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3 years ago

A car has a momentum of 20,000 kg. m/s. What would the car's momentum be if its velocity doubles?

Physics

1 answer:

Savatey [412]3 years ago

8 0

Answer:

40,000

Explanation:

Momentum is defined as mass*velocity, so a doubling of velocity means a doubling of momentum

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How many Joules in 41.87 kcal?

gladu [14]

The answer is 175184.08 joules

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3 years ago

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Imagine you derive the following expression by analyzing the physics of a particular system: v2=v20+2ax. The problem requires so

lisabon 2012 [21]

As per kinematics equation we are given that

$v^2 = v_o^2 + 2ax$

now we are given that

a = 2.55 m/s^2

$v_0 = 21.8 m/s$

$v = 0$

now we need to find x

from above equation we have

$0^2 = 21.8^2 + 2(2.55)x$

$0 = 475.24 + 5.1 x$

$x = 93.2 m$

so it will cover a distance of 93.2 m

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3 years ago

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A spacecraft with a proper length of 250 m passes by an observer on the earth. according to this observer, it takes 0.770 µs for

Sophie [7]

If 1 second = 10000 μs
what about how many seconds = 0.77 μs
the calculations would be done lke this
0.77 x 1/1000000 = 7.7 x 10 ^ -7
Speed = distance / time
distance = length of the spacecraft
therefore speed = 250/7.7 x 10^ -7
ans = 324675324.7 m/s

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3 years ago

A storage tank holds methane at 120 K, with a quality of 25 %, and it warms up by 5°C per hour due to a failure in the refrigera

lord [1]

One of the fundamental pillars to solve this problem is the use of thermodynamic tables to be able to find the values of the specific volume of saturated liquid and evaporation. We will be guided by the table B.7.1 'Saturated Methane' from which we will obtain the properties of this gas at the given temperature. Later considering the isobaric process we will calculate with that volume the properties in state two. Finally we will calculate the times through the differences of the temperatures and reasons of change of heat.

Table B.7.1: Saturated Methane

$T_1 = 120K$

$p_1 = 191.6kPa$

$v_f = 0.002439m^3/kg$

$v_{fg} = 0.30367 m^3/kg$

Calculate the specific volume of the methane at state 1

$v_1 = v_f+x_1v_{fg}$

$v_1 = 0.002439+ (0.25)(0.30367)$

$v_1 = 0.0783m^3/kg$

Assume the tank is rigid, specific volume remains constant

$v_2 = v_1$

$v_2 = 0.0783m^3/kg$

Now from the same table we can obtain the properties,

At $v_g = 0.0783m^3/kg$

$T_2 = 145K$

$p_2 = 823.7kPa$

We can calculate the time taken for the methane to become a single phase

$t = \frac{T_2-T_1}{\dot{T}}$

Here

$T_1 =$ Initial temperature of Methane

$\dot{T} =$ Warming rate

Replacing

$t = \frac{(145-273)-(120-273)}{5}$

$t = \frac{25}{5}$

$t = 5hr$

Therefore the time taken for the methane to become a single phase is 5hr

5 0

3 years ago

Adam drops a ball from rest from the top floor of a building at the same time Bob throws a ball horizontally from the same locat

guapka [62]

Answer:

Both balls hit the ground at the same time

Explanation:

Adam drops the ball from rest, so the ball just "<em>falls</em>" in vertical direction, being gravity its only acceleration, for cinematic movements we use that:

$y(t)=y_{0}+v_{0y}t+\frac{1}{2}gt^{2}$

In this case we have that gravity is negative, and as Adam drops the ball, $v_{0y}=0$

Bob throws the ball horizontally, so the movement will be a <em>parabola</em>, we can divide into horizontal direction, and vertical direction.

But we only need to analize the vertical movement, in wich again the only acceleration is gravity, and compare it with Adam's ball. Again we have that gravity is negative, and as the initial throw is horizontal, $v_{0y}=0$

Finally, we have that

$y(t)=h-\frac{1}{2}gt^{2}$