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MatroZZZ [7]
3 years ago
7

Find the velocity of the student using KE= 1/2 mv^2. A 50-kilogram student is running and has 225 joules of kinetic energy. The

student’s speed is meters/second
Physics
1 answer:
defon3 years ago
3 0
225 = 1/2 (50) (v2)
225 = 25 (v2)
225/25 = v2
9 = v2
√9 = v
v = 3 m/s
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Ionic compounds have high melting points because a lot of energy is needed to break the bonds between the ions.
velikii [3]
Its true the ionic compounds have a higher melting point
5 0
3 years ago
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A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h
Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

                    \tau = I \times \alpha

and,       I = \sum mr^{2}

So,      I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}

                       = 4 kg m^{2}

      \tau_{1} = 4 kg m^{2} \times \alpha_{1}

     \tau_{2} = I_{2} \alpha_{2}

So,      I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}

                     = 1 kg m^{2}

 as \tau_{2} = I_{2} \alpha_{2}

                   = 1 kg m^{2} \times \alpha_{2}        

Hence,     \tau_{1} = \tau_{2}

                  4 \alpha_{1} = \alpha_{2}

            \alpha_{1} = \frac{1}{4} \alpha_{2}

Thus, we can conclude that the new rotation is \frac{1}{4} times that of the first rotation rate.

8 0
3 years ago
"A block of metal weighs 40 N in air and 30 N in water. What is the buoyant force on the block due to the water? The density of
Alja [10]

Answer:

buoyant force on the block due to the water= 10 N

Explanation:

We know that

buoyant force(F_B) on a block= weight of the block in air (actual weight) - weight of block in water.

Given:

A block of metal weighs 40 N in air and 30 N in water.

F_B =  40-30= 10 N

therefore,  buoyant force on the block due to the water= 10 N

6 0
2 years ago
Read 2 more answers
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

<em>According to the question the final volume becomes twice of the initial volume.</em>

<u>Using ideal gas law:</u>

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

A.

<u>Work done by the gas during the initial isobaric expansion:</u>

W=P.dV

W=P_i\times (V_f-V_i)

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

W=600\ J

C.

<u>we have the specific heat capacity of oxygen at constant pressure as:</u>

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

<u>Now change in internal energy:</u>

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

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

\Delta U=1512\ J

B.

<u>Now heat added to the system:</u>

Q=W+\Delta U

Q=600+1512

Q=2112\ J

D.

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

W=0\ J

7 0
2 years ago
The upward velocity of a 2540kg rocket is v(t)=At + Bt2. At t=0 a=1.50m/s2. The rocket takes off and one second afterwards v=2.0
Alexeev081 [22]

Answer:

The value of A is 1.5m/s^2 and B is 0.5m/s^³

Explanation:

The mass of the rocket = 2540 kg.

Given velocity, v(t)=At + Bt^2

Given t =0  

a= 1.50 m/s^2

Now, velocity V(t) = A*t + B*t²

If,  V(0) = 0, V(1) = 2

a(t) = dV/dt = A+2B × t  

a(0) = 1.5m/s^²  

1.5m/s^²  =  A + 2B ×  0  

A = 1.5m/s^2

now,

V(1) = 2 = A× 1 + B× 1^²  

1.5× 1 +B× 1 = 2m/s

B = 2-1.5  

B = 0.5m/s^³

Now Check V(t) = A× t + B × t^²

So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² ×  1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s  

Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)

7 0
3 years ago
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