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2 years ago

7

A student shakes a rope such that 20 complete vibrations are made in 4.00 seconds. Determine the vibrational frequency of the ro

pe, along with the corresponding unit.

1 answer:

fgiga [73]2 years ago

8 0

Answer:

The vibrational frequency of the rope is 5 Hz.

Explanation:

Given;

number of complete oscillation of the rope, n = 20

time taken to make the oscillations, t = 4.00 s

The vibrational frequency of the rope is calculated as follows;

$Frequency = \frac{number \ of \ complete \ vibrations}{time \ taken} \\\\Frequency = \frac{20 }{4 \ s} \\\\Frequency = 5 \ Hz$

Therefore, the vibrational frequency of the rope is 5 Hz.

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A proton, initially traveling in the +x-direction with a speed of 5.05×10^5 m/s , enters a uniform electric field directed verti

Korvikt [17]

Answer:

The strength of the electric field is $1.35\times10^{4}\ N/C$ .

Explanation:

Given that,

Speed $v= 5.05\times10^{5}\ m/s$

Time $t= 3.90\times10^{-7}\ s$

Angle = 45°

We need to calculate the acceleration

Using equation of motion

$v = u+at$

$5.05\times10^{5}=0+a\times3.90\times10^{-7}$

$a =\dfrac{5.05\times10^{5}}{3.90\times10^{-7}}$

$a=1.29\times10^{12}\ m/s^2$

We need to calculate the strength of the electric field

Using relation of newton's second law and electric force

$F= ma=qE$

$ma = qE$

$E=\dfrac{ma}{q}$

Put the value into the formula

$E=\dfrac{1.67\times10^{-27}\times1.29\times10^{12}}{1.6\times10^{-19}}$

$E=1.35\times10^{4}\ N/C$

Hence, The strength of the electric field is $1.35\times10^{4}\ N/C$ .

3 0

2 years ago

A boy weighs 40 kilograms. He runs at a velocity of 4 meters per second north. Which is his momentum?

Artist 52 [7]

The formula for momentum is mass times velocity. Simply, we just multiply the given values:
p = mv
p = 40 kg x 4 m/s
p = 160 kg m/s

Other units for momentum is N s.
p = 160 N s

4 0

2 years ago

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by

Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, $\mu = 0.2$

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, $f_{y} = 40sin \theta$

$\sum f(y) = 0$

$N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N$

$\sum f(x) = 0$

$40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}$

$v^{2} = u^{2} + 2as\\u = 0 m/s\\v^{2} = 2 * 0.731 * 5\\v^{2} = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s$

Power, $P = Fvcos \theta$

$P = 40 *2.704 cos60\\P = 54.074 W$

7 0

3 years ago

PLEASE HELP I NEED THE CORRECT ANSWER TODAY

Natalija [7]

Answer:

The most common oxidation numbers for a given element

5 0

2 years ago

A coin released at rest from the top of a tower hits the ground after falling 5.7 s. What is the speed of the coin as it hits th

Bad White [126]

Given parameters:

Initial velocity of Coin = 0m/s

Time taken before coin hits ground = 5.7s

Unknown:

Final velocity of the coin = ?

Velocity is displacement with time. To solve this problem, we have to apply one of the equations of motion.

The fitting one of them here is shown below;

V = U + gt

where;

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

t is the time taken

Here we use positive value of acceleration due to gravity because the coin is falling with the effect of acceleration and not against it.

Now input the parameters and solve;

V = 0 + 9.81 x 5.7

V = 55.917m/s

Therefore, the final velocity is 55.917m/s.

8 0

2 years ago