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2 years ago

A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the work done to accomplish this task. 1.84 x 103 J

1 answer:

4 0

work will be equal to the potential energy gained by the person in climbing the stairs. work= potential energy gained = mgh W= 75kg*9.8m/s2*2.50m= 1837.5 J

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Two forces F1 and F2 act on a 5.00 kg object. Taking F1=20.0N and F2=15.00N, find the acceleration of the object for the configu

Anit [1.1K]

A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction

m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N ... y direction

Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 = 625N^2 =>

Net force = √625 = 25N

F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2

Answer: 5.00 m/s^2

b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal. 

m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N ... 60 degress above x direction

Components of F2

F2,x = F2*cos(60) = 15N / 2 = 7.5N

F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N

Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N

Total force in y = F2,y = 13.0 N

Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =

= 30.42N

a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2

Answer: 6.08 m/s^2

8 0

2 years ago

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Jess kicks a soccer ball, and it rolls across the ground. The force diagram shows all the forces acting on the ball at the momen

juin [17]

Arrow at the left side pointing towards right side represents the frictional force as it always acts opposite to motion

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2 years ago

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A cart is driven by a large propeller or fan, which can accelerate or decelerate the cart. The cart starts out at the position x

mash [69]

Answer:

The acceleration of the cart is 1.0 m\s^2 in the negative direction.

Explanation:

Using the equation of motion:

Vf^2 = Vi^2 + 2*a*x

2*a*x = Vf^2 - Vi^2

a = (Vf^2 - Vi^2)/ 2*x

Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.

Let x = Xf -Xi

Where Xf is the final position of the cart and Xi the initial position of the cart.

x = 12.5 - 0

x = 12.5

The cart comes to a stop before changing direction

Vf = 0 m/s

a = (0^2 - 5^2)/ 2*12.5

a = - 1 m/s^2

The cart is decelerating

Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.

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2 years ago

A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mir

Semmy [17]

Answer:

The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

Explanation:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{0.24}=\dfrac{1}{0.25}+\dfrac{1}{u}$

$\dfrac{1}{u}=\dfrac{1}{0.24}-\dfrac{1}{0.25}$

$\dfrac{1}{u}=\dfrac{1}{6}$

$u=6\ m$

We need to calculate the magnification

Using formula of magnification

$m=-\dfrac{v}{u}$

Put the value into the formula

$m=-\dfrac{0.24}{-6}$

$m=0.04$

We need to calculate the height of the object

Using formula of magnification

$m=\dfrac{h'}{h}$

$h=\dfrac{0.080}{0.04}$

$h=2\ m$

A convex mirror produce a virtual and upright image behind the mirror.

Hence, The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

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2 years ago

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You travel in a circle, whose circumference is 8 kilometers, at an average speed of 8 kilometers/hour. If you stop at the same p

Schach [20]

Velocity = (displacement) / (time)

Displacement = straight-line distance between start-point and end-point

If you stop at the same point you started from, then
your displacement for the trip is zero, and your average
velocity is also zero.

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2 years ago

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