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mixer [17]
3 years ago
5

Bonnie and Clyde are sliding a 323 kg bank safe across the floor to their getaway car. The safe slides with a constant speed if

Clyde pushes from behind with 383 N of force while Bonnie pulls forward on a rope with 343 N of force. What is the coefficient of friction between the safe and the floor?
Physics
1 answer:
Harrizon [31]3 years ago
4 0

Answer:

835/(320*9.8)=0.266

Explanation:

since the safe is moving at a constant speed then the net force is zero

in other words the friction force equals the sum of Bonnie and Clyde force

so the friction force must be 485+350=835N

the frictional force = mu*m*g

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What is the kinetic energy of a jogger with a mass of 70.8 kg traveling at a speed of 2.4 m/s?
BARSIC [14]
Kinetic energy equation is KE= 1/2mv^2. So plug in the numbers and you get a Kinetic Energy of 203.904 J
8 0
3 years ago
4 points
zubka84 [21]

(a) 25lx

(b) 11.11lx

<u>Explanation:</u>

Illuminance is inversely proportional to the square of the distance.

So,

I = k\frac{1}{r^2}

where, k is a constant

So,

(a)

If I = 100lx and r₂ = 2r Then,

I_2 = k\frac{1}{(2r)^2}

Dividing both the equation we get

\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(2r)^2}{k} \\\\\frac{I_1}{I_2} = 4\\\\I_2 = \frac{I_1}{4}\\\\I_2 = \frac{100}{4}  = 25lx

When the distance is doubled then the illumination reduces by one- fourth and becomes 25lx

(b)

If I = 100lx and r₂ = 3r Then,

I_2 = k\frac{1}{(3r)^2}

Dividing equation 1 and 3 we get

\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(3r)^2}{k} \\\\\frac{I_1}{I_2} = 9\\\\I_2 = \frac{I_1}{9}\\\\I_2 = \frac{100}{9}  = 11.11lx

When the distance is tripled then the illumination reduces by one- ninth and becomes 11.11lx

3 0
3 years ago
A force can give a certain mass a certain acceleration. if i double the force and double the mass, the acceleration of this obje
leva [86]

The acceleration of this item will remain constant if I double the force and double the mass.

<h3>What does acceleration mean ?</h3><h3 />

The rate at which a moving object's speed and direction change over time. When something moves faster or slower, it is considered to be accelerating. Motion on a circle accelerates even while the speed is constant because the direction is always changing.

  • If an object is accelerating and moving in the appropriate direction, it has positive acceleration. The speeding automobile in the first scenario showed positive acceleration. The car is moving forward and accelerating in the same direction as the acceleration.

Learn more about Acceleration here:

brainly.com/question/605631

#SPJ4

6 0
1 year ago
Convert 2 days into second​
eduard

Answer:

<em><u>172,000 second </u></em>

<em><u>I'M</u></em><em><u> </u></em><em><u>NOT</u></em><em><u> </u></em><em><u>SURE</u></em><em><u> </u></em><em><u>THAT</u></em><em><u> </u></em><em><u>THIS</u></em><em><u> </u></em><em><u>IS</u></em><em><u> </u></em><em><u>RIGHT</u></em><em><u> </u></em><em><u>OR</u></em><em><u> </u></em><em><u>WRONG</u></em><em><u> </u></em><em><u> </u></em><em><u>IF</u></em><em><u> </u></em><em><u>IT'S</u></em><em><u> </u></em><em><u>WRONG</u></em><em><u> </u></em><em><u>THEN</u></em><em><u> </u></em><em><u>SORRY</u></em><em><u> </u></em>

3 0
3 years ago
Read 2 more answers
The specialty of an athlete on the women's track team is the pole vault. She has a mass of 48 kg and her approach speed is 8.9 m
Lyrx [107]

Answer:

H = 3.9 m

Explanation:

mass (m) = 48 kg

initial velocity (initial speed) (U) = 8.9 m/s

final velocity (V) = 1.6 m/s

acceleration due to gravity (g) = 9.8 m/s^{2}

find the height she raised her self to as she crosses the bar (H)

from energy conservation, the change in kinetic energy = change in potential energy

0.5m(V^{2} - [test]U^{2}[/tex]) = mg(H-h)

where h = initial height = 0 since she was on the ground

the equation becomes

0.5m(V^{2} - [test]U^{2}[/tex]) = mgH

0.5 x 48 x (1.6^{2} - [test]8.9^{2}[/tex]) = 48 x 9.8 x H

-1839.6 = 470.4 H  (the negative sign indicates a decrease in kinetic energy so we would not be making use of it further)

H = 3.9 m

4 0
3 years ago
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