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Ivanshal [37]
2 years ago
13

Maria throws an apple vertically upward from a height of 1.3 m with an initial velocity of +2.8 m/s. Will the apple reach a frie

nd in a tree house 5.9 m above the ground? The acceleration due to gravity is 9.81 m/s 21. No, the apple will reach 5.27136 m below the tree house2. Yes, the apple will reach 5.27136 m above the tree house3. No, the apple will reach 1.43117 m below the tree house4. No, the apple will reach 1.5289 m below5. Yes, the apple will reach 1.5289 m above the tree house6. Yes, the apple will reach 1.43117 m above the tree house
Physics
1 answer:
evablogger [386]2 years ago
5 0

Answer:

No, the apple will reach 4.20041 m below the tree house.

Explanation:

t = Time taken

u = Initial velocity = 2.8 m/s

v = Final velocity = 0

s = Displacement

g = Acceleration due to gravity = -9.81 m/s² = a (negative as it is going up)

Equation of motion

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-2.8^2}{2\times -9.81}\\\Rightarrow s=0.39959\ m

The height to which the apple above the point of release will reach is 0.39959 m

From the ground the distance will be 1.3+0.39959 = 1.69959 m

Distance from the tree house = 5.9-1.69959 = 4.20041 m

No, the apple will reach 4.20041 m below the tree house.

The values in the option do not reflect the answer.

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To solve this problem it is necessary to apply Boyle's law in which it is specified that

P_1V_1 =P_2 V_2

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P_1 and V_1 are the initial pressure and volume values

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The final pressure here is the atmosphere, then

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h = 10m

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Pressure at the water is given by,

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P_1 = 1*10^5 +1000*9.8*10 =198000Pa

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V_2 = \frac{P_1V_1}{P_2}

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4 0
2 years ago
A person on a cruise ship is doing laps on the promenade deck. on one portion of the track the person is moving north with a spe
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The resultant motion is given by pithagoras, since the two components (north and east) are perpendicular to each other.
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In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r
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Answer:

W = 506.75 N

Explanation:

tension = 2300 N

Rider is towed at a constant speed means there no net force acting on the rider.

hence taking all the horizontal force and vertical force in consideration.

net horizontal  force:

F cos 30° - T cos 19° = 0

F cos 30° = 2300 × cos 19°

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8 0
2 years ago
A 5⁢kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant.
Umnica [9.8K]

Answer:

The gravitational force exerted on the object is 75 N (answer D)

Explanation:

Hi there!

The gravitational force is calculated as follows:

F = m · g

Where:

F = force of gravity.

m = mass of the object.

g = acceleration due to gravity (unknown).

For a falling object moving in a straight line, its height at a given time can be calculated using the following equation:

y = y0 + v0 · t + 1/2 · a · t²

Where:

y = position at time t.

y0 = initial position.

v0 = initial velocity.

t = time.

g = acceleration due to gravity.

Let´s place the origin of the frame of reference at the point where the object is released so that y0 = 0. Let´s also consider the downward direction as negative.

Then, after 2 seconds, the height of the object will be -30 m:

y = y0 + v0 · t + 1/2 · g · t²

-30 m = 0 m + 0 m/s · 2 s + 1/2 · g · (2 s)²

-30 m = 1/2 · g · 4 s²

-30 m = 2 s ² · g

-30 m/2 s² = g

g = -15 m/s²

Then, the magnitude of the gravitational force will be:

F = m · g

F = 5 kg · 15 m/s²

F = 75 N

The gravitational force exerted on the object is 75 N (answer D)

Have a nice day!

8 0
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