Ask question

Ask question

All categories

4 years ago

5

In some cases, electrons, which have negative charges, can move from one object to another. Move the balloon to the wall and the

n to the sweater. Describe any movement of electrons.

2 answers:

irina1246 [14]4 years ago

3 0

Answer:

on the wall thye ballons looses electrons due to earthing and on the sweater it gains electrons again

Explanation:

maria [59]4 years ago

3 0

Answer:

There’s no movement of electrons between the wall and the balloon, but electrons move from the sweater to the balloon.

Explanation:

plato

You might be interested in

mezya [45]

The advantage is that we do not run out of resources and a disadvantage is that is dangerous when a “human” gets too close and gets sick by the radiation.

5 0

3 years ago

What is the biggest similarity between sports in the United States and sports in other cultures?

leva [86]

Answer:

The first difference is fairly straightforward: other countries prefer other sports. While some aspects of American sports culture are universal there are other things that are sports-specific, such as the 7th-inning stretch at a baseball game or counting the steps of a basketball player after he fouls out.

Explanation:

4 0

3 years ago

Read 2 more answers

9. a. Determine the mass of a football which has a weight of 0.80 N on a planet where the gravitational field strength is 2 N/kg

krek1111 [17]

Answer:

m = 0.4 [kg]

Explanation:

Weight is considered as a force and this is equal to the product of mass by gravitational acceleration.

$W=m*g\\$

where:

W = weight = 0.8 [N]

m = mass [kg]

g = gravity acceleration 2[N/kg]

Therefore:

$m=W/g\\m = .8/2\\m = 0.4 [kg]$

5 0

3 years ago

A felt-covered beanbag is fired into a empty wooden crate that sits on a concrete floor and is open on one side. The beanba

Phantasy [73]

Answer:

31.42383 m/s

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

$\mu$ = Coefficient of kinetic friction = 0.48

s = Displacement = 0.935 m

$m_1$ = Mass of bean bag = 0.354 kg

$m_2$ = Mass of empty crate = 3.77 kg

$v_1$ = Speed of the bean bag

$v_2$ = Speed of the crate

Acceleration

$a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g$

$a=--9.81\times 0.48=4.7088\ m/s^2$

From equation of motion

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.7088\times 0.935+0^2}\\\Rightarrow v=2.96739\ m/s$

In this system the momentum is conserved

$m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_1=\frac{(m_1+m_2)v_2}{m_1}\\\Rightarrow u=\frac{(0.354+3.77)\times 2.69739}{0.354}\\\Rightarrow u=31.42383\ m/s$

The speed of the bean bag is 31.42383 m/s

8 0

4 years ago

The roller coaster car has a mass of 700 kg, including its passenger. If it is released from rest at the top of the hill A, dete

sweet [91]

Answer:

$h = 18.75 m$

Now when it will reach at point B then its normal force is just equal to ZERO

$N_B = 0$

$F_n = 1.72 \times 10^4$

Explanation:

Since we need to cross both the loops so least speed at the bottom must be

$v = \sqrt{5 R g}$

also by energy conservation this is gained by initial potential energy

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

$v = \sqrt{2gh}$

so we will have

$\sqrt{2gh} = \sqrt{5Rg}$

now we have

$h = \frac{5R}{2}$

here we have

R = 7.5 m

so we have

$h = \frac{5(7.5)}{2}$

$h = 18.75 m$

Now when it will reach at point B then its normal force is just equal to ZERO

$N_B = 0$

now when it reach point C then the speed will be

$mgh - mg(2R_c) = \frac{1}{2]mv_c^2$

$v_c^2 = 2g(h - 2R_c)$

$v_c = 13.1 m/s$

now normal force at point C is given as

$F_n = \frac{mv_c^2}{R_c} - mg$

$F_n = \frac{700\times 13.1^2}{5} - (700 \times 9.8)$

$F_n = 1.72 \times 10^4$

7 0

4 years ago