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Zinaida [17]
3 years ago
8

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

e ground? (Disregard air resistance. a = -g = -9.81 m/s.)
Physics
1 answer:
Bad White [126]3 years ago
8 0

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.

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What is the angle of the 2nd order bright fringe produced by two slits that are 8.25x10-5m apart if the wavelength of the incide
dexar [7]

In order to calculate the angle, we can use the formula below for a constructive interference (the interference is constructive because the fringe is bright):

d\sin\theta=m\lambda

Where d is the distance between the slits, m is the order of the interference and lambda is the wavelength.

So, using d = 8.25 * 10^-5, m = 2 and lambda = 4.5 * 10^-7, we have:

\begin{gathered} 8.25\cdot10^{-5}\cdot\sin\theta=2\cdot4.5\cdot10^{-7}\\ \\ \sin\theta=\frac{9\cdot10^{-7}}{8.25\cdot10^{-5}}\\ \\ \sin\theta=1.091\cdot10^{-2}\\ \\ \theta=0.625° \end{gathered}

Therefore the correct option is the second one.

5 0
1 year ago
The standard wave format for any wave is
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The standard wave format for any wave is transverse wave
8 0
3 years ago
A cannonball is fired horizontally at the same time a ball being dropped from the same height How do the times it takes them to
ipn [44]

Answer:

The cannonball and the ball will both take the same amount of time before they hit the ground.

Explanation:

For a ball fired horizontally from a given height, there is only a vertical acceleration on it towards the ground. This acceleration is equal to the acceleration due to gravity (g = 9.81 m/s^2). A ball dropped from a height will also only experience the same vertical acceleration downwards which is also equal to g = 9.81 m/s^2.

Therefore both the cannonball and the ball will take the same amount of time to hit the ground if they are released/fired from the same height.

3 0
3 years ago
Assignment 10 Coulombic Equation Practice Directions: Complete the following problems to calculate the electrostatic force that
Tatiana [17]

Answer:

Magnitude of the force between the charges is F = 1.92×10^20N

Explanation:

Given the magnitude of force according to coulombs law

F =K[(q1*q2)/r2]

Where q1 and q2 are the charges

r is the distance between the charges

K is the coulombs constant

Substituting the given values, we have;

F = 8.98×10^9 × 1.5×10^6 × 3.2×10^4/1.5²

F = 43.1×10^19/2.25

F = 19.16×10^19N

F = 1.92×10^20N

8 0
2 years ago
A textbook of mass 2.05 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose d
blsea [12.9K]

Answer:

a. 7.38 N b. 40.87 N c. 0.113 kg-m²

Explanation:

a. Let T be the tension in the cord. For the textbook, T = ma since no other force acts on it and it is an horizontal force, and m = mass = 2.05 kg and a = acceleration. We find the acceleration from s = ut + 1/2at² where u = initial speed = 0 (since it starts from rest),  s = distance moved = 1.30 m and t = time = 0.850 s.

Substituting these values into s,

1.30 m = 0 × 0.850 + 1/2a × 0.850² = 0 + 0.36125a

1.30 = 0.36125a

a = 1.30/0.36125 = 3.6 m/s²

Substituting this into T, we have

T = ma = 2.05 kg × 3.6 m/s² = 7.38 N

b.  Let T be the tension in the cord attached to the book. The book has the only vertical forces acting on it as the tension, T(acting upwards) and its weight mg (acting downwards). So the net force acting on it is

T - mg = ma

T = m(a + g)

substituting a = 3.6 m/s² and g = 9.8 m/s² and m = 3.05 kg

T = 3.05(3.6 + 9.8) = 3.05 × 13.4 = 40.87 N

c. Since the tangential acceleration of the pulley is also the acceleration of the masses, the a = rα where r = radius of pulley = 0.200 m/2 = 0.100 m and α = angular acceleration of the pulley.

α = a/r = 3.6 m/s² ÷ 0.100 m = 36 rad/s²

Now, the torque on the pulley τ = Tr = Iα where I = moment of inertia of pulley about its rotational axis and T = tension in cord attached to book and r = radius of pulley = 0.200 m/2 = 0.100 m

From the equation above, I = Tr/α

Substituting the variables we have

I = 40.87 N × 0.100 m ÷ 36 rad/s² = 0.113 kg-m²

4 0
3 years ago
Read 2 more answers
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