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evablogger [386]
3 years ago
10

A student rides a bicycle for 15 miles in 3 hours. What is the student's speed? What else would you need to know for the velocit

y?
Physics
1 answer:
kaheart [24]3 years ago
3 0

Answer:

5 miles per hour

Explanation:

if you divide 15 by 3 you get 5, therefore the student is going 5 miles per hour.

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Select the correct arrows. Identify the arrows that show the correct direction of heat transfer.
ohaa [14]

Answer: Is there suppose to be a picture?

Explanation:

4 0
3 years ago
Read 2 more answers
1. A girl drops a ball off the roof of her house. It takes the ball 2 seconds to hit the ground. How tall
Masja [62]

If the acceleration constant..

you can use the formula s = ut + 1/2at²

Known that :

s = ?

u = 0

t = 2s

a = 10ms-²

Then you can apply the formula

s = ut + 1/2at²

s = 0 + 1/2(10)(2)²

s = 5 × 4

s = 20m

Answer : 20m

Explanation :

The gravity can be 9,8 or 10. Also im not sure how people teach you but in my school, if the ball goes down the gravity is positive and not negative thats why i put 10ms-² and not -10ms-²

s = displacement/distance

u = initial speed

a = acceleration

t = time

sorry if im wrong

5 0
3 years ago
How much gravitational force do two lead balls with a mass of 8 kilograms, the centers of mass of which are 17 cm apart, affect
Arisa [49]

Answer:

1.48×10⁻⁷ Newtons

Explanation:

From the question,

According to newton's law of universal gravitation.

F = Gmm'/r²........................ Equation 1

F = gravitational force, G = gravitational constant, m = mass of the first ball, m' = mass of the second ball, r = distance between the balls.

Given: m = m' = 8 kg, r = 17 cm = 0.17 m,

Constant : G = 6.67×10⁻¹¹ Nm²/kg²

Substitute these values into equation 1

F =  (6.67×10⁻¹¹×8×8)/(0.17²)

F = 1.48×10⁻⁷ N

7 0
3 years ago
A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a
Zielflug [23.3K]

Answer:F_{v} =\mu_{k} mg

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to  start crate moving is F_{s} =\mu_{s} mg.

but once crate starts moving the force of friction is reduced  F_{v} =\mu_{k} mg.

Hence  to keep crate moving at constant velocity we have to reduce the  force pushing crate ie F_{v} =\mu_{k} mg.

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as  forces are balanced.

Magnitude of the force

F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9  N

4 0
3 years ago
A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 3300 kcal o
Tema [17]

Answer

given,

heat added to the gas,Q = 3300 kcal

initial volume, V₁ = 13.7 m³

final volume, V₂ = 19.7 m³

atmospheric pressure, P = 1.013 x 10⁵ Pa

a) Work done by the gas

    W = P Δ V

    W = 1.013 x 10⁵ x (19.7 - 13.7)

    W = 6.029 x 10⁵ J

b) internal energy of the gas = ?

  now,

 change in internal energy

  Δ U = Q - W

    Q = 3300 x 10³ cal

    Q = 3300 x 10³ x 4.186 J

    Q = 1.38 x 10⁷ J

now,

  Δ U = 1.38 x 10⁷  - 6.029 x 10⁵

  Δ U = 1.32 x 10⁷ J

6 0
3 years ago
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