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Mazyrski [523]
3 years ago
10

1. A 0.250 kg baseball sits on the ledge of a window in Treadwell hall. If the ball has 18.5

Physics
1 answer:
Lunna [17]3 years ago
4 0

Answer:

h = 7.54 m

t = 1.24 s

Explanation:

1.Let g = 9.81 m/s2 is the gravitational acceleration. Since the formula for potential energy is:

E_p = mgh

where m = 0.25 is the ball mass and h is the height. We can solve for h

h = \frac{E_p}{mg} = \frac{18.5}{0.25*9.81} = 7.54m

2. The time it take for the ball to reach a distance of 7.54m with a gravitational acceleration of 9.81m/s2:

h = \frac{gt^2}{2}

t^2 = \frac{2h}{g} = \frac{2*7.54}{9.81} = 1.54

t = \sqrt{1.54} = 1.24 s

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A horizontal force, F1 = 65 N, and a force, F2 = 12.4 N acting at an angle of θ to the horizontal, are applied to a block of mas
Nezavi [6.7K]

Answer:

(a) FN = 24.18 N

(b) a = 22.87 m/s²

Explanation:

Newton's second law of the  block:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the block on the surface   and the y-axis in the direction perpendicular to it.

F₁ : Horizontal force

F₂ : acting at an angle of θ to the horizontal,

W: Weight of the block  : In vertical direction

FN : Normal force : perpendicular to the direction the surface

fk : Friction force: parallel to the direction to the surface

Known data

m =3.1 kg : mass of the  block

F₁ = 65 N,  horizontal force

F₂ = 12.4 N acting at an angle of θ to the horizontal

θ = 30° angle θ of F₂ with respect to the horizontal

μk = 0.2 : coefficient of kinetic friction between the block and the surface

g = 9.8 m/s² : acceleration due to gravity

Calculated of the weight  of the block

W= m*g  = (3.1 kg)*(9.8 m/s²) = 30.38 N

x-y F₂ components

F₂x = F₂cos θ= (12.4)*cos(30)° = 10.74 N

F₂y = F₂sin θ= (12.4)*sin(30)° = 6.2 N

a)Calculated of the Normal force  (FN)

We apply the formula (1)

∑Fy = m*ay    ay = 0

FN+6.2-30.38 = 0

FN = -6.2+30.38

FN = 24.18 N

Calculated of the Friction force:

fk=μk*N=  0.2* 24.18 N = 4.836 N

b) We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax ,  ax= a  : acceleration of the block

F₁ + F₂x -fk = ( m)*a

65 N + 10.74 -4.836 = ( 3.1)*a

70.904 = ( 3.1)*a

a = (70.904 ) / ( 3.1)

a = 22.87 m/s²

4 0
2 years ago
How long will it take a car to accelerate from 15.2 to 23.5 m/s if the car has an average acceleration of 3.2 m/s?
Nuetrik [128]
 a =  ( V2 - V1)/( t2 - t1)
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t(3.2) = 8.3
t = 8.3/3.2
t = 2.59 seconds
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2 years ago
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The cheetah can maintain its maximum speed for only 7.5 s. What is the minimum distance the gazelle must be ahead of the cheetah
amm1812

Basically the cheetah is running 31.5km/h faster than the gazelle. So to determone how long it will take to cover 9mm at that speed, you have to a lot of work. If you skip all of that work, the answer is 1.60m seconds

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irga5000 [103]
14 m/s in the direction of the truck
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Power is defined as the rate at which the body is doing work:
P=\frac{W}{t}
Work is defined as displacement done by the force times that displacement:
W=F\cdot h
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W=62N\cdot10m=620J
of work.
Now we just divide that by 5s to get how much power is required:
P=\frac{620J}{5s}=124W
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2 years ago
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