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vlabodo [156]
2 years ago
8

What would the reaction force of a person be that stands completely still with a mass of 100 kg?

Physics
1 answer:
Anastaziya [24]2 years ago
4 0

-981N would be the reaction force of a person be that stands completely still with a mass of 100 kg.

An opposing force to an action force is referred to as a reaction force. The response force that results from surface engagement and adhesion while sliding is known as friction. Usually, the activities of applied forces result in reaction forces and reaction moments.

As,

F = ma

where, F = force

           m = mass of object

           a = acceleration due to gravity

Plugging in the values we get,

F = 100kg × 9.81m/s

F = 981 N

Since, reaction force acts opposite to action force so the correct answer is -981N.

Learn more about force here;

brainly.com/question/13191643

#SPJ4

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a stone is thrown vertically upward with a velocity of 29.4m/s from the top of tower 34.3m high.calculate the total time taken b
Alexus [3.1K]

Answer:

7 s

Explanation:

We'll begin by calculating the maximum height reached by the stone from the point of projection. This can be obtained as follow:

Initial velocity (u) = 29.4 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) = 0 m/s (at maximum height)

Maximum height (h) =?

v² = u² – 2gh (since the stone is going against gravity)

0² = 29.4² – (2 × 9.8 × h)

0 = 864.36 – 19.6h

Collect like terms

0 – 864.36 = –19.6h

–864.36 = –19.6h

Divide both side by –19.6

h = –864.36 / –19.6

h = 44.1 m

Next, we shall determine the time taken to reach the maximum height from the point of projection. This can be obtained as follow:

Initial velocity (u) = 29.4 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) = 0 m/s (at maximum height)

Time (t₁) take to reach the maximum height =?

v = u – gt (since the stone is going against gravity)

0 = 29.4 – (9.8 × t₁)

0 = 29.4 – 9.8t₁

Collect like terms

0 – 29.4 = – 9.8t₁

–29.4 = –9.8t₁

Divide both side by –9.8

t₁ = –29.4 / –9.8

t₁ = 3 s

Next, we shall determine the time taken for the stone to reach the ground from the maximum height reached by the stone. This can be obtained as follow:

Maximum height (H) from the ground = 34.3 + 44.1 = 78.4 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t₂) taken for the stone to reach the ground from the maximum height reached by the stone =?

H = ½gt₂²

78.4 = ½ × 9.8 × t₂²

78.4 = 4.9 × t₂²

Divide both side by 4.9

t₂² = 78.4 / 4.9

t₂² = 16

Take the square root of both side

t₂ = √16

t₂ = 4 s

Finally, we shall determine the total time taken for the stone to reach the ground. This can be obtained as follow:

Time (t₁) take to reach the maximum height = 3 s

Time (t₂) taken to reach the ground from the maximum height reached by the stone = 4s

Total time (T) taken to reach the ground =?

T = t₁ + t₂

T = 3 + 4

T = 7 s

Thus, it took the stone 7 s to reach the ground.

6 0
3 years ago
What is the mass of a man who accelerates 4 m/s2 under the action of a 200 N net force?
Over [174]

Answer:

\huge  \boxed{ \boxed{50 \:   kg }}

Explanation:

The mass of the man can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{200}{4}  \\

We have the final answer as

<h3>50 kg</h3>

Hope this helps you

7 0
3 years ago
Cosine law can be applied when .....................
jarptica [38.1K]

Answer:

<em>Two</em><em> </em><em>sides</em><em> </em><em>and</em><em> </em><em>angle</em><em> </em><em>between</em><em> </em><em>them</em><em> </em><em>is</em><em> </em><em>given</em><em> </em>

3 0
2 years ago
Why is a simple cell importable​
densk [106]
Simple cellll importable
Ikr
Answer:


Explanation:



8 0
3 years ago
An airplane accelerates from a speed of 88m/s to a speed of 132 m/s during a 15 second time interval. How far did the airplane t
Gelneren [198K]

Answer:

1650\:\mathrm{m}

Explanation:

We can use the following kinematics equations to solve this problem:

v_f=v_i+at,\\{v_f}^2={v_i}^2+2a\Delta x.

Using the first one to solve for acceleration:

132=88+a(15),\\15a=44,\\a=\frac{44}{15}=2.9\bar{3}\:\mathrm{m/s^2}.

Now we can use the second equation to solve for the distance travelled by the airplane:

132^2=88^2+2\cdot2.9\bar{3}\cdot \Delta x,\\\Delta x= \frac{9680}{2\cdot2.9\bar{3}},\\\Delta x =\fbox{$ 1650\:\mathrm{m}$}(three significant figures).

6 0
3 years ago
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