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Minchanka [31]
3 years ago
11

A strong gust of wind blows an apple off a tree. The apple has a mass of 0.25 kg. The gust of wind pushes the apple with a force

of 1.05 N to the right. What is the net force on the apple?
5.5 N
- 2.2 N
1.78 N
2.66 N

Physics
2 answers:
LenKa [72]3 years ago
8 0

Answer:

2.66N

Explanation:

Nesterboy [21]3 years ago
3 0

For this case we must draw the free body diagram of the apple.

It is observed in the attached figure that we have the force of the wind acting to the right, in addition to the force of the weight that goes down.

Mg: It is the force exerted by the weight of the apple (M: It is the mass and g: It is the acceleration of gravity)

F_ {1}: It is the force exerted by the wind

Then, the net force will be given by:

F_ {n} = \sqrt {(Mg) ^ 2 + (F_ {1}) ^ 2}

We have to:

Mg = 0.25kg * 9.8 \frac {m} {s ^ 2} = 2.45N\\F_ {1} = 1.05N

Substituting we have:

F_ {n} = \sqrt {(2.45) ^ 2 + (1.05) ^ 2}\\F_ {n} = \sqrt {6.0025 + 1.1025}\\F_ {n} = \sqrt {7.105}\\F_ {n} = 2.66

Thus, the net force acting on the apple is 2.66N

Answer:

Option D

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We know, speed = Distance / Time
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7 0
3 years ago
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A dart gun consists of a horizontal spring with k = 52 Newtons/m that is compressed 43
gulaghasi [49]

Answer:

299 m/s^2

Explanation:

When a spring is compressed, the force exerted by the spring is given by:

F=kx

where

k is the spring constant

x is the compression of the spring

In this problem we have:

k = 52 N/m is the spring constant

x = 43 cm = 0.43 m is the compression

Therefore, the force exerted by the spring on the dart is

F=(52)(0.43)=22.4 N

Now we can apply Newton' second law of motion to calculate the acceleration of the dart:

F=ma

where

F = 22.4 N is the force exerted on the dart by the spring

m = 75 g = 0.075 kg is the mass of the dart

a is its acceleration

Solving for a,

a=\frac{F}{m}=\frac{22.4}{0.075}=299 m/s^2

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3 years ago
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What is meant by Compression and Rarefaction of a longitudinal wave?
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(20) A rocket is launched vertically. At time t = 0 seconds, the rocket’s engine shuts down. At the time, the rocket has reached
Diano4ka-milaya [45]

Answer:

Explanation:

Given that,

h(t) = -9.8t² / 2 + 125t + 500

for t > 0.

At t = 0, the rocket is at height h = 500m, at a velocity of Vo = 125m/s.

We want to find the maximum height reached by rocket

Using mathematics maxima and minima

let find the turning point when dh/dt = 0

dh/dt = -9.8t + 125

dh / dt = 0 = -9.8t + 125

9.8t = 125

t = 125 / 9.8

t = 12.76s

Let find the turning point to know if this time t = 12.76 is maximum or minimum point

Let find d²h / dt²

d²h / dt² = -9.8

Since, d²h/dt² < 0, then, at t = 12.76s is the maximum points.

Then, the maximum height reached is

h =  -9.8t² / 2 + 125t + 500

h =  -9.8(12.76)² / 2 + 125(12.76) + 500

h = -797.80 + 1595 + 500

h = 1297.2 m

The maximum height reached is 1297.2 m

From the attachment, the maximum height is 1297.2m at t = 12.76sec.

Comment, the result are the same for both the analysis aspect and the graphical aspect.

3 0
3 years ago
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