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A box weighing 200N is pushed on a horizontal floor. What acceleration will result if a horizontal force of 100N is applied on t

Bogdan [553]

Answer:

the friction force in the reverse direction is 200 *0.4=80 N.

the net forward force acting on the box is therefore

Fnet= 100 - 80 N

= 20 N

acceleration = Fnet / mass

=Fnet *g/(weight)

=20 *9.8/200 = 0.98 m/s^2

Explanation:

4 0

3 years ago

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A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

$E_{A}=E_{B}$

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

$E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}$

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

$E_{B}=m*g*h+\frac{1}{2}*m*v^{2}$

Therefore we will have the following equation:

$(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]$

The kinetic energy can be easily calculated by means of the kinetic energy equation.

$KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]$

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

$E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]$

5 0

3 years ago

A lacrosse ball that is thrown straight upwards reaches a maximum height of 4.5 m. At what initial velocity was it thrown? (note

shtirl [24]

Answer:

The initial velocity was 9.39 m/s

Explanation:

<em>Lets explain how to solve the problem</em>

The ball is thrown straight upward with initial velocity u

The ball reaches a maximum height of 4.5 m

At the maximum height velocity v = 0

The acceleration of gravity is -9.8 m/s²

We need to find the initial velocity

The best rule to find the initial velocity is <em>v² = u² + 2ah</em>, where v is

the final velocity, u is the initial velocity, a is the acceleration of

gravity and h is the height

⇒ v = 0 , h = 4.5 m , a = -9.8 m/s²

⇒ 0 = u² + 2(-9.8)(4.5)

⇒ 0 = u² - 88.2

Add 88.2 to both sides

⇒ 88.2 = u²

Take square root for both sides

⇒ u = 9.39 m/s

<em>The initial velocity was 9.39 m/s</em>

5 0

3 years ago

When running a half marathon (13.1 miles), it took Kevin 8 minutes to run from mile marker 1 to mile marker 2, and 18 minutes to

Vsevolod [243]

Answer:

It took Kevin 26 minutes to run from markers 1 to 4

His average speed from mile markers 1 to 4 is 0.154 miles/minute

Kevin must run by average speed 0.1 miles/minute to finish the race

Explanation:

Lets explain how to solve the problem

A half marathon 13.1 miles

Kevin took 8 minutes to run from mile marker 1 to mile marker 2 and

18 minutes to run from mile marker 2 to mile marker 4

→ He took 8 minutes and 18 minutes to run from marker 1 to marker 4

→ The total time of the first 4 marker = 8 + 18 = 26 minutes

<em>It took Kevin 26 minutes to run from markers 1 to 4</em>

Average speed is total distance divided by total time

The average speed of Kevin as he ran from mile marker 1 to mile

marker 4 is the 4 miles divides by 26 minutes

→ Average speed = 4 ÷ 26 = $\frac{2}{13}$ = 0.154 miles/minute

<em>His average speed from mile markers 1 to 4 is 0.154 miles/minute</em>

It took Kevin 71 minutes to pass mile marker 9

Kevin need to complete the race in 112 minutes, then what must

Kevin's average speed be as he travels from mile marker 9 to the

finish line?

The total distance of the race is 13.1 miles, he ran 9 miles

→ The remaining distance = 13.1 - 9 = 4.1 miles

He must run 4.1 miles to complete the race

The total time is 112 minutes, he used 71 minutes to run the first 9 miles

→ The remaining time = 112 - 71 = 41 minutes

He must finish the 4.1 miles in 41 minutes

→ His average speed for the last part of the race = 4.1 ÷ 41 = 0.1 mi/min

<em>Kevin must run by average speed 0.1 miles/minute to finish the race</em>

7 0

3 years ago

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Consider a concave spherical mirr or that has focal length f = +19.5 cm.

lidiya [134]

The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

<h3>What is concave mirror?</h3>

A concave mirror has a reflective surface that is curved inward and away from the light source.

Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.

<h3>Object distance of the concave mirror</h3>

Apply mirrors formula as shown below;

1/f = 1/v + 1/u

where;

f is the focal length of the mirror
v is the object distance
u is the image distance

when image height = object height, magnification = 1

u/v = 1

v = u

Substitute the given parameters and solve for the distance of the object from the mirror's vertex

1/f = 1/v + 1/v

1/f = 2/v

v = 2f

v = 2(19.5 cm)

v = 39 cm

Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

Learn more about concave mirror here: brainly.com/question/27841226

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7 0

2 years ago