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3 years ago

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Suppose you are riding your bike along a path that is also used by in-line skaters. You pass a skater, and another biker passes

you, both going in the same direction you are going. You pass a family having a picnic on the grass. Describe your motion from the points of view of the skater, the other biker, and the family.

1 answer:

REY [17]3 years ago

5 0

Their all going to the same place your motion is moving

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The energy an object acquires when it is exposed to a force is called _____ energy

r-ruslan [8.4K]

I'm pretty sure the energy an object acquires when exposed to a force is known was potential energy.

4 0

2 years ago

A car traveled 112.0m in 24.500s, what was a car’s average speed?

kodGreya [7K]

Average speed of the car is 4.57 m/s

Explanation:

Speed is calculated by the rate of change of displacement.

It is given by the formula, Speed = Distance/Time

Here, distance = 112 m and time = 24.5 s

Speed of the car = 112/24.5 = 4.57 m/s

3 0

3 years ago

Can someone solve this problem and explain to me how you got it

evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒ $F\alpha\frac{q1*q2}{r^{2}}$

∴ $F=k\frac{q1*q2}{r^{2}}$

where $F$ is the force of attraction or repulsion

$k$ is Coulumb's constant= $9*10^{9}Nm^{2}C^{-2}$

$q1$ and $q2$ are the magnitude of the charges

$r$ is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

$F=9*10^{9}*\frac{0.041*0.029}{12^{2}}$

$F=74312.5N$

Question6:

Given: q1= $3*10^{-18}C$ , r=5 m, F= $4*10^{-11}N$

therefore by Coulumb's law,

$4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}$

⇒ $q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C$

4 0

3 years ago

A racetrack curve has radius 70.0 m and is banked at an angle of 12.0 ∘. The coefficient of static friction between the tires an

zubka84 [21]

Answer:

See attachment below

Explanation:

7 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

2 years ago