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2 years ago

How much force (in Newtons) acts on a 1700 kg car going around a

Physics

1 answer:

Daniel [21]2 years ago

5 0

Answer:

b) 4781 N

Explanation:

Because there is a redius do this question is talking about the acceleration force which= mv^2/r

so a=15^2/80=2.8125 m^2/s

so the force will be = m.a

F =1700×2.8125=4781.25 N

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During intercourse, Lydia constantly worries whether her partner thinks she is

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Answer is c spectatoring

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2 years ago

Consider the two vectors A = 3 î − ĵ and B = − î − 2 ĵ.

LUCKY_DIMON [66]

Answer:

(a) A+B = 2i-3j

(B) A-B = 4i + j

Explanation:

We have given two vectors A = 3i-j and B = -1-2j

We have to find the two vectors that is A+B and A-B

(A) In first art we have calculate A+B for this we have to add simply vector A and v ector B

So A+B = 3i-j-i-2j = 2i-3j

(B) In this part we have to find A-B for this we have to simply subtract B from A so A-B = 3i-j-(-i-2j) =3i-j+i+2j =4i+j

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3 years ago

A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball

Aleksandr [31]

Answer:

the velocity (magnitude and direction) of the ball) just before the collision is $v_i = 5.144 \ m/s$

The velocity (magnitude and direction) of the ball) just after the collision is $v_f = - 1.129 \ m/s$

Explanation:

According to the law of conservation of energy;

$m__{ball}} gh = \frac{1}{2}m__{ball}}v_i^2\\\\2* m__{ball}} gh = m__{ball}}v_i^2\\\\v_i^2 = \frac{2*m_{ball}gh}{m_{ball}}\\\\v_i^2 = 2gh\\\\v_i = \sqrt{2gh} \\\\v_i = \sqrt{2*9.8*1.35}\\\\$

$v_i = 5.144 \ m/s$

Thus; the velocity (magnitude and direction) of the ball) just before the collision is $v_i = 5.144 \ m/s$

Since, Air resistance is negligible, and the collision is elastic.

The equation for the conservation of momentum and energy can be expressed as:

$v_f = [\frac{m_1 -m_2}{m_1+m_2}]v_i\\\\v_f = [\frac{m_{ball} -m_{block}}{m_{ball}+m_{block}}]v_i\\\\v_f = [\frac{1.6 -2.5}{1.6+2.5}]*5.144\\\\$

$v_f = - 1.129 \ m/s$

The velocity (magnitude and direction) of the ball) just after the collision is $v_f = - 1.129 \ m/s$

8 0

2 years ago

A rotating fan completes 1160 revolutions every minute. Consider the tip of a blade, at a radius of 19.0 cm. (a) Through what di

lakkis [162]

For the rotating fan that completes 1160 revolutions every minute, with a tip's radius of 19.0 cm, we have:

a) The tip will move a distance of 1.19 m in one revolution.

b) The tip's speed is 23.08 m/s.

c) The magnitude of the tip's acceleration is 2803.6 m/s².

d) The period of the motion is 0.052 s.

a) The distance to which the tip move in one revolution is given by the circumference of the circle:

$d = 2 \pi r$

Where:

r: is the radius of the blade = 19.0 cm

$d = 2 \pi r = 2\pi*0.19 m = 1.19 m$

Hence, the tip will move a distance of 1.19 m in one revolution.

b) The speed of the tip can be calculated with the following equation:

$v = \omega r$

Where:

ω: is the angular speed = 1160 rev/min

Then, the tip's speed is:

$v = \omega r = 1160 \frac{rev}{min}*\frac{2\pi rad}{1 rev}*\frac{1 min}{60 s}*0.19 m = 23.08 m/s$

c) The acceleration of the tip is given by:

$a = \frac{v^{2}}{r} = \frac{(23.08 m/s)^{2}}{0.19 m} = 2803.6 m/s^{2}$

Therefore, the magnitude of the acceleration is 2803.6 m/s².

d) The period of the motion is:

$T = \frac{1}{f} = \frac{2\pi}{\omega}$

Where:

f: is the frequency

$T = \frac{2\pi}{\omega} = \frac{2\pi}{1160 \frac{rev}{min}*\frac{2\pi rad}{1 rev}*\frac{1 min}{60 s}} = 0.052 s$

Hence, the period of the motion is 0.052 s.

You can find more about the period and angular acceleration here: brainly.com/question/9708010?referrer=searchResults

I hope it helps you!

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2 years ago

Newton's second law states that the force on a drag racer is equal to the mass of the car times its acceleration. One of the mos

Oxana [17]

The answer is Friction

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3 years ago

Read 2 more answers