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Cerrena [4.2K]
3 years ago
14

A ball with a weight of 2 N is attached to the end of a cord of length 2m . The ball is whirl in a vertical circle counterclockw

ise. The tension in the cord at the top of the circle is 7N, and at the bottom it is 15 N. (The speed of the ball is not the same in all these points).
a) Three students discuss the net forces on the ball at the top:
Deante: The net force on the ball at the top position is 7 N since the net force is the same as the tension.
Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.
Fiona: No, you are both wrong. You need to figure out the centripetal force (mv^2/r) and include it in the net force.

Which, if any, of these students do you agree with?

Deante: _____
Edgar: _____
Fiona: _____
None of them: _____
Physics
1 answer:
ZanzabumX [31]3 years ago
8 0

Answer:

we agree with

Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.

Explanation:

Weight of the ball is given as

W = 2N

so we have

m = \frac{W}{g}

m = 0.204 kg

now tension force at the top is given as

T_{top} = 7 N

T_{bottom} = 15 N

Now at the top position by force equation we can say that ball will have two downwards forces

1) Tension force

2) Weight of the ball

so net force on the ball is given as

F_{net} = T + W

F_{net} = 7 + 2 = 9 N

So we agree with

Edgar: The net force on the ball at the top position is 9 N. Both the tension and the weight are acting downward so you have to add them.

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kirill115 [55]
The observation point on Earth and the two stars form a triangle. The two sides of the triangle are 23.3 ly and 34.76 ly and their included angle is 76.04°. We can use the cos rule to find the third side, which is the distance between the two stars.
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c² = (23.3)² + (34.76)² - 2(23.3)(34.76)Cos(76.04)
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4 0
3 years ago
A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves o
aniked [119]

Correct question is;

A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves offers a damping force numerically equal to √2 times the instantaneous velocity. Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 7 ft/s. (Use g = 32 ft/s²)

Answer:

x(t) = 7te^(-2t√2)

Explanation:

We are given;

Weight; W = 8 lbs

mass; m = W/g

g = 32 ft/s²

Thus;

m = 8/32

m = ¼ slugs

From Newton's second law we can write the equation as;

m(d²x/dt²) = -kx - β(dx/dt)

Rearranging this, we have;

(d²x/dt²) + (β/m)(dx/dt) + (k/m)x = 0

Where;

β is damping constant = √2

k is spring constant = W/s

Where s = 8ft - 4ft = 4ft

k = 8/4

k = 2

Thus,we now have;

(d²x/dt²) + (√2/(¼))(dx/dt) + (2/(¼))x = 0

>> (d²x/dt²) + (4√2)dx/dt + 8x = 0

The auxiliary equation of this is;

m² + (4√2)m + 8 = 0

Using quadratic formula, we have;

m1 = m2 = -2√2

The general solution will be gotten from;

x_t = c1•e^(mt) + c2•t•e^(mt)

Plugging in the relevant values gives;

x_t = c1•e^(mt) + c2•t•e^(mt)

At initial condition of t = 0, x_t = 0 and thus; c1 = 0

Also at initial condition of t = 0, x'(0) = 7 and thus;

Since c1 = 0, then c2 = 7

Thus,equation of motion is;

x(t) = 7te^(-2t√2)

8 0
3 years ago
What happens when a sound wave passes from oil into air? A. the sound wave slows down and bounces back B. the sound wave slows d
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Oil is optically denser than water. When sound/light goes from optically denser medium to optically rarer medium, their velocity increase and they moves away for normal.

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The sound wave speeds up and bends

\Large{ \underline{ \boxed{ \pink{ \bf{Option \: (D)}}}}}

As, In optics we learnt that light undergoes refraction when travels from medium of different densities. Similarly, Sound also follows the law of refraction.

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<u>━━━━━━━━━━━━━━━━━━━━</u>

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3 years ago
A vector has an x component of 25.0 units and a y
sammy [17]
Here is the answer to your question

8 0
3 years ago
Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.3 m/s parallel to the gro
pashok25 [27]

Answer:

Explanation:

cSep 20, 2010

well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.

Anonymous

Sep 20, 2010

First you need to solve for time by using

d=(1/2)(a)(t^2)+(vi)t

1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s

t=.45 sec

Then you find the horizontal distance traveled by using

v=d/t

1.3m/s=d/.54sec

d=.585m

Then you need to find the time of player B by using

d=(1/2)(a)(t^2)+(vi)t

1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0

t=.61 sec

Finally to find player Bs initial horizontal velocity you use the horizontal equation

v=d/t

v=.585m/.61 sec

so v=.959m/s

5 0
3 years ago
Read 2 more answers
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