All categories

3 years ago

Which change increases the electric force between objects?

Physics

2 answers:

sdas [7]3 years ago

7 0

the answer to your question is b broskey

Brrunno [24]3 years ago

3 0

Well, the force is proportional to the product of the charges
on the two objects. So if the objects are already negatively
charged distance between them is unchanged, then adding
electrons to either or both objects would increase the forces
between them.

You might be interested in

Without the wheels, a bicycle frame has a mass of 8.29 kg. Each of the wheels can be roughly modeled as a uniform solid disk wit

trapecia [35]

Answer:

69.66 Joule

Explanation:

mass of bicycle frame, mf = 8.29 kg

mass of wheel, mw = 0.820 kg

radius, r = 0.343 m

velocity, v = 3.6 m/s

There are two wheels in the bicycle.

There are two types of kinetic energy of the system one is kinetic energy of rotation and another is rotational kinetic energy.

$K = \frac{1}{2}m_{f}v^{2}+ 2\times \frac{1}{2}m_{w}v^{2}+ 2\times \frac{1}{2}I_{w}\omega^{2}$

$K = \frac{1}{2}m_{f}v^{2}+m_{w}v^{2}+ \frac{1}{2}\times m_{w}v^{2}$

$K = \frac{1}{2}m_{f}v^{2}+ \frac{3}{2}\times m_{w}v^{2}$

$K = \frac{1}{2}\times 8.29\times 3.6^{2}+ \frac{3}{2}\times 0.820\times 3.6^{2}$

K = 69.66 J

3 0

4 years ago

A cyclist does work at the rate of 500 w while riding. how much force does her foot push with when she is traveling at 8.0 m/s?

NikAS [45]

P=F*V
F=P/V
F=500W/8=62.5N

7 0

3 years ago

Read 2 more answers

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$