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3 years ago

5. What is the velocity of a 0.5 kg ball that has a momentum of 3.00 kg m/s?

Physics

1 answer:

kow [346]3 years ago

3 0

Answer:

6 m/s

Explanation:

p=mv

p=3 kg.m/s

m=0.5 kg

rearrange and substitute to find v:

(3)=(0.5)v

v=3/0.5

v=6 m/s

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The temperature and pressure at the surface of Mars during a Martian spring day were determined to be -41 oC and 900 Pa, respect

Fittoniya [83]

Answer:

Part a)

$\rho = 0.0205 kg/m^3$

Part b)

$\rho = 1.22 kg/m^3$

So density of atmosphere at Martian Surface is very less than the density at Earth.

Explanation:

Part a)

As per ideal gas equation we know that

$PM = \rho RT$

here we know that Martian atmosphere is equivalent to that of carbon

so we will have

$P = 900 Pa$

$T = 273 - 41 = 232 K$

now we will have

$(900)(0.044) = \rho (8.31)(232)$

$\rho = 0.0205 kg/m^3$

Part b)

Now for the earth surface the density of air is given for

$P = 101.6 kPa$

$T = 18 ^oC$

so we will have

$PM = \rho RT$

$(101.6\times 10^3)(0.029) = \rho(8.31)(273 + 18)$

$\rho = 1.22 kg/m^3$

So density of atmosphere at Martian Surface is very less than the density at Earth.

8 0

2 years ago

A particle moves in a velocity field V(x, y) = x2, x + y2 . If it is at position (x, y) = (7, 2) at time t = 3, estimate its loc

MArishka [77]

Answer:

New location at time 3.01 is given by: (7.49, 2.11)

Explanation:

Let's start by understanding what is the particle's velocity (in component form) in that velocity field at time 3:

$V_x=x^2=7^2=49\\V_y=x+y^2=7+2^2=11$

With such velocities in the x direction and in the y-direction respectively, we can find the displacement in x and y at a time 0.01 units later by using the formula:

$distance=v\,*\, t$

$distance_x=49\,(0.01)=0.49\\distance_y=11\,(0.01)=0.11$

Therefore, adding these displacements in component form to the original particle's position, we get:

New position: (7 + 0.49, 2 + 0.11) = (7.49, 2.11)

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Work = force x distance

200 Newtons x 20 meters

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klio [65]

The truck has more KE than the bike

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Thepotemich [5.8K]

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