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Sveta_85 [38]
3 years ago
11

A m=7.2 gram object is accelerated at a rate of a=2.9 m/s^2. What force (in millinewtons) does the object experience? No need to

add the unit (already given).
Physics
1 answer:
yulyashka [42]3 years ago
8 0

Answer:

20.88 mN

Explanation:

given data:

mass of object = 7.2 gm

acceleration of object = 2.9 m/s2

we know that force is given as

F = ma

where m is mass of object and a is acceleration of moving object.

putting all value to get required force

    = 7.2*10^{-3}\ kg *2.9 m/s2

   = 20.88*10^{-3} N

force in milli newton is

  = 20.88*10^{-3} * 1000 = 20.88 mN

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A person fires a 38 gram bullet straight up into the air. It rises, then falls straight back down, striking the ground with a sp
Y_Kistochka [10]

Answer:

Force exerted = 25.41 kN

Explanation:

We have equation of motion

      v² = u²+2as

u = 345 m/s, s = 8.9 cm = 0.089 m, v = 0 m/s

     0² = 345²+2 x a x 0.089

       a = -668679.78 m/s²

Force exerted = Mass x Acceleration

Mass of bullet = 38 g = 0.038 kg

Acceleration = 668679.78 m/s²

Force exerted = 25409.83 N = 25.41 kN

4 0
3 years ago
Normally, jet engines push air out the back of the engine, resulting in forward thrust, but commercial aircraft often have thrus
ANEK [815]

Answer:

When the ejected air is moving in the downward direction then the thrust force acts in the upward direction, due to reversal thrust, the jets can take off vertically without needing a runway this way.

Explanation:

Newton’s third law motion states that for every action there will be an equal and opposite reaction.

Thrust reversal is also known as reverse thrust. It acts opposite to the motion of the aircraft by providing the deceleration.

Commercial aircraft moves the ejected air in the forward direction means that the thrust will acts opposite to the motion of the aircraft that is backward direction due to thrust reversal. This thrust force might be used to decelerate the craft.

Uses of thrust reversal in practice:

When the ejected air is moving forward direction then the thrust force moving backward direction due to reversal thrust the speed of the craft slows down.

When the ejected air is moving in the downward direction then the thrust force acts in the upward direction, due to reversal thrust, the jets can take off vertically without needing a runway this way.

6 0
3 years ago
Sylvia and Jadon now want to work a problem. Imagine a puck of mass 0.5 kg moving in a circular simulation. Suppose that the ten
leva [86]

Answer:

Tangential speed, v = 2.64 m/s

Explanation:

Given that,

Mass of the puck, m = 0.5 kg

Tension acting in the string, T = 3.5 N

Radius of the circular path, r = 1 m

To find,

The tangential speed of the puck.

Solution,

The centripetal force acting in the string is balanced by the tangential speed of the puck. The expression for the centripetal force is given by :

F=\dfrac{mv^2}{r}

v=\sqrt{\dfrac{Fr}{m}}

v=\sqrt{\dfrac{3.5\ N\times 1\ m}{0.5\ kg}}

v = 2.64 m/s

Therefore, the tangential speed of the puck is 2.64 m/s.

3 0
3 years ago
Oliver weighs 600. N. He climbs a flight of stairs that is 3.0 meters tall in 4.0 seconds.
BlackZzzverrR [31]
L=G·d
L=600N·3m=1800J
P=L/Δt=1800J/4s=450W
8 0
3 years ago
A book rest on a table which it's face having a sides 30cm by 25cm . if it exerts apressure of 200pa then determine the mass of
Simora [160]

Answer:

Approximately 1.5\; \rm kg. (Assuming that this table is level, and that the gravitational field strength is g = 9.8\; \rm N \cdot kg^{-1}.

Explanation:

Convert the dimensions of this book to standard units:

\displaystyle 30\; \rm cm = 30\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm}  = 0.30\; \rm m.

\displaystyle 25\; \rm cm = 25\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm}  = 0.25\; \rm m.

Calculate the surface area of this book:

0.30\; \rm m  \times 0.25\; \rm m = 0.075\; \rm m^{2}.

Pressure is the ratio between normal force and the area over which this force is applied.

\displaystyle \text{Pressure} = \frac{\text{normal Force}}{\text{contact Area}}.

Equivalently:

(\text{normal Force}) = \text{Pressure} \cdot (\text{contact Area}).

In this question, \text{Pressure} = 200\; \rm Pa = 200\; \rm N \cdot m^{-2}.

It was found that (\text{contact Area}) = 0.075\; \rm m^{2} (assuming that the entire side of this book is in contact with the table.

Hence:

\begin{aligned}& (\text{normal Force}) \\ &= \text{Pressure} \cdot (\text{contact Area}) \\ &= 200\; \rm N \cdot m^{-2} \times 0.075\; \rm m^{2} \\ &= 15\; \rm N \end{aligned}.

If that the table is level, this normal force would be equal to the weight of this book:

\text{weight} = 15\; \rm N.

Assuming that the gravitational field strength is g = 9.8\; \rm N \cdot kg^{-1}. The mass of this book would be:

\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{15\; \rm N}{9.8\; \rm N \cdot kg^{-1}}\approx 1.5\; \rm kg\end{aligned}.

4 0
2 years ago
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