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disa [49]
3 years ago
8

In which of these situations is convection most likely the main form of heat transfer? A.Warm air from a heater on the first flo

or of a house moves to the upper floors.
B.The sides of a metal pan become hot when the pan is placed on a stove burner.
C.A person gets a sunburn from lying on the beach too long.
D.An ice cube melts when a person holds it in his hand.
Physics
2 answers:
Sergeeva-Olga [200]3 years ago
6 0
All points ate true

But option A is perfect
andreyandreev [35.5K]3 years ago
6 0

Answer:

A.   Warm air from a heater on the first floor of a house moves to the upper floors.

Explanation:

As we know that in convection mode of heat transfer the thermal energy is transferred from one point to other point which the actual movement of the medium molecules.

When thermal energy is given to the medium particles then due to difference in the density medium molecules will move from one point to other point along with the thermal energy.

So here we can say that correct answer will be

A.Warm air from a heater on the first floor of a house moves to the upper floors.  

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A 110 kg football player running with a velocity of 5.0 m/s hits another stationary football
KengaRu [80]

Answer:

The final velocity of the second player is 6.1 m/s.

Explanation:

The final velocity of the second player can be calculated by conservation of linear momentum (p):

p_{i} = p_{f}  

m_{a}v_{a_{i}} + m_{b}v_{b_{i}} = m_{a}v_{a_{f}} + m_{b}v_{b_{f}}  (1)

Where:

m_{a}: is the mass of the first football player = 110 kg

m_{b}: is the mass of the second football player = 90 kg

v_{a_{i}}: is the initial velocity of the first football player = 5.0 m/s

v_{b_{i}}: is the initial velocity of the second football player = 0 (he is at rest)

v_{a_{f}}: is the final velocity of the first football player = 0 (he stops after the impact)

v_{b_{f}}: is the final velocity of the second football player =?

By solving equation (1) for v_{b_{f}} we have:

110 kg*5.0 m/s + 0 = 0 + 90 kg*v_{b_{f}}

v_{b_{f}} = \frac{110 kg*5.0 m/s}{90 kg} = 6.1 m/s

Therefore, the final velocity of the second player is 6.1 m/s.

I hope it helps you!

8 0
3 years ago
The interest rate that the Fed charges banks that borrow money from it is called A. the discount rate. B. the federal rate. C. t
son4ous [18]
The answer is A. the discount rate.
7 0
3 years ago
Read 2 more answers
A horizontal force of 25 N is required to push a wagon across a sidewalk at a constant speed.
lions [1.4K]

Answer:

a. The net force acting on the wagon is zero

b. The value of the force of friction is 25 N

c. The effect would be is the frictional force must be increase

Explanation:

Lets explain how to solve the problem

A horizontal force of 25 N is required to push a wagon across a sidewalk

at a constant speed

a.

→ Constant speed means acceleration is zero

→ Force = mass × acceleration

→ Force = mass × 0

→ Force = 0

<em>The net force acting on the wagon is zero</em>

b.

The wagon moves with constant speed, then

→ ∑ force = 0

There are two forces acting on the wagon external force of 25 N,

and the frictional force which is opposite to the direction of motion

→ 25 - Frictional force = 0

→ Frictional force = 25 N

<em>The value of the force of friction is 25 N</em>

c.

The force increased to 30 N

If the wagon still moves with constant speed

Due to Newton's law ∑ force in the direction of motion = mass ×

acceleration, but acceleration is 0 because the wagon moves with

constant speed, then ∑ F = 0

So the frictional force must increase to 30 N

<em>The effect would be is the frictional force must be increase</em>

<em></em>

<em>If the frictional force does not increase the wagon will accelerate</em>

6 0
4 years ago
A voltmeter reads 200 V between parallel plates that are 2 cm apart. what is the strength of the electric field between the two
Misha Larkins [42]
I think that is c) l’m not sure
3 0
3 years ago
A spacecraft in the shape of a long cylinder has a length of 100 m, and its mass with occupants is 1 480 kg. It has strayed too
maks197457 [2]

Answer:

2352645198509.9604 m/s²

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of black hole = 90\times 1.989\times 10^{30}\ kg

R_{f} = 10000+100 m

R_{sb} = Distance between the nose and the center of the black hole = 10000 m

The difference in the gravitational field in this system is given by

\Delta F=\dfrac{GMm}{R_{f}^2}-\dfrac{GMm}{R_{sb}^2}\\\Rightarrow \Delta g=GM\left(\frac{1}{R_f^2}-\frac{1}{R_{sb}^2}\right)\\\Rightarrow g=6.67\times 10^{-11}\times 90\times 1.989\times 10^{30}\left(\frac{1}{(10000+100)^2}-\frac{1}{10000^2}\right)\\\Rightarrow \Delta g=-2352645198509.9604\ m/s^2

The acceleration is 2352645198509.9604 m/s²

4 0
3 years ago
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