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nydimaria [60]
3 years ago
7

I've got a test tomorrow morning help!

Physics
1 answer:
topjm [15]3 years ago
3 0

Answer:

The efficiency is just 0.016

Explanation:

The efficiency is given by the useful energy ÷ by the total energy.

1 W is the same that 1 joule per second

If you do the math:

1.6/100=0.016

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Consider the data table charting the speed of a toy car moving across the floor. The line graph representing this data would BES
MrMuchimi

Answer:

The answer is D) diagonal line with varying slope, from 3 to 5 on USATestprep

Explanation:

4 0
3 years ago
Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant
Reika [66]

Answer:

v = 21 m / s

Explanation:

We can solve this exercise with the kinematics equations, let's start by finding the acceleration of the train with the initial data

            v = v₀ + a t

the initial speed is the speed within the city 6 m / s, the final speed is v = 11 m / s and the time is t = 8 s

             a = (v-v₀) / t

             a = (11 - 6) / 8

             a = 0.625 m / s²

when it leaves the city with speed vo = 11 m / s it accelerates for t = 16 s

            v = v₀ + a t

            v = 11 + 0.625 16

             v = 21 m / s

4 0
4 years ago
How would you describe a solution?
liberstina [14]

Answer: A solution is the end of a task.

Explanation:

4 0
3 years ago
Find the acceleration for a force of 2 N acting on 2 kg.
igomit [66]

Answer:

The acceleration of a 2 kg box acted on by a net force of 2 N is solved by using Newton’s second law of motion in the form acceleration = F_net / mass. Since the mass of the box is 2 kg and the net force is 2 N, the acceleration of the 2 kg box is 1 m/s^2.

4 0
3 years ago
To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T
Fantom [35]

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

8 0
3 years ago
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