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3 years ago

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A 4.5 kg box is being pulled by a girl using a string that makes an angle of 20° horizontal. Find the force of friction if the f

orce exerted on the string is 15 N and the coefficient of friction is 0.5.

1 answer:

Mashutka [201]3 years ago

5 0

Answer:

Fcoso-f =ma. F= ma 15 = 4.5 a then get acceleration. Fcoso- umg = ma then get friction

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A tennis player serves a tennis ball such that it is moving horizontally when it leaves the racquet. When the ball travels a hor

finlep [7]

Answer:

The initial velocity is 38.46 m/s.

Explanation:

The horizontal distance travel by the tennis ball = 13 m

The height at which the tennis ball dropped = 56 cm

Now calculate the initial speed of tennis ball.

The vertical velocity is zero.

Below is the calculation. Here, first convert centimetre into kilometre. So, height at which ball dropped is 0.56 km.

$v = \sqrt{2 \times 9.8 \times 0.56} = 3.32 m/s \\$

$t = \frac{3.32}{9.8} = 0.338s \\$

$Ux \times t = 13 \\$

$Ux = \frac{13}{0.338} = 38.46 m/s = Initial velocity.$

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4 years ago

Two balloons with the same weight and vol-

yanalaym [24]

I would say the expandable one

I’m not sure if I’m right, if I am I hope this helped you, have a wonderful day!

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3 years ago

Find the position of the center of mass of the system of the sun and Jupiter? (Since Jupiter is more massive than the rest of th

8090 [49]

Answer:

$r_{cm} = 0.074 m$ from the position of the center of the Sun

Explanation:

As we know that mass of Sun and Jupiter is given as

$M_s = 1.98 \times 10^{30} kg$

$M_j = 1.89 \times 10^{27} kg$

distance between Sun and Jupiter is given as

$r = 7.78 \times 10^{11} m$

now let the position of Sun is origin and position of Jupiter is given at the position same as the distance between them

so we will have

$r_{cm} = \frac{M_s r_1 + M_j r_2}{M_s + M_j}$

$r_{cm} = \frac{1.98 \times 10^{30} (0) + (1.89 \times 10^{27})(7.78 \times 10^{11})}{1.98 \times 10^{30} + 1.89 \times 10^{27}}$

$r_{cm} = 0.074 m$ from the position of the center of the Sun

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4 years ago

A car has a mass of 1.00 × 103 kilograms, and it has an acceleration of 4.5 meters/second 2. What is the net force on the car?

Elodia [21]

Newton #2: F = M a

(1 x 10³ kg) (4.5 m/s²) = 4,500 newtons

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3 years ago

I'll mark brainless pictures down below

julsineya [31]

The net force is (40 Newtons) (down the road).

But 40 Newtons is not going to move a piano very enthusiastically.

8 0

3 years ago