Objects 1 and 2 attract each other with a electrostatic force of 18.0 units. If the charge

Physics

1 answer:

xxMikexx [17]3 years ago

5 0

Answer:

new force is 6 times of the initial force.

Explanation:

Let the charges on two objects is q₁ and q₂. The electric force between charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

Objects 1 and 2 attract each other with a electrostatic force of 18.0 units

If the charge of Object 1 is doubled and the charge of object 2 is tripled, it means, $q_1'=2q_1$ and $q_2'=3q_2$ . New force is given by :

$F'=\dfrac{kq_1'q_2'}{r^2}\\\\F'=\dfrac{k(2q_1)(3q_2)}{r^2}\\\\F'=6\dfrac{kq_1q_2}{r^2}\\\\F'=6F$

So, the new electrostatic force between objects will become 6 times of the initial force.

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The skateboarder has a mass of 100 kg. When traveling downward from E to D, he reaches a velocity of 11 m/s. Calculate his kinet

patriot [66]

6050 J is the kinetic energy at D

<u>Explanation:</u>

In physics, the object's kinetic energy (K.E) defined as the energy it possesses during movement. It can be defined as the required work to accelerate a certain body weight in order to rest at a certain speed. When the body receives this energy as it speeds up (accelerates), it retains this energy unless speed varies. The equation is given as,

$K . E=\frac{1}{2} \times m \times v^{2}$

Where,

m - mass of an object

v - velocity of the object

Here,

Given data:

m = 100 kg

v = 11 m/s

By substituting the given values in the above equation, we get

$K . E=\frac{1}{2} \times 100 \times(11)^{2}=\frac{1}{2} \times 100 \times 121=\frac{12100}{2}=6050\ \mathrm{J}$

6 0

4 years ago

A bicyclist starting at rest produces a constant angular acceleration of 1.30 rad/s2 for wheels that are 35.5 cm in radius.

Debora [2.8K]

Answer:

a) 0.462 m/s^2

b) 31.5 rad/s

c) 381 rad

d) 135m

Explanation:

the linear acceleration is given by:

$a=\alpha *r\\a=1.30rad/s^2*(35.5*10^{-2}m)\\a=0.462m/s^2$

the angular speed is given by:

$\omega=\frac{v}{r}\\\\\omega=\frac{11.2m/s}{35.5*10^{-2}m}\\\\\omega=31.5rad/s$

to calculate how many radians have the wheel turned we need the apply the following formula:

$\theta=\frac{1}{2}\alpha*t^2\\\\t=\frac{\omega}{\alpha}\\\\t=\frac{31.5rad/s}{1.30rad/s^2}\\\\t=24.2s\\\\\theta=\frac{1}{2}*1.30rad/s^2*(24.2s)^2\\\\\theta=381rad$