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2 years ago

When a block slides a certain distance down an incline, the work done by

1 answer:

7 0

That is -300 J, because the work = the force times the distance. When the distance is a positive value when it is going down, the distance is a negative value when it is going up. The only thing that change is the minus, the absolute value of the work remains the same

You might be interested in

how long would it take a 3.4 kg bike with 79 kg rider traveling at 18.6 m/s to stop if 867 N of force is applied to the brakes?

nalin [4]

Answer: 1.76 s

Explanation:

We have the following data:

$m=3.4 kg+79 kg=82.4 kg$ is the total mass of the bike and the rider

$V_{o}=18.6 m/s$ is the initial velocity

$F=867 N$ is the force applied to the brakes

Firstly, we will find the acceleration $a$ with the following equation:

$F=m.a$ (1)

Isolating $a$ :

$a=\frac{F}{m}$ (2)

$a=\frac{867 N}{82.4 kg}$ (3)

$a=10.52 m/s^{2}$ (4) This is the magnitude of the acceleration, however, since the final velocity is 0 m/s, this means the direction is negative

Hence:

$a=-10.52 m/s^{2}$ (5)

On the other hand, with the following equation we can find the time $t$ :

$V=V_{o}+at$ (6)

Where:

$V=0 m/s$ is the final velocity (the bike stops)

Isolating $t$ :

$t=-\frac{V_{o}}{a}$ (7)

$t=-\frac{18.6 m/s}{-10.52 m/s^{2}}$ (8)

Finally:

$t=1.76 s$ This iste time it takes to the bike to stop

7 0

3 years ago

Which scale of measurement measures the magnitude or strength of an earthquake based on seismic waves?

Pachacha [2.7K]

The answer is Richter

7 0

3 years ago

Read 2 more answers

A student wants to draw a diagram to show the path of a light wave. Which

wariber [46]

Answer:

Explanation:

light moves in a straight line (sorry if I'm wrong)

8 0

3 years ago

What is the escape velocity from a planet with a mass that is 10.8% Earths mass and a size that is 53% Earths radial size?

DENIUS [597]

The escape velocity from the planet is 5052 m/s

Explanation:

The formula to calculate the escape velocity from a planet is:

$v=\sqrt{\frac{2GM}{R}}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

M is the mass of the planet

R is the radius of the planet

For Earth,

$M_E=5.98\cdot 10^{24} kg$ is the mass

$R_E=6.37\cdot 10^6 m$ is the radius

Here we have a planet that:

- its mass is 10.8% of that of Earth, so

$M=0.108M_E$

- its radius is 53% of that of Earth, so

$R=0.53 R_E$

Substituting everything into the first equation, we find the escape velocity from this planet:

$v=\sqrt{\frac{2G(0.108M_E)}{0.53R_E}}=\sqrt{\frac{2(6.67\cdot 10^{-11})(0.108)(5.98\cdot 10^{24})}{(0.53)(6.37\cdot 10^6)}}=5052 m/s$

#LearnwithBrainly

4 0

3 years ago

A man driving a car is exactly 5000 meters due west from the line marking the eastern time zone. He

laiz [17]

The eastern time zone would the opposite side of the angle made by his car. So this means he would be traveling 30cos30 m/s completely in the east direction. Or approximately 26 m/s in the East direction. 5000 m divided by this answer results in approximately 192 seconds. I would use a calculator to get a more exact answer

8 0

3 years ago