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sveticcg [70]
3 years ago
11

Why is moving a box with a mass of 37 kilograms a distance of 37 meters why it did 360 joules of work in two minutes when moving

the Box what was his power output?
a) 9.72W
b) 3W
c) 180W
d) 323W

Physics
1 answer:
alexira [117]3 years ago
6 0
Power= energy/time= 360/(2*60)=3 W thats 2nd option
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Answer:

V_1= 3.4*10^7m/s

Explanation:

From the question we are told that

Nucleus diameter d=5.50-fm

a 12C nucleus

Required kinetic energy K=2.30 MeV

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

            K_2 +U_2=K_1+U_1

where

K_1 =initial kinetic energy

K_2 =final kinetic energy

U_1 =initial electric potential

U_2 =final electric potential

mathematically

   U_2 = \frac{Kq_pq_c}{r_2}

where

r_f=distance b/w charges

q_c=nucleus charge =6(1.6*10^-^1^9C)

K=constant

q_p=proton charge

Generally kinetic energy is know as

         K=\frac{1}{2}  mv^2

Therefore

         U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2}  mv_1^2 +U_1

Generally equation for radius is d/2

Mathematically solving for radius of nucleus

         R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})

         R=2.75*10^-^1^5m

Generally we can easily solving mathematically substitute into v_1

   q_p=6(1.6*10^-^1^9C)

   K_1=9.0*10^9 N-m^2/C^2

   U_1= 0

   R=2.75*10^-^1^5m

   K=2.30 MeV

   m= 1.67*10^-^2^7kg

   V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }

    V_1= 3.4*10^7m/s

Therefore the proton must be fired out with a speed of V_1= 3.4*10^7m/s

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3 years ago
A car whose total mass is 800kg travelling with a uniform velocity of 20m/s suddenly observes a stationary dog 50m ahead on its
maxonik [38]

Answer:

The driver hits the stationery dog because the applied force is less than required force

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Substituting 800 Kg for m and 20 m/s for v we obtain

KE=0.5*800*(20 m/s)^{2}=160,000

Frictional force by vehicle pads is given by

Fr=\frac {KE}{d} where d is the distance moved

Substituting 160000 for KE and 50 m for d we obtain

Fr=\frac {160000}{50}=3200 N

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A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w
trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

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Rearranging to find the Amplitude we have,

A = \sqrt{\frac{mv^2}{k}}

Replacing,

A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}

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The Period is defined as

T = 2\pi \sqrt{\frac{m}{k}}

Replacing,

T = 2\pi \sqrt{\frac{0.750}{13}}

T= 1.509s

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v = \frac{2\pi}{T} * \sqrt{A^2-x^2}

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Nostrana [21]

Answer:

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