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grin007 [14]
2 years ago
10

Mike is standing on the roof of a building looking at the roof of the neighboring building that is 15 meters away and 10 meters

below him. He wants to throw a canister horizontally from one building to the other. His maximum throwing speed is 24.5 m/s. On this planet g is % of that on Earth. A. How long does it take the canister to reach the height of the neighboring building? B. How far does the canister travel horizontally in the time you got in part A? Does the canister make it to the other building? C. What is the velocity (magnitude and direction) of the canister after it has gone down 10m?
Physics
1 answer:
amid [387]2 years ago
6 0

Answer:

Part a)

t = 1.65 s

Part b)

x = 40.4 m

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

v = 27.3 m/s

direction of velocity is given as

[tex]\theta = 26.35 degree

Explanation:

Part a)

acceleration due to gravity on this planet is 3/4 times the gravity on earth

So the acceleration due to gravity on this new planet is given as

a = \frac{3}{4}(9.81)

a = 7.36 m/s^2

now the vertical displacement covered by the canister is given as

y = 10 m

now by kinematics we have

y = \frac{1}{2}gt^2

10 = \frac{1}{2}(7.36)t^2

t = 1.65 s

Part b)

Horizontal speed of the canister is given as

v_x = 24.5 m/s

now the distance moved by it

x = v_x t

x = 24.5 (1.65)

x = 40.4 m

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

Final velocity in X direction will remains the same

v_x = 24.5 m/s

final velocity in Y direction

v_y = v_i + at

v_y = 0 + (7.36)(1.65)

v_y = 12.14 m/s

now magnitude of velocity is given as

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

v = \sqrt{24.5^2 + 12.14^2}

v = 27.3 m/s

direction of velocity is given as

\theta = tan^{-1}\frac{v_y}{v_x}

\theta = tan^{-1}\frac{12.14}{24.5}

[tex]\theta = 26.35 degree

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describe what is meant by "a constant change of direction".Identify whether the examples provided show a constant change of dire
liraira [26]

"Constant change of direction" simply means on a curve.

The examples provided don't show a durn thing.

7 0
2 years ago
Determine the magnitude of the force between two 42 m-long parallel wires separated by 0.03 m, both carrying 6.3 A in the same d
xz_007 [3.2K]

Answer:

The magnitude of the force between the two parallel wires is 0.0111 N.

Explanation:

Given;

length of the two parallel wires, L = 42 m

distance between the two wires, r = 0.03 m

current in both wires, I₁, I₂ = 6.3 A

Therefore, the magnitude of the repulsive force between the two parallel wires is given by;

F = \frac{\mu_0 I_1I_2l}{2\pi r}\\\\where;\\\mu_0 \ is \ permeability \ of \ free \ space = 4\pi *10^{-7} \ T.m/A \\\\F = \frac{(4\pi *10^{-7})(6.3)^2(42)}{2\pi (0.03)}\\\\F =   0.0111 \ N

Therefore, the magnitude of the force between the two parallel wires is 0.0111 N.

6 0
3 years ago
What is the driving force for (a) heat transfer, (b) electric current, and (c) fluid flow?
lara31 [8.8K]

Answer:

The driving force for (a) heat transfer is temperature difference. (b) electric current is voltage difference. (c) fluid flow is pressure or hydraulic head difference.

Explanation: (a) The driving force for heat transfer is temperature difference. Heat transfer between two mediums is possible only if the two mediums are at different temperature, the higher the temperature, the higher the heat transfer.

(b) The driving force for electric current is voltage difference. Voltage difference is defined as the potential difference in charge between two points in electrical field. For electric current to occur,the voltage must be high.

(c) The driving force for fluid flow is pressure difference or hydraulic head difference. For fluid to move upward,it requires energy.

3 0
3 years ago
Other quanto
Alex73 [517]

I'm not sure what you were trying to put here

5 0
3 years ago
You're driving your new sports car at 80 mph over the top of a hill that has a radius of curvature of 540 m. What fraction of yo
luda_lava [24]

Answer:

75.84%

Explanation:

We were given Speed of the sports car, v as 80 mph , we can convert to m/s for unit consistency.

v=80mph= 35.76 m/s

The radius of curvature is given as , r = 540 m

✓ the normal weight can be denoted as Wn

✓ the apparent weight of the person can be denoted as Wa

Wn= normal weight= mg

Wa=apparent weight = (mg - mv^2/r)

g= acceleration due to gravity= 9.8m/s^2

The apparent weightand normal weight has a ratio of

Mn/Ma= [mg - mv^2/r]/mg ........eqn(1)

If we simplify eqn(1) we have

Mn/Ma=[g - vr^2/g].............eqn(2)

Then substitute the given values

Mn/Ma=9.8 - [(35.76^2)/540]/ 9.8

=0.758×100%

Mn/Ma=75.84%

Hence, the required fraction is 75.84%

4 0
2 years ago
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