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klio [65]
3 years ago
12

A ball attached to a string is whirled around in a horizontal circle having a radius r. If the radius of the circle is changed t

o 4r and the same centripetal force is applied by the string, the new speed of the ball is which of the following?a. One-quarter the original speed
b. Twice the original speed
c. One-half the original speed
d. Four times the original speed
Physics
1 answer:
LenKa [72]3 years ago
5 0

Answer:

Option D: Four times the original speed.

Explanation:

A centripetal force accelerates a body by changing the direction of the body's  velocity without changing the body's speed.

The speed(v) is therefore constant, thereby making the magnitudes of the of the acceleration and the force constant.

The formula used to calculate the Centripetal force is given below:

F = \frac{mv^2}{r}

where F represents the Centripetal force, m represents the mass of the moving body, v represents the speed or velocity at which the body is moving and r represents the radius.

Making the speed the subject of the formula: v^{2} = \frac{rF}{m} \\

Therefore, when the radius (r) is changed to 4r, i.e r = 4r

speed(v) becomes v^2 = \frac{4rF}{m}

After comparing, the difference between the speeds is Four times the original speed.

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How to solve these two questions? ​
hammer [34]

1) See attached figure

The relationship between charge and current is:

i = \frac{Q}{t}

where

i is the current

Q is the charge

t is the time

Therefore, the current is the rate of change of the charge passing through a given point over time.

This means that for a graph of charge over time, the current is just equal to the slope of the graph.

For the graph in this problem:

- Between t = 0 and t = 2 s, the slope is

\frac{50-0}{2-0}=25 C/s

therefore the current is

i = 25 A

- Between t = 2 s and t = 6 s, the slope is

\frac{-50-(50)}{6-2}=-25 C/s

therefore the current is

i = -25 A

- Between t = 6 s and t = 8 s, the slope is

\frac{0-(-50)}{8-6}=25 C/s

therefore the current is

i = 25 A

The figure attached show these values plotted on a graph.

2) 15 \mu C

The previous equation can be rewritten as

Q = i t

This equation is valid if the current is constant: if the current is not constant, then the total charge is simply equal to the area under a current vs time graph.

Here we have the current vs time graph, so we gave to find the area under it.

The area of the first triangle is:

A_1 = \frac{1}{2}(0.001 s)(0.010 A)=5\cdot 10^{-6} C

While the area of the second square is

A_2 = (0.002 s - 0.001 s)(0.010 A)=1\cdot 10^{-5}C

So, the total area (and the total charge) is

Q=A_1 +A_2 = 5\cdot 10^{-6} + 1\cdot 10^{-5} = 1.5\cdot 10^{-5}C=1.5 \mu C

3 0
3 years ago
A 286-kg motorcycle is accelerating up along a ramp that is inclined 31.6° above the horizontal. The propulsion force pushing th
kkurt [141]

Answer:

The acceleration motorcycle

a = 5.13 m / s²

Explanation:

Now to determine the acceleration of the motorcycle

Use the force to analysis motion

∑ F = m * a

∑ F = E - D - m*g * sin ( β ) = m * a

E = 3168 N

D = 230 N

β = 31.6 °

3168 N - 230 N - 286 kg * 9.8 m / s² * sin ( 31.6° ) = 286 kg * a

Now solve to a'

a = [ 3168 N - 230 N - 286 kg * 9.8  m / s² * sin ( 31.6° ) ] / (286 kg)

a = 5.13 m / s²

7 0
3 years ago
How are mass and inertia related?
aleksandr82 [10.1K]
The greater the mass the greater is inertia.
7 0
3 years ago
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 m/s, and
Leya [2.2K]

Answer:

A) 0.9844 s

B) x2 = 0.4587 m

C) v = 6.657 m/s

Explanation:

We are given;

Height of take off point above pool; x1 = 1.8 m

Initial take off velocity; u = 3 m/s

Final velocity at highest point before free fall; v = 0 m/s

B) To find the highest point above the board her feet reaches means the distance from take off to the top of the motion just before free fall.

Thus, we will be using equation of motion and we have;

v² = u² + 2gs

Now, let s = x2 which will be the distance between take off and the top before free fall.

So;

v² = u² + 2g(x2)

Now,since the motion is against gravity, g will be negative.

Thus;

v² = u² + 2(-9.81)(x2)

Plugging in the relevant values to give;

0² = 3² - (19.62x2)

19.62(x2) = 9

x2 = 9/19.62

x2 = 0.4587 m

A) We want to find how long her feet is in air.. It means we want to find out the time to get to a distance of x1 and also the time to achieve the distance (x1 + x2) on free-fall.

Thus, using equation of motion;

v = u + gt

Again, g = -9.81

Thus;

0 = 3 - 9.81t1

9.81t1 = 3

t1 = 3/9.81

t1 = 0.3058 s

Now, for the time taken to achieve the distance (x1 + x2) on free-fall, we will use the formula;

s = ut + ½gt²

Where s = (x1 + x2) = 1.8 + 0.4587 = 2.2587 m

And now, u = 0 m/s because the start of the free fall is from maximum height with velocity of 0 m/s. Again, g = - 9.81 m/s²

Thus;

2.2587 = 0 - ½(-9.81)(t2)²

2.2587 = 4.905(t2)²

(t2)² = 2.2587/4.905

(t2)² = 0.4605

t2 = √0.4605

t2 = 0.6786 s

Thus, total time of feet in air = t1 + t2 = 0.3058 + 0.6786 = 0.9844 s

C) Velocity when feet hit the water would be given by;

v = u + gt

Where u = 0 m/s and t = t2 = 0.6786

Since it's in direction of gravity, g = 9.81 m/s

v = 0 + (0.6786 × 9.81)

v = 6.657 m/s

4 0
3 years ago
In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co
labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m

a)

The final velocity of cart 1 after collision is given as:

v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\  Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s

The final velocity of cart 2 after collision is given as:

v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\  Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s

b) Using the law of conservation of energy:

\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm

7 0
3 years ago
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