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3 years ago

11

After a projectile is fired into the air, what is the magnitude of the acceleration

1 answer:

sergiy2304 [10]3 years ago

7 0

Answer: Option A; 9.8 m/s^2

Explanation:

When an object is in the air, and there is no air resistance acting on the object, the only force that will act on the object is the gravitational force (on the vertical axis).

Then, if the only force acting on the object is the gravitational force, the acceleration of the object will be equal to the gravitational acceleration.

We know that the gravitational acceleration is equal to:

g = 9.8m/s^2

Then the acceleration on the vertical axis will be equal to:

a(t) = 9.8m/s^2

The correct option is the first one:

A. 9.8 m/s^2

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When you jump from an elevated position you usually bend your knees upon reaching the ground. By doing this, you make the time o

dsp73

Answer:

c. about 1/10 as great.

Explanation:

While jumping form a certain height when we bend our knees upon reaching the ground such that the time taken to come to complete rest is increased by 10 times then the impact force gets reduced to one-tenth of the initial value when we would not do so.

This is in accordance with the Newton's second law of motion which states that the rate of change in velocity is directly proportional to the force applied on the body.

Mathematically:

$F\propto\frac{d}{dt} (p)$

$\Rightarrow F=\frac{d}{dt} (m.v)$

since mass is constant

$F=m\frac{d}{dt}v$

when $dt=10t$

then,

$F'=m.\frac{v}{10\times t}$

$F'=\frac{1}{10} \times \frac{m.v}{t}$

$F'=\frac{F}{10}$ the body will experience the tenth part of the maximum force.

where:

$\frac{d}{dt} =$ represents the rate of change in dependent quantity with respect to time

$p=$ momentum

$m=$ mass of the person jumping

$v=$ velocity of the body while hitting the ground.

7 0

3 years ago

The figure above shows 4forces 3N, 10N, 3√3N, and 6N acting on a particle P. The resultant of the four forces is.

Brilliant_brown [7]

As you gave no pic I took them on one lined

F_1=3N
F_2=10N
F_3=3root 3 N
F_4=6N

$\\ \sf\longmapsto F_{net}=F_1+F_2\dots$

$\\ \sf\longmapsto F_{net}=3+10+6+3\sqrt{3}$

$\\ \sf\longmapsto F_{net}=19+3(1.732)$

$\\ \sf\longmapsto F_{net}=19+5.196$

$\\ \sf\longmapsto F_{net}=24.196N$

6 0

3 years ago

Can anyone please answer this cq..

ZanzabumX [31]

Hqisgshsjsvg udyfudueuduutuufududuydd

6 0

2 years ago

An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. how high does it rise (v = 0 cm

Aliun [14]

Do you need more info?

5 0

3 years ago

Read 2 more answers

Two forces act on a 6.00- kg object. One of the forces is

Stella [2.4K]

Answer:

Fx = 22N

Explanation:

There are 2 possible scenarios for this problem:

1.- The 10N force is in the same direction of the acceleration. In this case the other force would be:

$F1 - Fx = m*a$ where F1 = 10N, m=6kg, a = 2m/s2

$Fx = F1 - m*a = -2N$ The negative result tells us that this is not possible.

2.- The 10N force is in the opposite direction of the acceleration. In this case the other force would be:

$Fx - F1 = m*a$ $Fx = m*a + F1 = 22N$

3 0

3 years ago