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sineoko [7]
3 years ago
13

a baby carriage is sitting at the top of a hill that is 21 m high. The carriage with the baby weighs 12 Kg. how much energy does

the carriage have?
Physics
1 answer:
Tcecarenko [31]3 years ago
8 0

Answer:

The carriage has the energy, W = 2469.6 J

Explanation:

Given data,

The height of the hill, h = 21 m

The carriage with the baby weighs, m = 12 kg

The energy possessed by the body due to its position is the potential energy,

                                      <em>W = P.E = mgh joules</em>

Substituting the values,

                                       W = 12 x 9.8 x 21

                                            = 2469.6 J

Hence, the carriage has the energy, W = 2469.6 J

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_____ is the frictional force needed to slow an object in motion
s2008m [1.1K]

Answer:

<u>Drag force</u> is the frictional force needed to slow an object in motion

Explanation:

8 0
3 years ago
Read 2 more answers
Six new refrigerator prototypes are tested in the laboratory. For each refrigerator, the electrical power P needed for it to ope
Mandarinka [93]

Answer:

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s.

the rate at which they raise the temperature of the room.

2.1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

Explanation:

A refrigerator is a device that uses work to remove heat energy from a cold reservoir and deposit it into a hot reservoir. .A good refrigerator (with a large performance coefficient) will remove a large amount of heat energy from the cold reservoir for a small amount of work input

The performance coefficient  of a refrigerator is defined as the ratio of the heat energy removed from the cold reservoir  to the work  input to the refrigerator:

k=QC/W

power is defined as work per unit time

1.k=1500/750=2

2. 1200/400=3

3.2000/500=4

4.1000/250=4

5.1500/500=3

6.3000/1000=3

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s

2, Rate at which they raise the temperature of the room.

rate at which temperature rises in the inner chamber of the refrigerator is proportional to the rate of energy used to dispel heat from the refrigerator

1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

5 0
3 years ago
Collar P slides outward at a constant relative speed along rod AB, which rotates counterclockwise with a constant angular veloci
diamong [38]

Answer:

a= 23.65 ft/s²

Explanation:

given

r= 14.34m

ω=3.65rad/s

Ф=Ф₀ + ωt

t = Ф - Ф₀/ω

= (98-0)×\frac{\pi}{180}/3.65

98°= 1.71042 rad

1.7104/3.65

t= 0.47 s

r₁(not given)

assuming r₁ =20 in

r₁ = r₀ + ut(uniform motion)

u = r₁ - r₀/t

r₀ = 14.34 in= 1.195 ft

r₁ = 20 in = 1.67 ft

= (1.667 - 1.195)/0.47

0.472/0.47

u= 1.00ft/s

acceleration at collar p

a=rω²

= 1.67 × 3.65²

a = 22.25ft/s²

acceleration of collar p related to the rod = 0

coriolis acceleration = 2ωu

= 2× 3.65×1 = 7.3 ft/s²

acceleration of collar p

= 22.5j + 0 + 7.3i

√(22.5² + 7.3²)

the magnitude of the acceleration of the collar P just as it reaches B in ft/s²

a= 23.65 ft/s²

4 0
3 years ago
Please help ASAP please
Alex_Xolod [135]

Answer:

14

Explanation:

6 0
3 years ago
Two balloons are charged with an identical quantity and type of charge: -6.25 x 10-6 C. They are held apart at a separation dist
steposvetlana [31]

Answer:

The electric force between them is 878.9 N

Explanation:

Given:

Identical charge q = -6.25 \times 10^{-6}C

Separation between two charges r = 0.02 m

For finding the electrical force,

According to the coulomb's law

   F = \frac{k q^{2} }{r^{2} }

Here, force between two balloons are repulsive because both charges are same.

Where k = 9 \times 10^{9}

   F = \frac{9 \times 10^{9} \times (-6.25 \times 10^{-6} )^{2}  }{4 \times 10^{-4} }

   F = 878.9 N

Therefore, the electric force between them is 878.9 N

8 0
3 years ago
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