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Sonbull [250]
3 years ago
14

The Grand Canyon is 1800 meters deep at its deepest point. A rock is dropped from the rim above this point. Write the height of

the rock as a function of the time in seconds. How long will it take the rock to hit the canyon floor?
Physics
1 answer:
snow_tiger [21]3 years ago
3 0

Answer:

Explanation:

Given

height of grand canyon is h=1800\ m

Rock is dropped i.e. initial velocity is zero

u=0

Using Equation of motion

h=ut+\frac{1}{2}at^2

h=height

u=initial velocity

a=acceleration

t=time

here a=acceleration due to gravity

1800=0+\frac{1}{2}\times 9.8\times t^2

t^2=\frac{2\times 1800}{9.8}

t=19.166\ s

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A student examines a 20-meter long rectangular stream channel and takes the following measurements: width of stream = 4 meters,
Scrat [10]

Answer:

The discharge of the stream at this location is 40 cubic meters per second.

Explanation:

The discharge is the volume flow rate of the water in the stream. For this purpose we can use the following formula:

Discharge = Volume Flow Rate = (Cross-Sectional Area)(Velocity of Stream)

Volume Flow Rate = (Width of Stream)(Depth of Stream)(Velocity of Stream)

Volume Flow Rate = (4 meters)(2 meters)(5 meters per second)

<u>Volume Flow Rate = 40 cubic meters per second</u>

Therefore, the discharge of the stream at this location is found to be <u>40 cubic meters per second</u>

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3 years ago
Please help me
Grace [21]

Answer:

a. 60 N*s

b. 60 (kg*m)/s

c. 3 m/s

Explanation:

Givens:

m = 20 kg

v_i = 0 m/s

t = 10 s

F = 6 N

a) Impulse:

I = F*t

I = 6 N*10 s

I = 60 N*s

b) Momentum:

p = v*m

F = m(a)

a = F/m

a = 6 N/20 kg

a = 0.3m/s^2

a = (v_f -v_i)/t

v_f = (0.3 m/s^2)*10 s

v_f = 3.0 m/s

p = 3 m/s*20 kg

p = 60 (kg*m)/s

c. Final velocity

a = (v_f -v_i)/t

v_f = (0.3 m/s^2)*10 s

v_f = 3.0 m/s

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What items can be classified as matter?
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A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th
son4ous [18]

Answer:

Part a)

v = 7.57 km/h

Part b)

\theta = 67.5 degreeNorth of East

Explanation:

Speed of train towards East = 60 km/h

displacement towards East is given as

d_1 = 40 km

now it turns towards 50 degree East of North

so its distance is given as

d_2 = 20 km(sin50 \hat i + cos50\hat j)

d_2 = 15.3 \hat i + 12.8 \hat j

then finally it moves towards west for 50 min

d_3 = -50 \hat i

Now the total displacement of the train is given as

d = d_1 + d_2 + d_3

d = (40 + 15.3 - 50)\hat i + 12.8 \hat j

d = 5.3\hat i + 12.8 \hat j

now total time duration of the motion is given as

T = 40 min + 20 min + 50 min

T = 1.83 h

now average velocity is given as

v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}

v_{avg} = 2.89\hat i + 6.99\hat j

Part a)

magnitude of the average velocity is given as

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{2.89^2 + 6.99^2}

v = 7.57 km/h

Part b)

Direction of the velocity is given as

tan\theta = \frac{v_y}{v_x}

tan\theta = \frac{6.99}{2.89}

\theta = 67.5 degreeNorth of East

6 0
4 years ago
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