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8_murik_8 [283]
3 years ago
11

Q11) If you were standing at the top of a building and you dropped a rock.

Physics
1 answer:
Dafna1 [17]3 years ago
8 0

Answer:

Part A

The distance travel by the rock is approximately 132.496 m

Part B

The speed when the rock hits the ground is approximately 50.96 m/s

Explanation:

Part A

The question is focused on the kinetics equation of a free falling object

The given parameter is the time it takes the rock to hit the ground, t = 5.2 s

For an object in free fall, we have;

h = 1/2·g·t²

Where;

h = The height from which the object is dropped

g = The acceleration due to gravity ≈ 9.8 m/s²

t = The time taken to travel the distance, h = 5.2 s

∴ h = 1/2 × 9.8 m/s² × (5.2 s)² ≈ 132.496 m

The distance travel by the rock, h ≈ 132.496 m

Part B

The speed, 'v', when the rock hits the ground, is given by the following kinematic equation,

v = g·t

∴ v = 9.8 m/s² × 5.2 s = 50.96 m/s

The speed when the rock hits the ground, v ≈ 50.96 m/s.

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Answer:

Explanation:

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In the second attached image, we can find graphically the sum of each of the vectors.

The resultant vector has a length of 57.36[m], and goes 48° north of east.

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What is the frequency of a photon with an energy of 4. 56 x 10^-19 j
Sauron [17]

The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

<h3>What is a frequency?</h3>

The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.

Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma  = 6.88 \times 10^{14}\ s^{-1}

Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

Learn more about Frequency:

brainly.com/question/5102661

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5 0
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During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 33.5 rad/s. Find the angular displ
zavuch27 [327]

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angular displacement in rev = 483.589 rev

Explanation: mathematically

Angular velocity = angular displacement / time taken.

Angular velocity = 33.5 rad/s, time taken = 90.7s

33.5 = angular displacement /90.7

Angular displacement = 33.5 * 90.7 = 3038.45 rad

But 1 rev =2π

Hence 3038.45 rad to rev is

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Let the displacement function of a particle is x(t)=(20t^2-15t+200). Find the total displacement, instantaneous velocity and ins
irga5000 [103]

Answer:

A.) 39.5 m

B.) 0

C.) 60m/s^2

Explanation:

Given that a displacement function of a particle is x(t)=(20t^2-15t+200).

To Find the total displacement,

Reduce everything by dividing them by 5

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For instantaneous velocity, differentiate x(t). That is,

dy/dt = 60t - 15 ...... (2)

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If dy/dt = 0, then

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60t = 15

t = 15/60

t = 0.25s

Substitutes t in equation (1)

Total displacement will be

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X(t) = 0.25 - 0.75 + 40

Total displacement = 39.5 m

To calculate instantaneous velocity, substitute t into equation (2)

V = 60 (0.25) - 15

V = 0.

and to find instantaneous acceleration, differentiate dv/dt

dv/dt = 60

Therefore, acceleration = 60 m/s^2

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3 years ago
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