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2 years ago

14

Consider a pair of forces, one having a magnitude of 20 n and the other a magnitude of 12 n. what is the strongest possible net

force for these two forces? what is the weakest possible net force?

1 answer:

AysviL [449]2 years ago

3 0

Accroding to head to tail rule
For maximum force
F=F1+F2
F=20+12
F=32N
Minimum force
F=F1-F2
F=20-12
F=8N

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A 1130-kg car is held in place by a light cable on a smooth (frictionless) ramp. The cable makes an angle of 31.0° above the sur

zubka84 [21]

Answer:

T = 5163.89 N

Explanation:

Newton's first law:

∑F =0 Formula (1)

∑F : algebraic sum of the forces in Newton (N)

We define the x-axis in the direction parallel to the movement of the car on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the car

W: Weight of the car : In vertical direction

FN : Normal force : perpendicular to the ramp

T :Tension force: at angle of 31.0° above the surface of the ramp

Calculated of the Weight of the car (W)

W = m*g m: mass g:acceleration due to gravity

W = 1130-kg* 9.8 m/s² = 11074 N

x-y weight components

Wx = 11074 N*sin 25.0° = 4680.07 N

Wy = 11074 N*cos 25.0° = 10036.45 N

x-y Tension components

Tx = T*cos 25.0°

Ty = T*sin 25.0°

Newton's first law:

∑Fx =0 Formula (1)

Tx-Wx = 0

T*cos 25.0° - 4680.07 = 0

T*cos 25.0° = 4680.07

T = 4680.07 / cos 25.0°

T = 5163.89 N

4 0

2 years ago

An unfortunate 18 kg monkey falls from a 40 m tall tree. What is the monkeys final velocity just befor he impacts the ground.? a

zalisa [80]

The correct answer is c) 28 m/s.
Let's find the step-by-step solution. The motion of the monkey is an uniformly accelerated motion, with acceleration equal to $g=9.81 m/s^2$ . The initial velocity of the monkey is zero, while the distance covered is S=40 m. Therefore, we can use the following relationship to find vf, the final velocity of the monkey:
$2aS=v_f^2-v_i^2=v_f^2$
from which
$v_f= \sqrt{2aS}= \sqrt{2\cdot 9.81 m/s^2 \cdot 40 m}=28 m/s$

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2 years ago

Calculate the GPE of a rock with a mass of 55kg on a cliff that is 27m high

GaryK [48]

Answer:

<em>P=mgh</em>

<em>P=mghm=55</em>

<em>P=mghm=55g=9.8 or ~10</em>

<em>P=mghm=55g=9.8 or ~10h=27</em>

Explanation:

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2 years ago

A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.680 Hz. The pendulum ha

sineoko [7]

Answer:

Therefore, the moment of inertia is:

$I=0.37 \: kgm^{2}$

Explanation:

The period of an oscillation equation of a solid pendulum is given by:

$T=2\pi \sqrt{\frac{I}{Mgd}}$ (1)

Where:

I is the moment of inertia
M is the mass of the pendulum
d is the distance from the center of mass to the pivot
g is the gravity

Let's solve the equation (1) for I

$T=2\pi \sqrt{\frac{I}{Mgd}}$

$I=Mgd(\frac{T}{2\pi})^{2}$

Before find I, we need to remember that

$T = \frac{1}{f}=\frac{1}{0.680}=1.47\: s$

Now, the moment of inertia will be:

$I=2*9.81*0.340(\frac{1.47}{2\pi})^{2}$

Therefore, the moment of inertia is:

$I=0.37 \: kgm^{2}$

I hope it helps you!

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3 years ago

9. How does the length of the hypotenuse in a right triangle relate to the lengths of the legs? (2 points)

konstantin123 [22]

<span>The pythagorean theorem addresses the length of the hypotenuse in relation to the length of the legs. The square root of the length of the hypotenuse is equal to the sum of one leg squared plus the other leg squared. In other words, A squared plus B squared equals C squared where A and B are the lengths of the legs of the triangle and C is the length of the hypotenuse.</span>

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3 years ago

Read 2 more answers