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3 years ago

Please help!! For a science quiz

Physics

1 answer:

klio [65]3 years ago

4 0

1.Velocity (v) is a vector quantity that measures displacement (or change in position, Δs) over the change in time (Δt), represented by the equation v = Δs/Δt. Speed (or rate, r) is a scalar quantity that measures the distance traveled (d) over the change in time (Δt), represented by the equation r = d/Δt.

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What is the speed of a bobsled whose distance-time graph indicates that it traveled 117m in 25s

Scilla [17]

Speed= distance/time
speed=117/25
speed=4.68m/sec

3 0

3 years ago

A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its potential energy (PE) when it is 2.00 m above the grou

Colt1911 [192]

The potential energy of the lemming is 1.53 J

Explanation:

The potential energy (PE) of an object is the energy possessed by the object due to its position in the Earth's gravitational field, and it is given by:

$PE=mgh$

where:

m is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity

h is the height of the object relative to the ground

In this problem:

m = 0.0780 kg is the mass of the lemming

We want to find the potential energy when the height is

h = 2.00 m

Therefore, we find:

$PE=(0.0780)(9.8)(2.00)=1.53 J$

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

7 0

3 years ago

PLEASE HELP!!! I'm dumb lol.

Len [333]

Answer:

A theory or hypothesis does not necessarily provide an accurate scientific explanation to any topic but predicts what can happen.

Explanation:

7 0

3 years ago

Read 2 more answers

A SMART car can accelerate from rest to a speed of 28 m/s in 20s. What distance does it travel in this time?

sashaice [31]

Speed= distance/ time so distance = speed * time= 28 * 20= 560m .. so the answer is 560m.. l hope it helped :)

6 0

2 years ago

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