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3 years ago

You are standing in a moving bus, facing forward, and you suddenly fall

Physics

2 answers:

RideAnS [48]3 years ago

7 0

The velocity decreases

natulia [17]3 years ago

4 0

A) velocity decreased.

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A 15.0-kg object sitting at rest is struck elastically in a head-on collision with a 10.5-kg object initially moving at 3.0 m/s.

DIA [1.3K]

Answer:

The final velocity of the 15.0-kg object after the collision is 2.47 m/s in forward direction.

Explanation:

Given;

mass of the object, m₁ = 15 kg

initial velocity of this object, u₁ = 0

mass of the second object, m₂ = 10.5 kg

initial velocity of this object, u₂ = 3.0 m/s

let the final velocity of the first object = v₁

also, let the final velocity of the second object = v₂

Apply the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(15 x 0) + (10.5 x 3) = 15v₁ + 10.5v₂

31.5 = 15v₁ + 10.5v₂ ----- (1)

One directional velocity;

u₁ + v₁ = u₂ + v₂

0 + v₁ = 3 + v₂

v₂ = v₁ - 3 ------(2)

Substitute (2) into (1);

31.5 = 15v₁ + 10.5v₂

31.5 = 15v₁ + 10.5(v₁ - 3)

31.5 = 15v₁ + 10.5v₁ - 31.5

63 = 25.5v₁

v₁ = 63 / 25.5

v₁ = 2.47 m/s

Therefore, the final velocity of the 15.0-kg object after the collision is 2.47 m/s in forward direction.

7 0

3 years ago

Can a body having larger mass be hotter than another of smaller mass.expalain

Rom4ik [11]

Answer:

it would make sense because a larger body could produce more body heat.

8 0

3 years ago

a body of a mass 1kg suspended from a spring is found to stretch the spring by 10cm. (A) what is the spring constant? what is pe

Nikitich [7]

Answer:

(A) The spring constant is <u>98 N/m.</u>

(B) The period of oscillation is <u>0.635 s.</u>

Explanation:

Given:

Mass of the body is, $m=1\ kg$

Extension length of the spring is, $x=10\ cm=0.1\ m$

Now, let 'k' be the spring constant.

The force acting on the body is due to gravity only and is equal to its weight.

So, weight of the body is given as:

$F_g=mg\\F_g=1\times 9.8\\F_g=9.8\ N$

Now, we know that for a spring-mass system, the net force acting on the body is equal to the product of the spring constant and extension length. Therefore,

$F_g=kx\\9.8=k(0.1)\\k=\frac{9.8}{0.1}=98\ N/m$

Hence, the spring constant is 98 N/m.

(B)

Period of oscillation of a body of spring-mass system in SHM is given as:

$T=2\pi\sqrt{\frac{m}{k}}$

Plug in the given values and solve for period 'T'. This gives,

$T=2\pi\sqrt{\frac{1}{98}}\\T=2\pi\times 0.101\\T=0.635\ s$

Therefore, the period of oscillation is 0.635 s.

5 0

4 years ago

Pls help i have 5 minutes

anyanavicka [17]

Number order:

5, 4, 3, 2, 1

7 0

4 years ago

Read 2 more answers

395,000 meters in 9000

marta [7]

Answer:

43.88 meters per second

Explanation:

The computation of the speed is shown below:

As we know that

$Speed = \frac{Distance}{time}$

where,

Distance is 395,000 meters

Time is 9,000 seconds

Now placing these values to the formula

So, the speed is

$= \frac{395,000}{9,000}$

= 43.88 meters per second

As speed shows the relation between the distance and time and the same is to be considered i.e by applying the formula

6 0

4 years ago