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4 years ago

What is the kinetic energy of a toy truck with a mass of 0.75 kg and a velocity of 4 m/s? (Formula: )

Physics

2 answers:

Dvinal [7]4 years ago

6 0

Answer: Option (b) is the correct answer.

Explanation:

Kinetic energy is the energy of an object or particle when it is in motion.

Mathematically, Kinetic energy = $\frac{1}{2}mass \times (velocity)^{2}$

It is given that toy truck has mass = 0.75 kg, and velocity = 4 m/s.

Calculate the kinetic energy as follows.

Kinetic energy = $\frac{1}{2}mass \times (velocity)^{2}$

= $\frac{1}{2}0.75 kg \times (4 m/s)^{2}$

= $\frac{1}{2}0.75 kg \times 16 m^{2}/s^{2}$

= $0.75 kg \times 8 m^{2}/s^{2}$

= $6 kg m^{2}/s^{2}$

or, Kinetic energy = 6 joules

Thus, we can conclude that kinetic energy of toy truck is 6 J.

eduard4 years ago

6 0

Thank you for posting your question here at brainly. Feel free to ask more questions. The best and most correct answer among the choices provided by the question is 6 J . 

KE = (1/2)(m)(v^2)

Hope my answer would be a great help for you.

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4 years ago

A rifle is aimed horizontally at a target 49 m away. the bullet hits the target 2.3 cm below the aim point. part a what was the

Nezavi [6.7K]

We use the equation of motion for vertical component,

$s_{y} = u_{y} t+\frac{1}{2} gt^2.$

Here, $s_{y}$ is displacement of bullet, $u_{y}$ is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.

Therefore,

$s_{y} =\frac{1}{2}gt^2$

Given, $s_{y} =2.3 \ cm = 2.3 \times 10^{-2} \ m$

Substituting the values, we get time of flight

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4 years ago

A uranium nucleus is traveling at 0.94 c in the positive direction relative to the laboratory when it suddenly splits into two p

Anettt [7]

Answer:

A u = 0.36c B u = 0.961c

Explanation:

In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains

u ’= (u-v) / (1- uv / c²)

Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory

The data give is u ’= 0.43c and the initial core velocity v = 0.94c

Let's clear the speed with respect to the observer (u)

u’ (1- u v / c²) = u -v

u + u ’uv / c² = v - u’

u (1 + u ’v / c²) = v - u’

u = (v-u ’) / (1+ u’ v / c²)

Let's calculate

u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)

u = 0.51c / (1 + 0.4042)

u = 0.36c

We repeat the calculation for the other piece

In this case u ’= - 0.35c

We calculate

u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)

u = 1.29c / (1- 0.329)

u = 0.961c

6 0

3 years ago