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3 years ago

when the mass of an object is increased it would decrease its acceleration in the force is left alone

Physics

1 answer:

mojhsa [17]3 years ago

4 0

Answer:

See the explanation below.

Explanation:

This analysis can be easily deduced by means of Newton's second law which tells us that the sum of the forces or the total force on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = total force [N]

m = mass [kg]

a = acceleration [m/s²]

We must clear the acceleration value.

$a=\frac{F}{m}$

We see that the term of the mass is in the denominator, so that if the value of the mass is increased the acceleration decreases, since they are inversely proportional.

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Yuki888 [10]

I think the answer is -2.9m/s2.

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3 years ago

A balloon contains 0.04m3 of air at a pressure of 1.20 x 105Pa. Calculate the pressure required to reduce its volume to 0.025m3

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Answer:

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3 years ago

Two particles each have the same mass but particle #1 has four times the charge of particle #2. Particle #1 is accelerated from

marin [14]

Answer:

v_2 = 2*v

Explanation:

Given:

- Mass of both charges = m

- Charge 1 = Q_1

- Speed of particle 1 = v

- Charge 2 = 4*Q_1

- Potential difference p.d = 10 V

Find:

What speed does particle #2 attain?

Solution:

- The force on a charged particle in an electric field is given by:

F = Q*V / r

Where, r is the distance from one end to another.

- The Net force acting on a charge accelerates it according to the Newton's second equation of motion:

F_net = m*a

- Equate the two expressions:

a = Q*V / m*r

- The speed of the particle in an electric field is given by third kinetic equation of motion.

v_f^2 - v_i^2 = 2*a*r

Where, v_f is the final velocity,

v_i is the initial velocity = 0

v_f^2 - 0 = 2*a*r

Substitute the expression for acceleration in equation of motion:

v_f^2 = 2*(Q*V / m*r)*r

v_f^2 = 2*Q*V / m

v_f = sqrt (2*Q*V / m)

- The velocity of first particle is v:

v = sqrt (20*Q / m)

- The velocity of second particle Q = 4Q

v_2 = sqrt (20*4*Q / m)

v_2 = 2*sqrt (20*Q / m)

v_2 = 2*v

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3 years ago

A 2 kg object traveling at 5 m s on a frictionless horizontal surface collides head-on with and sticks to a 3 kg object initiall

Svetlanka [38]

Answer: (d)

Explanation:

Given

Mass of object $m=2\ kg$

Speed of object $u=5\ m/s$

Mass of object at rest $M=3\ kg$

Suppose after collision, speed is v

conserving momentum

$\Rightarrow mu+0=(m+M)v\\\\\Rightarrow v=\dfrac{2\times 5}{2+3}\\\\\Rightarrow v=2\ m/s$

Initial kinetic energy

$k_1=\dfrac{1}{2}\times 2\times 5^2\\\\k_1=25\ J$

Final kinetic energy

$k_2=\dfrac{1}{2}\times (2+3)\times 2^2\\\\k_2=10\ J$

So, it is clear there is decrease in kinetic energy . Thus, energy decreases and velocity becomes 2 m/s.

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3 years ago

A hiker walked a hiking trail according to the chart below. What is a possible explanation of the hiker's movement?

joja [24]

<span>a.The hiker had an easy, level trail from 11:00-12:00 and was able to travel the fastest during that time period.---> may be because this was indeed fastest stage

b.The hiker got tired and walked the slowest from 1:00-2:00.---> no, because this was not the slowest stage

c.The hiker stopped for lunch from 11:00-12:00 and that slowed him down.---> no because this was the fastest stage

d.The hiker ended up in the same place that he started.---> no, because the hiker walked more toward east than toward west and more toward south than toward north.

Answer: option a) </span>

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3 years ago