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ddd [48]
3 years ago
15

A heavy ball with a weight of 100 NN is hung from the ceiling of a lecture hall on a 4.4-mm-long rope. The ball is pulled to one

side and released to swing as a pendulum, reaching a speed of 5.7 m/sm/s as it passes through the lowest point
Physics
1 answer:
allochka39001 [22]3 years ago
7 0

Answer:

175.3 N

Explanation:

The motion of the ball is a uniform circular motion, therefore the net force on it must be equal to the centripetal force.

There are two forces acting on the ball at the lowest point of motion:

- The tension in the string, T , upward

- The weight of the ball, mg, downward

The net force (centripetal force) has the same direction as the tension (upward, towards the centre of the circular path), so we can write:

T-mg=m\frac{v^2}{r}

where the term on the right is the expression for the centripetal force, and where:

T is the tension in the string

mg=100 N is the weight of the ball

m=\frac{mg}{g}=\frac{100}{9.8}=10.2 kg is the mass of the ball

v = 5.7 m/s is the speed of the ball at the lowest point

r = 4.4 m is the length of the rope, so the radius of the circle

Solving for T, we find the tension in the string:

T=mg+m\frac{v^2}{r}=(100)+(10.2)\frac{5.7^2}{4.4}=175.3 N

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VikaD [51]

Answer : a=10\ m/s^2

Explanation :

It is given that,

Mass of the engine, m = 30 kg

Thrust is equivalent to the force acting perpendicularly and it is F = 300 N

According to Newton's second law of motion :

F = m\times a

a is the acceleration of the engine.

a=\dfrac{F}{m}

a=\dfrac{300\ N}{30\ Kg}

a=10\ m/s^2

So, the acceleration of the engine is 10\ m/s^2.

Hence, this is the required solution.

5 0
3 years ago
Read 2 more answers
Constants A capacitor is connected across an ac source that has voltage amplitude 59 0 V and frequency 77 0 Hz Part C What is th
Vladimir79 [104]

Answer:

C = 1.77 \times 10^{-4} F

Explanation:

As we know that in AC circuit we have

V = i x_c

here we have

V = 59 V

i = 5.05 A

so we will have

x_c = \frac{59}{5.05}

x_c = 11.68 ohm

also we know that

x_c = \frac{1}{\omega C}

here we will have

11.68 = \frac{1}{(2\pi 77)C}

C = 1.77 \times 10^{-4} F

7 0
3 years ago
A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a
blsea [12.9K]

We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

E=7*10^{9}N/C

From the question we are told that

A solid metal ball of radius 1.5 cm

bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing

uniformly distributed charge of -7 nC

The distance between the centers of the balls is 9 cm

Generally the equation for the electric field  is mathematically given as

E=\frac{kq_2}{d^2}\\\\E=\frac{(9*10^9)7*10^{-2}}{9*10^{-2}}\\\\

E=7*10^{9}N/C

For more information on this visit

brainly.com/question/21811998

4 0
3 years ago
Two cannonballs are dropped from a second-floor physics lab at height h above the ground. Ball B has four times the mass of ball
denpristay [2]

Answer:

1:4

Explanation:

The formula for calculating kinetic energy is:

KE=\dfrac{1}{2}mv^2

If the mass is multiplied by 4, then, the kinetic energy must be increased by 4 as well. Since they will be travelling at the same speed when they are at the same point, the relation between KA and KB must be 1:4 or 1/4. Hope this helps!

3 0
3 years ago
At 9:00 on Saturday morning, two bicyclists heading in opposite directions pass each other on a bicycle path.The bicyclist headi
notka56 [123]
Between 9:00 am and 10:45 am, there have been 1 hour and 45 minutes or 1.75 hours have passed. Let x be the speed of the slower cyclist and x+ 5 be the rate of the second cyclist. The given situation is best represented through the equation below,
                                 x(1.75) + (x + 5)(1.75) = 47.25 km
The value of x from the equation is 11. Thus, the two bicyclists' rates are 11 km/h and 16 km/h. 
3 0
3 years ago
Read 2 more answers
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