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zubka84 [21]
3 years ago
14

Sandy is whirling a ball attached to a string in a horizontal circle over his head. If Sandy doubles the speed of the ball, what

happens to the tension in the string?
A) It doubles.
B) It quadruples.
C) It is cut in half.
D) It remains constant.
Physics
1 answer:
jeka57 [31]3 years ago
6 0

The tension in the string B) It quadruples.

Explanation:

The ball is in uniform circular motion in a horizontal circle, so the tension in the string is providing the centripetal force that keeps the ball in circular motion. So we can write:

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

where:

T is the tension in the string

m is the mass of the ball

v is the speed of the ball

r is the radius of the circle (the lenght of the string)

In this problem, we are told that the speed of the ball is doubled, so

v' = 2v

Substituting into the previous equation, we find the new tension in the string:

T' = m \frac{(2v)^2}{r}=4(m\frac{v^2}{r})=4T

Therefore, the tension in the string will quadruple.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

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A 1100 kgkg safe is 2.4 mm above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compres
Drupady [299]

Answer:

191.36 N/m

Explanation:

From the question,

The Potential Energy of the safe = Energy of the spring when it was compressed.

mgh = 1/2ke²............... Equation 1

Where m = mass of the safe, g = acceleration due to gravity, h = height of the save above the heavy duty spring , k = spring constant, e = compression

Making k the subject of the equation,

k =2mgh/e²................ Equation 2

Given: m = 1100 kg, h = 2.4 mm = 0.0024 m, e = 0.52 m

Constant: g = 9.8 m/s²

Substitute into equation 2

k = 2(1100)(9.8)(0.0024)/0.52²

k = 51.744/0.2704

k = 191.36 N/m

Hence the spring constant of the heavy-duty spring = 191.36 N/m

3 0
3 years ago
Sea floor spreading could produce A) an ocean B) a mountain range C) a volcano D) a single fault-block mountain
lesya692 [45]
I think it is A. An Ocean
5 0
3 years ago
The engineer of a passenger train traveling at 25.0m/s sights a freight train whose caboose is 200m ahead on the same track. The
Vika [28.1K]

Answer:

No, there won't be a collision.

Explanation:

We will use the constant acceleration formulas to calculate,

v = u + a*t

0 = 25 + (-0.1)*t

t = 250 seconds (the time taken for the passenger train to stop)

v^2 = u^2 + 2*a*s

0 = (25)^2  + 2*(-0.1)*s

s = 3125 m (distance traveled by passenger train to stop)

If the distance traveled by freight train in 250 seconds is less than (3125-200=2925 m) than the collision will occur

Speed*time = distance

Distance = (15)*(250)

Distance = 3750 m

As the distance is way more, there won’t be a collision

5 0
3 years ago
Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250x10-7C (charges)
SIZIF [17.4K]

The electric field at one corner of a square is 1614217 N/C.

Explanation:

The distance between x and y direction diagonals.

As per the given details the distance between diagonals is calculated as

0.5² + 0.5² = c²  =>  c = 0.707 m

Charge to the right:  In x direction

In order to find the electric charge towards x direction

we use e = kq/r² formula

As 'k' is coulomb's constant it's value is 9 x 10^{9} N m²/C²

e = (9 x 10^{9})(250 x 10^{-7}) / (0.5)²

e = 9 x 10^{5} N/C

Charge diagonal:

e = kq/r²

e = [(9 x 10^{9})(250 x 10^{-7}) / (0.707)²] cos 45

e = 225000√2 N/C

X direction sum = 1218198 N/C.

Similarly as shown in x direction the charge is same for y direction also

Charge below:  For y direction

e = kq/r²

e = (9 x 10^{9})(250 x 10^{-7}) / (0.5)²

e = 9 x 10^{5} N/C

Charge diagonal:

e = kq/r²

e = [(9 x 10^{9})(250 x 10^{-7}) / (0.5)²] sin 45

e = 159099 N/C

Y direction sum = 1059099 N/C

Resultant electric field strength:

1218198 ² + 1059099² = e²

e = 1614217 N/C [45 degrees below the horizontal]

4 0
4 years ago
A transformer has two sets of coils, the primary with N1 = 160 turns and the secondary with N2 = 1400 turns. The input rms volta
vovikov84 [41]

To solve the problem it is necessary to apply the concepts related to the voltage in a coil, through the percentage relationship that exists between the voltage and the number of turns it has.

So things our data are given by

N_1 = 160

N_2 = 1400

\Delta V_{1rms} = 62V

PART A) Since it is a system in equilibrium the relationship between the two transformers would be given by

\frac{N_1}{N_2} = \frac{\Delta V_{1rms}}{\Delta V_{2rms}}

So the voltage for transformer 2 would be given by,

\Delta V_{2rms} = \frac{N_2}{N_1} \Delta V_{1rms}

PART B) To express the number value we proceed to replace with the previously given values, that is to say

\Delta V_{2rms} = \frac{N_1}{N_2} \Delta V_{1rms}

\Delta V_{2rms} = \frac{1400}{160} 62V

\Delta V_{2rms} = 1446.66V

7 0
3 years ago
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