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polet [3.4K]
3 years ago
10

A package is pushed across the floor a distance of 90 feet by exerting a force of 32 lbs downward at an angle of 24 ∘ with the h

orizontal. How much work is done? Round to the nearest foot pound.
Physics
1 answer:
Crazy boy [7]3 years ago
4 0

Answer:

2632 foot-pound

Explanation:

Work done: Work is said to be done when ever a force moves a body through a given distance. The S.I unit of force is Newton (N).

From the question,

The expression for work done is given as,

W = Fdcos∅......................... Equation 1

Where W = work done, F = force, d = distance, ∅ = angle between the force and the horizontal.

Given: F = 32 lbs, d = 90 feet, ∅ = 24°

substitute into equation 1

W = 32×90×cos24

W = 2880(0.914)

W = 2632.32

W = 2632 foot-pound

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At a separation distance of 0.500-m, two like-charged balloons experience a repulsive force of 0.320 N. If the distance is is de
soldi70 [24.7K]

Answer:

2.9 N

Explanation:

When the separation distance, r, is 0.5 m, the electrostatic force is 0.32 N. Electrostatic force is given as:

F = (k * q1 * q2) / r²

Where F = force acting on the balloons

k = Coulombs constant

Therefore:

0.32 = (k * q1 * q2) / 0.5²

=> k * q1 * q2 = 0.32 * 0.5² ------------(1)

When the distance is decreased by 3, that is r = r/3 = 0.5/3

F = (k * q1 * q2) / (0.5/3)² ------------(2)

Putting (1) into (2):

=> F = (0.32 * 0.5²) / (0.5/3)²

F = (0.32 * 0.5² * 3²) / 0.5²

F = 2.9 N

Therefore, the force would be 2.9 N

3 0
3 years ago
if you take one hour drive at an average speed of 65 mph, is it possible for another car with an average speed of 55 mph to pass
SVETLANKA909090 [29]
Nope, this is impossible because in order for a car to pass another, they the 55 mph car would have to be behind the 65 mph car (meaning that starting ahead of the 65 mph doesn't count as pass). Assuming that they both drive for one hour, it is impossible because the first will cover a distance of 65 mi and the second would cover a distance of 55 mi. One is obviously ahead of the other and is therefore impossible for the slow one to pass the first one unless the slow car keeps driving after an hour. In that case, it would take approximately 11 minutes for it to pass the other car. This was found by finding the distance needed to pass the first car : 65 - 55 = 10 mi and converted using 1 hr/ 55 mi = .18181818 hr x 60 min/ 1 hr = 11 seconds
I hope this helps :)
7 0
3 years ago
Read 2 more answers
What is the impact of arteriosclerosis?
solong [7]
It makes the artery walls hard and thicc.
5 0
3 years ago
Read 2 more answers
A couple of astronauts agree to rendezvous in space after hours. Their plan is to let gravity bring them together. She has a mas
Oduvanchick [21]

Answer:

(a) Acceleration of female astronaut = 9.33*10^-12 m/s^2; Acceleration of male astronaut = 7.56*10^-12 m/s^2

(b) 27.013 days

(c) No, their accelerations would not be constant.

Explanation:

In the given question, we have:

Mass of female astronaut M_{1} = 60.0 kg

Mass of male astronaut M_{2} = 74.0 kg

Distance between them (S) = 23.0 m

(a) The free-body diagram is shown in the attached figure.

Using the equations below, the initial accelerations of the two astronauts can be calculated:

Force of gravity (F) = \frac{G*M_{1}*M_{2}}{S^{2} }

G = 6.67*10^-11 \frac{m^{3} }{kg*s^{2} }

F = (6.67*10^-11 *60*74)/23^2 = 5.598*10^-10 N

For the female astronaut, her initial acceleration = F/M_{1} = 5.598*10^-10/60 = 9.33*10^-12 m/s^2

For the male astronaut, his acceleration = F/M_{2} = 5.598*10^-10/74 = 7.56*10^-12 m/s^2

(b) Since the different between their mass is not much, we can deduce that:

a_{average} = \frac{a_{1}+a_{2}}{2} = (9.33*10^-12 + 7.56*10^-12)/2 = 8.445*10^-12 m/s^2

Using the equation below, we can calculate the the time:

S = ut + 1/2 (at^2)   where u = 0

23 = 1/2 (8.445*10^-12)*t^2

t^2 = 5.447*10^12

t = 2333883.044 s = 27.013 days

(c) No, their accelerations will not be constant. It will increase because their radii would be decreasing.

8 0
3 years ago
A snowball starting at rest rolls down a hill and reaches 5 m/s. If the hill is
Lostsunrise [7]

Answer:

The acceleration of the snowball is 0.3125

Explanation:

The initial speed of the snowball up the hill, u = 0

The speed the snowball reaches, v = 5 m/s

The length of the hill, s = 40 m

The equation of motion of the snowball given the above parameters is therefore;

v² = u² + 2·a·s

Where;

a = The acceleration of the snowball

Plugging in the values, we have;

5² = 0² + 2 × a × 40

∴ 2 ×  40 × a  = 5² = 25

80 × a = 25

a = 25/80 = 5/16

a = The acceleration of the snowball = 5/16 m/s².

The acceleration of the snowball = 5/16 m/s² = 0.3125 m/s² .

4 0
3 years ago
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