22. The mass of a brick is 5 kg and it is fired into the air at 100 m/s. Thinking

A charge of 4.5x10^-5 C is placed in an electric field with a strength of 2.0x10^4 N/C. What is the electric force acting on the

Alex

Answer:

0.9 N

Explanation:

The electric force acting on a charge is given by:

$F=qE$

where

q is the magnitude of the charge

E is the strength of the electric field

In this problem, we have

$q=4.5\cdot 10^{-5}C$ is the charge

$E=2.0\cdot 10^4 N/C$ is the strength of the electric field

Substituting into the equation, we find

$F=(4.5\cdot 10^{-5}C)(2.0\cdot 10^4 N/C)=0.9 N$

6 0

3 years ago

URGENT!!!!!!!<br><br> PLEASE HELP WITH THIS PHYSICS PROBLEM

Levart [38]

Explanation:

Let

$x_1$ = distance traveled while accelerating

$x_2$ = distance traveled while decelerating

The distance traveled while accelerating is given by

$x_1 = v_0t + \frac{1}{2}at^2 = \frac{1}{2}at^2$

$\:\:\:\:\:= \frac{1}{2}(2.5\:\text{m/s}^2)(30\:\text{s})^2$

$\:\:\:\:\:= 1125\:\text{m}$

We need the velocity of the rocket after 30 seconds and we can calculate it as follows:

$v = at = (2.5\:\text{m/s}^2)(30\:\text{s}) = 75\:\text{m/s}$

This will be the initial velocity when start calculating for the distance it traveled while decelerating.

$v^2 = v_0^2 + 2ax_2$

$0 = (75\:\text{m/s})^2 + 2(-0.65\:\text{m/s}^2)x_2$

Solving for $x_2,$ we get

$x_2 = \dfrac{(75\:\text{m/s})^2}{2(0.65\:\text{m/s}^2)}$

$\:\:\:\:\:= 4327\:\text{m}$

Therefore, the total distance x is

$x = x_1 + x_2 = 1125\:\text{m} + 4327\:\text{m}$

$\:\:\:\:= 5452\:\text{m}$

3 0

3 years ago

The reactants of a certain chemical reaction contain 26 kJ of potential

Anon25 [30]

Answer:c

Explanation:

3 0

3 years ago

Read 2 more answers

An Olympic high jumper, with a mass of 82 kg, has a

Digiron [165]

Answer:

I don't really know

Explanation:

I really wanted to help you, but then I realized i didnt know how to

8 0

2 years ago

During heavy rain, a section of a mountainside measuring 2.5 km horizontally, 0.80 km up along the slope, and 2.0 m deep slips i

Crank

Answer:

The mass of the mud is 3040000 kg.

Explanation:

Given that,

length = 2.5 km

Width = 0.80 km

Height = 2.0 m

Length of valley = 0.40 km

Width of valley = 0.40 km

Density = 1900 Kg/m³

Area = 4.0 m²

We need to calculate the mass of the mud

Using formula of density

$\rho=\dfrac{m}{V}$

$m=\rho\times V$

Where, V = volume of mud

$\rho$ = density of mud

Put the value into the formula

$m=1900\times4.0\times0.40\times10^{3}$

$m =3040000\ kg$

Hence, The mass of the mud is 3040000 kg.

4 0

3 years ago