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3 years ago

7

A conveyer belt moves a 40 kg box at a velocity of 2 m/s. What is the kinetic energy of the box while it is on the conveyor belt

?

2 answers:

ankoles [38]3 years ago

4 0

Kinwtic energy = (40kg*2*2)/2=40*2=80J

Lorico [155]3 years ago

3 0

The answer to the Question is 80 Joules.

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MaRussiya [10]

Answer:

1- experimental

2-descriptive

3-comparative

Explanation:

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Which object has the most Kinetic Energy, and why?

Ilya [14]

A small rock quickly rolling down a hill because as the velocity or speed increases the particles in the rock start to increase kinetic energy. The particles start to act up and create more energy. Also because the small rock would go faster than a giant rock because of Newton's second law. Can I have brainliest pls?

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3 years ago

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What is the mass of an object which has a force of 600 N acting on it and is travelling

Paha777 [63]

Answer:

<em>The mass of the object is 40 Kg</em>

Explanation:

<u>Net Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object is:

F = m.a

Where:

a = acceleration of the object.

m = mass of the object.

The mass can be calculated by solving for m:

$\displaystyle m=\frac{F}{a}$

The object has a net force of F=600 N acting on it and travels at $a=15\ m/s^2$ , thus the mas is:

$\displaystyle m=\frac{600}{15}$

m = 40 Kg

The mass of the object is 40 Kg

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3 years ago

A train travels 120 km in 2 hours and 30 minutes. What is its average speed?

ivann1987 [24]

Answer:

13.33 or 13 1/3m/s (meters per second)

Explanation:

In physics, we use the basic units of meters and seconds. So first convert (km) into meters (m) and also hours and minutes into seconds (s). We end up with 120000m and 9000s. Then divide the 120000m by the 9000s and you end up with 13.33 or 13 1/3 m/s.

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3 years ago

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A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d

prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

$V_{t}= V_{o}e^{\frac{-t}{R*C} }$

Voltage in this case is the energy dissipated so

$E_{t}= E_{o}e^{\frac{-t}{R*C} }$

$\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }$

$\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }$

Using the equation to find capacitance

$ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }$

$C= 5.32x10^{-6}$ F

C= 5.32 uF because u is the symbol for micro that is equal to $10^{-6}$

8 0

3 years ago