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Nataliya [291]
2 years ago
5

A truck drives to a rock quarry at a speed of 20 m/s. The truck takes on a load of rocks, which doubles its mass, and leaves at

the same speed of 20 m/s. Quantitatively compare the kinetic energy of the truck as it left the quarry with its kinetic energy on the way to the quarry. Explain your answer.
Physics
1 answer:
Rus_ich [418]2 years ago
8 0

Given that,

A truck drives to a rock quarry at a speed of 20 m/s. The truck takes on a load of rocks, which doubles its mass, and leaves at the same speed of 20 m/s.

To find,

Comparison of the kinetic energy of the truck as it left the quarry with its kinetic energy on the way to the quarry.

Solution,

Kinetic energy of an object is given by:

K=\dfrac{1}{2}mv^2

m is the mass of the truck

K=\dfrac{1}{2}m\times (20)^2\\\\K=200m\ J

If mass is doubled and speed is same i.e. 20 m/s, neew kinetic energy is given by :

K=\dfrac{1}{2}\times (2m)\times 20^2\\\\K=400m\ J

It can be seen that when mass of truck is doubled, its kinetic energy gets doubled.

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In the absence of air resistance, at what other angle will a thrown ball go the same distance as one thrown at an angle of 75 de
snow_tiger [21]

As we know that range of the projectile motion is given by

R = \frac{v^2 sin(2\theta)}{g}

here we know that range will be same for two different angles

so here we can say the two angle must be complementary angles

so the two angles must be

\theta, 90 - \theta

so it is given that one of the projection angle is 75 degree

so other angle for same range must be 90 - 75 = 15 degree

so other projection angle must be 15 degree

5 0
2 years ago
You buy a AA battery in the store, and it is marked 1.5 V. If this marking is strictly accurate, while this battery is fresh its
vitfil [10]

Answer:

Option A is correct.

when it is used in a circuit. its terminal voltage will be less than 1.5 V.

Explanation:

The terminal voltage of the battery when it is in use in circuits drops lower than the 1.5 V rating given to it due to internal resistance.

All batteries give internal resistances when used in circuits. The internal resistance (though very small) is usually modelled as connected in series with the battery. It is due to some form of interference from the chemical makeup of the battery.

Normally, while the battery is fresh, the voltage (V) obtained at its terminals when connected in series with a resistor of resistance R is V = IR; where I is the current flowing in this circuit.

But once the interenal resistance (r) of the battery comes into play,

V = I₁ (r + R)

The current in the circuit evidently drops (that is I₁ < I) and V = (I₁r + I₁R)

The voltage across the terminals of the battery is no longer V but is now (V) × [R/(R+r)] which is less than the initial V and it reduces as the internal resistance, r, increases.

Hope this Helps!!!

3 0
2 years ago
The alpha particles leave visible tracks in the cloud chamber because
julia-pushkina [17]
<h2>Answer: Ionization </h2>

The inner atmosphere of a <u>cloud chamber</u> is composed of an easily ionizable gas, this means that little energy is required to extract an electron from an atom. <u>This gas is maintained in the supercooling state, so that a minimum disturbance is enough to condense it</u> in the same way as the water is frozen.

<h2>Then, when a charged particle with enough energy interacts with this gas, it <u>ionizes</u> it. </h2>

This is how alpha particles are able to ionize some atoms of the gas contained inside the chamber when they cross the cloud chamber.

These ionized atoms increase the surface tension of the gas around it allowing it to immediately congregate and condense, making it easily distinguishable inside the chamber like a <u>small cloud</u>. In this way, it is perfectly observable the path the individual particles have traveled, simply by observing the cloud traces left in the condensed gas.

6 0
2 years ago
A Micro –Hydro turbine generator is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity
Salsk061 [2.6K]

Answer:

Angular displacement of the turbine is 234.62 radian

Explanation:

initial angular speed of the turbine is

\omega_i = 2\pi f_1

\omega_1 = 2\pi(\frac{610}{60})

\omega_1 = 63.88 rad/s

similarly final angular speed is given as

\omega_f = 2\pi f_2

\omega_2 = 2\pi(\frac{837}{60})

\omega_2 = 87.65 rad/s

angular acceleration of the turbine is given as

\alpha = 5.9 rad/s^2

now we have to find the angular displacement is given as

\theta = \omega t + \frac{1}{2}\alpha t^2

\theta = (63.88)(3.2) + (\frac{1}{2})(5.9)(3.2^2)

\theta = 234.62 radian

3 0
2 years ago
The world record for pole vaulting is 6.15 m. If the pole vaulter's gravitational potential energy is 4942 J, what is his mass?
navik [9.2K]
The gravitational potential energy is calculated by multiplying the mass of the object to the height and the gravitational acceleration which is 9.8 m/s^2. We do as follows:

GPE = mgh
GPE = 4942
4942 = m (9.8)(6.15)
m = 82 kg 

Hope this helps.
7 0
2 years ago
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