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3 years ago

Which force requires contact?

Physics

2 answers:

Minchanka [31]3 years ago

8 0

Answer:

Explanation: A is a example of Air resistance force, which is a contact force.

rosijanka [135]3 years ago

6 0

Answer is A. The force exerted by air as an object moves through it.

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Sound, light, feelings and ideas are not

taurus [48]

I think it’s A, I’m so sorry if I’m wrong.

6 0

3 years ago

Q: The picture above shows a frying pan. Why are frying pans usually made out of metal?

Natasha_Volkova [10]

A. Because metal is a good conductor of heat.

If a frying pan wasn't a good conductor of heat it's purpose of cooking things wouldn't really be resolved. Without heat it couldn't cook well. I think lol that's my way of thinking.

Hope I helped :)

4 0

2 years ago

Read 2 more answers

The magnitude of Earth’s magnetic field is about 0.5 gauss near Earth’s surface. What’s the maximum possible magnetic force on a

vampirchik [111]

Answer:

$F = 1.5 \times 10^{-16} N$

this force is $1.68 \times 10^{13}$ times more than the gravitational force

Explanation:

Kinetic Energy of the electron is given as

$KE = 1 keV$

$KE = 1 \times 10^3 (1.6 \times 10^{-19}) J$

$KE = 1.6 \times 10^{-16} J$

now the speed of electron is given as

$KE = \frac{1}{2}mv^2$

now we have

$v = \sqrt{\frac{2 KE}{m}}$

$v = 1.87 \times 10^7 m/s$

now the maximum force due to magnetic field is given as

$F = qvB$

$F = (1.6\times 10^{-19})(1.87 \times 10^7)(0.5 \times 10^{-4})$

$F = 1.5 \times 10^{-16} N$

Now if this force is compared by the gravitational force on the electron then it is

$\frac{F}{F_g} = \frac{1.5 \times 10^{-16}}{9.1 \times 10^{-31} (9.8)}$

$\frac{F}{F_g} = 1.68 \times 10^{13}$

so this force is $1.68 \times 10^{13}$ times more than the gravitational force

4 0

3 years ago

Air resistance is an example of what type of friction?

Sav [38]

Well, Air resistance is a special type of friction (you cannot classify it in other categories). That force of air-resistance is often observed to oppose the motion of the object,( like every other frictional forces)

Hope this helps!

4 0

3 years ago

A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$

Thus, the final velocity of thrower is $3.88~m/s$ and that for the catcher is $0.029~m/s$ .

8 0

2 years ago