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rosijanka [135]

3 years ago

8

A 1500 kg car traveling at 80.0 km/h comes to a screeching halt in a time of 4.00 seconds. Calculate the force of friction exper

ienced by the car.

1 answer:

amm18123 years ago

7 0

Answer:

F=m x a

(F is force ,M is mass and A is acceleration)

in thisncase the Mass is given but we need to find ou the acceleration

Formula for acceleration-

a=(v - u)/t

(v is final velocity , u is initiatal velocity and t is time)

a = (0 - 80)/4

a= -80/4

a= -20

By substituting the values-

F= m x a

F= 1500 x -20

F=-30000N

Thus the force acted is -30000N

hope this helps

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DENIUS [597]

Answer:

Notebook paper makes paper airplane fly farther

Explanation:

It's because paper airplane made of notebook is lighter and can fly far.

4 0

1 year ago

Read 2 more answers

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

2 years ago

A 750-kg automobile is moving at 16.8 m/s at a height of 5.00 m above the bottom of a hill when it runs out of gasoline. The car

anygoal [31]

Answer:h=19.4 m

Explanation:

Given

mass of automobile $m=750\ kg$

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Thus conserving total energy we get

$K_i+U_i=K_f+U_f$

$\frac{1}{2}mv^2+mgh_o=\frac{1}{2}m(0)^2+mgh$

$\frac{v^2}{2g}+h_o=h$

$h=5+\frac{16.8^}{2\times 9.8}$

$h=5+14.4$

$h=19.4\ m$

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2 years ago

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3 years ago

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dexar [7]

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7 0

3 years ago