × v1) + (45 × 0.50)
                       0 = 32V1 + 22.5
                     32 V1 = -22.5
                          V1 = - 0.703 ( negative indicates difference in direction)
There, Tracey's speed will be 0.703 m/s
8 0
3 years ago
In this electric motor, an electric current causes the coil to rotate, changing
mojhsa [17]

Answer:

D

Explanation:

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7 0
4 years ago
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When describing a maritime polar air mass, you might say it formed over _____ and is _____.
Vilka [71]
The correct answer for the question that is being presented above is this one: "an air mass that formed over land. it is cold and dry." When describing a maritime polar air mass, you might say it formed over land and is cold and dry. It's temperature c<span>an range from around freezing to near 20 degrees Celsius</span>
3 0
3 years ago
A rocket rises vertically, from rest, with an acceleration of 3.6m/s^2 until it runs out of fuel at an altitude of 1500m. After
Montano1993 [528]

A. 103.9 m/s

The motion of the rocket until it runs out of fuel is an accelerated motion with constant acceleration a = 3.6 m/s^2, so we can use the following equation

v^2 -u^2 = 2ad

where

v is the velocity of the rocket when it runs out of fuel

u = 0 is the initial velocity of the rocket

a = 3.6 m/s^2 is the acceleration

d = 1500 m is the distance covered during this first part

Solving for v, we find

v=\sqrt{u^2+2ad}=\sqrt{(0^2+2(3.6 m/s^2)(1500 m)}=103.9 m/s

B. 28.9 s

We can calculate the time taken for the rocket to reach this altitude with the formula

d=\frac{1}{2}at^2

where

d = 1500 m

a = 3.6 m/s^2

Solving for t, we find

t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(1500 m)}{3.6 m/s^2}}=28.9 s

C. 2050.8 m

We can calculate the maximum altitude reached by the rocket by using the law of conservation of energy. In fact, from the point it runs out of fuel (1500 m above the ground), the rocket experiences the acceleration due to gravity only, so all its kinetic energy at that point is then converted into gravitational potential energy at the point of maximum altitude:

K_i = U_f\\\frac{1}{2}mu^2 = mgh

where h is distance covered by the rocket after it runs out of fuel, and v=103.9 m/s is the velocity of the rocket when it starts to decelerate due to gravity. Solving for h,

h=\frac{v^2}{2g}=\frac{(103.9 m/s)^2}{2(9.8 m/s^2)}=550.8 m

So the maximum altitude reached by the rocket is

h' = d+h=1500 m +550.8 m=2050.8 m

D. 39.5 s

The time needed for the second part of the trip (after the rocket has run out of fuel) can be calculated by

h=\frac{1}{2}gt^2

where

h = 550.8 m is the distance covered in the second part of the trip

g = 9.8 m/s^2

Solving for t,

t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(550.8 m)}{9.8 m/s^2}}=10.6 s

So the total time of the trip is

t'=28.9 s+10.6 s=39.5 s

E. 200.5 m/s

When the rocket starts moving downward, it is affected by gravity only. So the gravitational potential energy at the point of maximum altitude is all converted into kinetic energy at the instant the rocket hits the ground:

\frac{1}{2}mv^2 = mgh

where

v is the final velocity of the rocket

h = 2050.8 m is the initial altitude of the rocket

Solving for v,

v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(2050.8 m)}=200.5 m/s

F. 60 s

We need to calculate the time the rocket takes to fall down to the ground from the point of maximum altitude, and that is given by

h'=\frac{1}{2}gt^2

where

h' = 2050.8 m

g = 9.8 m/s^2

Solving for t,

t=\sqrt{\frac{2h'}{g}}=\sqrt{\frac{2(2050.8 m)}{9.8 m/s^2}}=20.5 s

So the total time of the trip is

t''=39.5 s+20.5 s=60 s

6 0
4 years ago
Assume you have a person with a mass of 65 kg riding a skateboard down a ramp. While they are
aliina [53]

Answer:

a) Eg = 3,060.72 j

b) Ek = 2,080 j

c) Etotal = 5,140.72 j

Explanation:

The given parameters are;

The mass of the person, m = 65 kg

The height of the person, h = 4.8 m

The speed of the person, v = 8.0 m/s

a) The gravitational potential energy, E_g = m·g·h

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

∴ Eg = 65 kg × 9.81 m/s² × 4.8 m = 3,060.72 j

Eg = 3,060.72 j

b) The kinetic energy, Ek = 1/2·m·v²

∴ Ek = 1/2 × 65 kg × (8.0 m/s)² = 2,080 j

Ek = 2,080 j

c) The constant total Mechanical Energy, Etotal = Eg + Ek

∴ Etotal = 3,060.72 j + 2,080 j = 5,140.72 j

Etotal = 5,140.72 j.

8 0
3 years ago
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