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gulaghasi [49]
1 year ago
6

A horizontal force, F, acts on a crate weighing 8 N, which moves on a horizontal floor at constant speed. The coefficient of fri

ction between the crate and the surface is 0.25. The work done by F when the crate has moved 12 m is: Group of answer choices 2 J None of the above 24 J 96 J 0 J 10 J
Physics
1 answer:
RoseWind [281]1 year ago
7 0

The answer is 24 J

F K =.25*8 N

     = 2N

F = f k = 2 N

Since a = 0

W = f * s

 2 N * 12 m = 24 J

The coefficient of friction is a ratio used to quantify the friction force among two gadgets when it comes to the everyday pressure this is keeping them collectively. The coefficient of friction is critical attention at some stage in material selection and floor requirement determination.

For instance, ice on steel has a low coefficient of friction – the 2 materials slide past each different without problems – whilst rubber on the pavement has an excessive coefficient of friction – the substances no longer slide past each other without difficulty.

The coefficient of friction is dimensionless and it does not have any unit. it is a scalar, meaning the direction of the force does not have an effect on the physical quantity. The coefficient of friction depends on the gadgets that are causing friction.

Learn more about the coefficient of friction here brainly.com/question/20241845

#SPJ4

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A person travels by car from one city to another with differem constan1 speeds between pairs of cities. She drives for 30.0 min
Montano1993 [528]

Answer:

a.52.9 km/h

b.90 km

Explanation:

We are given that

v_1=89km/h

t_1=30min

v_2=100km/h

t_2=12min

v_3=40km/h

t_3=45 min

Time spend on eating lunch and buying ga=15 min.

a.Total time=30+12+45+15=102 minute=\frac{102}{60}=1.7 hour

1 hour=60 minutes

Distance=speed\times time

d_1=v_1\times t_1=80\times\frac{30}{60}=40km

d_2=100\times \frac{12}{60}=20 km

d_3=40\times \frac{45}{60}=30 km

Total distance=d_1+d_2+d_3=40+20+30=90km

Average speed=\frac{total\;speed}{total\;time}

Using the formula

Average speed=\frac{90}{1.7}=52.9Km/h

b.Total distance between the initial and final city lies along the route=90 km

4 0
3 years ago
A ball is thrown from a rooftop with an initial downward velocity of magnitude vo = 2.9 m/s. The rooftop is a distance above the
Step2247 [10]

Answer:

a) The velocity of the ball when it hits the ground is -20.5 m/s.

b) To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

Explanation:

I´ve found the complete question on the web:

<em />

<em>A ball is thrown from a rooftop with an initial downward velocity of magnitude v0=2.9 m/s. The rooftop is a distance above the ground, h= 21 m. In this problem use a coordinate system in which upwards is positive.</em>

<em>(a) Find the vertical component of the velocity with which the ball hits the ground.</em>

<em>(b) If we wanted the ball's final speed to be exactly 27, 3 m/s from what height, h (in meters), would we need to throw it with the same initial velocity?</em>

<em />

The equation of the height and velocity of the ball at any time "t" are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the ball at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the ball at a time "t".

First, let´s find the time it takes the ball to reach the ground (the time at which h = 0)

h = h0 + v0 · t + 1/2 · g · t²

0 = 21 m - 2.9 m/s · t - 1/2 · 9.8 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 1.8 s  ( the other solution of the quadratic equation is rejected because it is negative).

Now, using the equation of velocity, let´s find the velocity of the ball at

t = 1.8 s:

v = v0 + g · t

v = -2.9 m/s - 9.8 m/s² · 1.8 s

v = -20.5 m/s

The velocity of the ball when it hits the ground is -20.5 m/s.

b) Now we have the final velocity and have to find the initial height. Using the equation of velocity we can obtain the time it takes the ball to acquire that velocity:

v = v0 + g · t

-27.3 m/s = -2.9 m/s - 9.8 m/s² · t

(-27.3 m/s + 2.9 m/s) / (-9.8 m/s²) = t

t = 2.5 s

The ball has to reach the ground in 2.5 s to acquire a velocity of 27.3 m/s.

Using the equation of height, we can obtain the initial height:

h = h0 + v0 · t + 1/2 · g · t²

0 = h0 -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

-h0 = -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

h0 = 38 m

To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

6 0
3 years ago
4. What is the magnitude and direction of the gravitational force that acts on a man
Cloud [144]

The gravitational force acting on the man is 800 N towards the Earth's centre

Explanation:

The weight of an object on the Earth is exactly the gravitational force exerted by the Earth on the object.

The gravitational force exerted by the Earth on an object located at the Earth's surface is given by:

F=G\frac{Mm}{R^2}

where

G is the gravitational constant

M is the Earth's mass

m is the mass of the object

R is the radius of the Earth

And the direction of the force is towards the Earth's centre.

Since G, M and R are constant, they are grouped into a single constant called g, acceleration of gravity:

g=\frac{GM}{R^2}

therefore the gravitational force can be rewritten as

F=mg

And this is the usual equation that we use to calculate the weight of an object.

Therefore, weight and gravitational force acting on an object on Earth are the same thing: so, the gravitational force acting on the man is equal to his weight, 800 N, and it acts towards the Earth's center.

Learn more about weight and gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0
3 years ago
Which of the following is the number of cycles per unit of time? infrasonis wave,frequency,ultrasonic wave or pitch
Allushta [10]
Frequency
Hope this helped :)
4 0
3 years ago
Read 2 more answers
Ask Your Teacher Cam Newton of the Carolina Panthers throws a perfect football spiral at 6.9 rev/s. The radius of a pro football
faltersainse [42]

Answer:

a=159.32\ m/s^2

Explanation:

It is given that,

Angular speed of the football spiral, \omega=6.9\ rev/s=43.35\ rad/s

Radius of a pro football, r = 8.5 cm = 0.085 m

The velocity is given by :

v=r\omega

v=0.085\times 43.35

v = 3.68 m/s

The centripetal acceleration is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(3.68)^2}{0.085}

a=159.32\ m/s^2

So, the centripetal acceleration of the laces on the football is 159.32\ m/s^2. Hence, this is the required solution.

6 0
3 years ago
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