If two objects with static charge are attracted what do you know about them?

A builder drops a brick from a height of 15 m above the ground. The gravitational field strength g is 10 N/ kg. What is the spee

Basile [38]

The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Given the data in the question;

Since the brick was initially at rest before it was dropped,

Initial Velocity; $u = 0$

Height from which it has dropped; $h = 15m$

Gravitational field strength; $g = 10N/kg = 10 \frac{kg.m/s^2}{kg} = 10m/s^2$

Final speed of brick as it hits the ground; $v = \ ?$

<h3>Velocity</h3>

velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:

$v^2 = u^2 + 2gh$

Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.

To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.

$v^2 = u^2 + 2gh\\\\v^2 = 0 + ( 2\ *\ 10m/s^2\ *\ 15m)\\\\v^2 = 300m^2/s^2\\\\v = \sqrt{300m^2/s^2}\\ \\v = 17.32m/s$

Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Learn more about equations of motion: brainly.com/question/18486505

8 0

2 years ago

A 23 g bullet traveling at 230 m/s penetrates a 2.0 kg block of wood and emerges cleanly at 170 m/s. If the block is stationary

Ann [662]

The distance traveled by the wood after the bullet emerges is 0.16 m.

The given parameters;

mass of the bullet, m = 23 g = 0.023 g
speed of the bullet, u = 230 m/s
mass of the wood, m = 2 kg
final speed of the bullet, v = 170 m/s
coefficient of friction, μ = 0.15

The final velocity of the wood after the bullet hits is calculated as follows;

$m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\0.023(230) + 2(0) = 0.023(170) + 2v_2\\\\5.29 = 3.91 + 2v_2\\\\2v_2 = 1.38\\\\v_2 = \frac{1.38}{2} = 0.69 \ m/s$

The acceleration of the wood is calculated as follows;

$\mu = \frac{a}{g} \\\\a = \mu g\\\\a = 0.15 \times 9.8\\\\a = 1.47 \ m/s^2$

The distance traveled by the wood after the bullet emerges is calculated as follows;

$v^2 = v_0^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\s = \frac{v^2}{2a} \\\\s = \frac{(0.69)^2}{2(1.47)} \\\\s = 0.16 \ m$

Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.

Learn more here:brainly.com/question/15244782

7 0

3 years ago

Which of the following is NOT a function of the lens in the eye?

tatyana61 [14]

Answer:

it is a light receptor that generates nerve signals that are sent to the brain

Explanation:

the lens are like the glasses, this means that is used to see things better. You just put them in your eye and that's all it's not connected to the brain

8 0

3 years ago

A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. its initial position

Nesterboy [21]

The position of the object at time t =2.0 s is 6.4 m.

Velocity vₓ of a body is the rate at which the position x of the object changes with time.

Therefore,

$v_x= \frac{dx}{dt}$

Write an equation for x.

$dx=v_xdt\\ x=\int v_xdt$

Substitute the equation for vₓ =2t² in the integral.

$x=\int v_xdt\\ =\int2t^2dt\\ =\frac{2t^3}{3} +C$

Here, the constant of integration is C and it is determined by applying initial conditions.

When t =0, x = 1. 1m

$x= \frac{2t^3}{3} +C\\ x_0=1.1\\ x= (\frac{2t^3}{3} +1.1)m$

Substitute 2.0s for t.

$x= (\frac{2t^3}{3} +1.1)m\\ =\frac{2(2.0)^3}{3} +1.1\\ =6.43 m$

The position of the particle at t =2.0 s is 6.4m

5 0

3 years ago

Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed v at a distance r from the center o

jolli1 [7]

Answer:

c)At a distance greater than r

Explanation:

For a satellite in orbit around the Earth, the gravitational force provides the centripetal force that keeps the satellite in motion:

$G\frac{Mm}{r^2}=m\frac{v^2}{r}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance between the satellite and the Earth's centre

v is the speed of the satellite

Re-arranging the equation, we write

$r = \frac{GM}{v^2}$

so we see from the equation that when the speed is higher, the distance from the Earth's centre is smaller, and when the speed is lower, the distance from the Earth's centre is larger.

Here, the second satellite orbit the Earth at a speed less than v: this means that its orbit will have a larger radius than the first satellite, so the correct answer is

c)At a distance greater than r

7 0

3 years ago