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Finger [1]
3 years ago
10

What are the ways of heat transfer?

Physics
2 answers:
AnnZ [28]3 years ago
7 0

Answer:

Conduction, Convection and Radiation

Explanation:

Hope this helps!

Illusion [34]3 years ago
5 0
Conduction radiation and convection
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A tennis ball is dropped from 1.3 m above
Dominik [7]

Answer:

2.65m/s

Explanation:

Using the equation of motion:

v² = u²+2a∆S where

v is the final velocity

u is the initial velocity

∆S is the change in distance

a is the acceleration

Given

u = 0m/s

a = 9.8m/s²

∆S = 1.3-0.943

∆S = 0.357m

Substituting the given parameters into the formula

v² = 0²+2(9.8)(0.357)

v² = 0+6.9972

v² = 6.9972

v=√6.9972

v = 2.65m/s

Hence the velocity at which it hit the ground is 2.65m/s

4 0
3 years ago
How does the density of water change when: (a) it is heated from 0o
Radda [10]

Answer:

[b] it id heated from 4o

Explanation:

3 0
3 years ago
A 10 Kg ball is rolling at 2.5 m/s. It is then hit from behind with a bat that puts a 300 N force on the ball for a quick .3 sec
vazorg [7]

Answer: g. gg g rfrcdv

Explanation:

vfv g bygyb

7 0
3 years ago
Ninais ba ni floranteng mapalapit ang kanyang loob kay adolfo
luda_lava [24]

Answer:

?

Explanation:

8 0
2 years ago
A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15
Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

T-mg = m\frac{v^2}{R}

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

m \frac{v^2}{R} is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

T=m\frac{v^2}{R}

And by substituting the numerical values, we find

T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

T+mg=m\frac{v^2}{R}

And substituting the numerical values and re-arranging it, we find

T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

T=0\\mg = m\frac{v^2}{R}

and re-arranging the equation, we find

v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s

7 0
3 years ago
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