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2 years ago

If the kinetic energy of the 40kg box is 784 J, what is the velocity before it strikes the ground?

Physics

1 answer:

mamaluj [8]2 years ago

4 0

Answer:

Explanation:

$KE=\frac{1}{2}mv^2$

$784=\frac{1}{2}(40)v^2$

$784=20v^2$

$39.2=v^2$

$v=6.26m/s$

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True or False: Objects that have more mass also have more gravity.

Fantom [35]

Answer:

True

Explanation:

The smaller object usually moves towards the larger object, because the larger object has more mass than the smaller object, so it has a larger gravitational pull.

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2 years ago

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A lab cart with a mass of 15 kg is moving with constant

bagirrra123 [75]

Answer:c

Explanation: i guessed but got it right bc im a bad bleep lol

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2 years ago

A cylinder with a piston contains 0.200 mol of oxygen at 2.00×105 Pa and 360 K . The oxygen may be treated as an ideal gas. The

ikadub [295]

Answer:

A. $W=600\ J$

B. $Q=2112\ J$

C. $\Delta U=1512\ J$

D. $W=0\ J$

Explanation:

Given:

no. of moles of oxygen in the cylinder, $n=0.2$

initial pressure in the cylinder, $P_i=2\times 10^5\ Pa$

initial temperature of the gas in the cylinder, $T_i=360\ K$

According to the question the final volume becomes twice of the initial volume.

Using ideal gas law:

$P.V=n.R.T$

$2\times 10^5\times V_i=0.2\times 8.314\times 360$

$V_i=0.003\ m^3$

Work done by the gas during the initial isobaric expansion:

$W=P.dV$

$W=P_i\times (V_f-V_i)$

$W=2\times 10^5\times (0.006-0.003)$

$W=600\ J$

we have the specific heat capacity of oxygen at constant pressure as:

$c_v=21\ J.mol^{-1}.K^{-1}$

Now we apply Charles Law:

$\frac{V_i}{T_i} =\frac{V_f}{T_f}$

$\frac{0.003}{360} =\frac{0.006}{T_f}$

$T_f=720\ K$

Now change in internal energy:

$\Delta U=n.c_p.(T_f-T_i)$

$\Delta U=0.2\times 21\times (720-360)$

$\Delta U=1512\ J$

Now heat added to the system:

$Q=W+\Delta U$

$Q=600+1512$

$Q=2112\ J$

Since during final cooling the process is isochoric (i.e. the volume does not changes). So,

$W=0\ J$

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2 years ago

A 30 ton rail car and a 90 ton rail car, initially at rest,are connected together with a giant but massless compressed springbet

Liula [17]

Answer:

V2 = 1.33m/s

Explanation:

M1 = 30 ton

M2 = 90 ton

V1 = 4 m/s

V2 = ?

Assumption: momentum conserved, no friction

initial momentum = final momentum = 0

momentum : p = MV

(M1*V1) - (M2*V2) = 0

V2 = (30*4)/90 = 1.33m/s

6 0

3 years ago

Problem 28.3 Suppose that you are planning a trip in which a spacecraZ is to travel at a constant velocity for exactly six month

iragen [17]

Answer:

v = 0.999981c m/s

Explanation:

Using the time dilation equation

$T = \frac{T_{0} }{\sqrt{1 - \frac{v^{2} }{c^{2} } } }$

T = stationary time = 100 years

T₀ = 11/12 years = 0.917 years

v = speed of travel in the space = ?

c = speed of light = 3 * 10⁸ m/s

$100 = \frac{0.917 }{\sqrt{1 - \frac{v^{2} }{(3*10^{8} )^{2} } } }\\$

$(0.917/100) ^{2} = 1 - \frac{v^{2} }{ 9 * 10^{16} }$

v = 299987395.57 m/s

v = 2.99 * 10⁸ m/s

v = 0.999981c m/s

3 0

2 years ago