A child kicks a ball horizontally with a speed of 2.8 m/s from the end of a deck that is 8.5 m high.

You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s

densk [106]

Here when car in front of us applied brakes then it is slowing down due to frictional force on it

So here we can say that friction force on the car front of our car is given as

$F_f = \mu m g$

So the acceleration of car due to friction is given as

$F_{net} = - \mu mg$

$a = \frac{F_{net}}{m}$

$a = -\mu g$

now it is given that

$\mu = 0.868$

$g = 9.81 m/s^2$

so here we have

$a = -0.868 * 9.81$

$a = -8.52 m/s^2$

so the car will accelerate due to brakes by a = - 8.52 m/s^2

4 0

4 years ago

A car is moving at 25.5 m/s when it accelerates at 1.94 m/s^2 for 2.3 s. What is the car's final speed? (Keep in mind direction

Stolb23 [73]

Answer:

29.96m/s

Explanation:

Given parameters:

Initial speed = 25.5m/s

Acceleration = 1.94m/s²

Time = 2.3s

Unknown:

Final speed of the car = ?

Solution:

To solve this problem, we are going to apply the right motion equation:

v = u + at

v is the final speed

u is the initial speed

a is the acceleration

t is the time taken

Now insert the parameters and solve;

v = 25.5 + (1.94 x 2.3) = 29.96m/s

3 0

3 years ago

Which of the following statements are true concerning the creation of magnetic fields? Check all that apply. Check all that appl

Andreas93 [3]

<h3><u>Answer;</u></h3>

A moving electric charge creates a magnetic field at all points in the surrounding region.
An electric current in a conductor creates a magnetic field at all points in the surrounding region.
A permanent magnet creates a magnetic field at all points in the surrounding region.

<h3><u>Explanation;</u></h3>

A magnetic field can be created by running electricity through a wire. All magnetic fields are created by moving charged particles. it is important to also note that charged particles create magnetic fields only when they are moving.
The strength of the magnetic field generated or created is proportional to the amount of current flowing through the wire. Thus, increasing the current increases the strength of the magnetic field.

8 0

3 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

3 years ago

How much energy is required to move an electron through a potential difference of

Tresset [83]

I think it’s going to be the 2nd one

3 0

3 years ago