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saw5 [17]
4 years ago
5

A bar extends perpendicularly from a vertical wall. The length of the bar is 2 m, and its mass is 10 kg. The free end of the rod

is attached to a point on the wall by a light cable, which makes an angle of 30° with the bar. Find the tension in the cable.
Physics
1 answer:
Simora [160]4 years ago
6 0

Answer:

T = 98.1 N

Explanation:

Given:

- mass of bar m = 10 kg

- Length of the bar L = 2 m

- Angle between Cable and wall Q = 30 degres

Find:

- Find the tension in the cable.

Solution:

- Take moments about intersection of bar and wall, to be zero (static equilibrium)

                                    (M)_a = 0

                                    T*sin( 30 )*L - m*g*L/2 = 0

                                    T*sin(30) - m*g / 2 = 0

                                    T = m*g / 2*sin(30)

                                    T = 10*9.81 / 2*sin(30)

                                    T = 98.1 N

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kupik [55]

Answer:

<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2>v_f = 35.3 m/s</h2>

Explanation:

Part a)

As we know that drag force is given as

F = \frac{C_d \rho A v^2}{2}

C_d = 0.35

A = \frac{\pi d^2}{4}

A = \frac{\pi(0.074)^2}{4}

A = 4.3 \times 10^{-3} m^2

v = 40.2 m/s

so we have

F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}

F = 1.46 N

So acceleration of the ball is

a = \frac{F}{m}

a = \frac{1.46}{0.145}

a = 10.1 m/s^2

Part B)

As per kinematics we know that

v_f^2 - v_i^2 = 2 a d

v_f^2 - 40.2^2 = 2(-10.1)(18.4)

v_f = 35.3 m/s

4 0
3 years ago
Imagine a sunny day at the pool. The sun is out and you are thinking about how the light travels from the sun and then hits the
Dovator [93]

Answer:

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Explanation:

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3 years ago
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A tug boat pulls a ship with a constant net horizontal force of 5.00•10*3 N and causes the ship to move through a harbor. How mu
Murljashka [212]

The work done on the ship is 1.5\cdot 10^7 J

Explanation:

The work done by a force on an object is given by:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

In this problem, we have:

F=5.00\cdot 10^3 N (force acting on the ship)

d = 3.00 km = 3000 m (displacement of the ship)

\theta=0^{\circ} (because the force is horizontal, and the displacement is horizontal as well)

Therefore, the work done on the ship is

W=(5.00\cdot 10^3)(3000)(cos 0^{\circ})=1.5\cdot 10^7 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

8 0
3 years ago
What is the name of the space telescope in low earth orbit that was launched in 1990?
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The Hubble Space Telescope is a joint ESA/NASA project and was launched in 1990 by the Space Shuttle mission STS-31 into a low-Earth orbit 569 km above the ground. During its lifetime Hubble has become one of the most important science projects ever. Hope this helps! ~ Autumn :)

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3 years ago
Remember to include your data, equation, and work when solving this problem.
andrezito [222]

Answer:

F = 0.00156[N]

Explanation:

We can solve this problem by using Newton's proposed universal gravitation law.

F=G*\frac{m_{1} *m_{2} }{r^{2} } \\

Where:

F = gravitational force between the moon and Ellen; units [Newtos] or [N]

G = universal gravitational constant = 6.67 * 10^-11 [N^2*m^2/(kg^2)]

m1= Ellen's mass [kg]

m2= Moon's mass [kg]

r = distance from the moon to the earth [meters] or [m].

Data:

G = 6.67 * 10^-11 [N^2*m^2/(kg^2)]

m1 = 47 [kg]

m2 = 7.35 * 10^22 [kg]

r = 3.84 * 10^8 [m]

F=6.67*10^{-11} * \frac{47*7.35*10^{22} }{(3.84*10^8)^{2} }\\ F= 0.00156 [N]

This force is very small compare with the force exerted by the earth to Ellen's body. That is the reason that her body does not float away.

6 0
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