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2 years ago

7

A car of mass 750kg accelerates away from traffic lights. At the end of the first 100m it has reached a speed of 12 m/s. During

this time, its engine provides an average forward force of 780 N, and the average force of friction on the car is 240 N. Calculate the increase in the car's kinetic energy at the end of the first 100 m

2 answers:

xenn [34]2 years ago

5 0

Answer:

the answer according to me is 54000j

Explanation:

im not sure why the force and friction are given inthe question because the the formula of ke is 1/mv^2

aliina [53]2 years ago

3 0

Answer:

Hruhriehrvdi8d8edeuebueufbrhddvuszu

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3 years ago

The lung capacity of the average adult is 5.5 liters at the surface and only .37 liters at a depth of 100 meters. The formula to

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2 years ago

A jet plane is launched from a catapult on an aircraft carrier. In 2.0 s it reaches a speed of 42 m/s at the end of the catapult

Dvinal [7]

Answer:

Explanation:

Given

time taken $t=2\ s$

Speed acquired in 2 sec $v=42\ m/s$

Here initial velocity is zero $u=0$

acceleration is the rate of change of velocity in a given time

$a=\frac{v-u}{t}$

$a=\frac{42-0}{2}=21\ m/s^2$

Distance travel in this time

$s=ut+0.5at^2$

where

s=displacement

u=initial velocity

a=acceleration

t=time

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$s=42\ m$

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5 0

3 years ago

A -kilogram car travels at a constant speed of 20. meters per second around a horizontal circular track. The diameter of the tra

Tatiana [17]

Answer:

The centripetal acceleration of the car is $8\ m/s^2$ .

Explanation:

Let the mass of the car, $m=10^3\ kg$

Diameter of the circular path, d = 100 m

Speed of car, v = 20 m/s

Radius, r = 50 m

When an object moves in a circular path, the centripetal acceleration acts on it. It is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(20\ m/s)^2}{50\ m}$

$a=8\ m/s^2$

So, the centripetal acceleration of the car is $8\ m/s^2$ . Hence, this is the required solution.

4 0

3 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

4 0

2 years ago