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1 year ago

What object requires less force to overcome inertia B.80 kg piece of furniture or A.1000 kg vehicle

Physics

1 answer:

Tju [1.3M]1 year ago

8 0

Less force will be necessary to overcome inertia for the 80 kg piece of furniture.

Force is a factor that has the power to alter an object's motion. A massed object's velocity can be changed or accelerated by a force. A push or a pull is a straightforward method to explain forces.

The term "moment of inertia" refers to the quantity that describes how a body resists angular acceleration and is calculated by multiplying each particle's mass by its square of the distance from the rotational axis.

I = mr², where m is the mass of the object and r is the distance to the rotation axis.

Therefore, The inertia is directly proportional to the mass of the object.

So, as the mass increases the inertia increases.

Therefore, 80 kg piece of furniture will require less force to overcome inertia.

Learn more about inertia here:

brainly.com/question/14460640

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Answer:

8.162

Explanation:

W=mg

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= 8.162 N

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3 years ago

Calculate the mass (in g) of 346 cm³ of polythene. The density of polythene is 0.95 g/cm³. Give your answer to 2 decimal places.

IceJOKER [234]

Answer:

M = 328.70g

Explanation:

From the given values:

V = 346 cm³

M of 1 cm³ of Polythene = 0.95g or 95/100g

Solve:

M = (95×346)

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2 years ago

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0

2 years ago

A 3.50 cm tall object is held 24.8 cm from a lens of focal length 16.0 cm. What is the image height?

Mandarinka [93]

Answer:

Explanation:

Using the formula to first get the image distance

1/f = 1/u+1/v

f = focal length of the lens

u = object distance

v = image distance

Given f = 16.0 cm, u = 24.8 cm

1/v = 1/16 - 1/24.8

1/v = 0.0625-0.04032

1/v = 0.02218

v = 1/0.02218

v = 45.09 cm

To get the image height, we will us the magnification formula.

Mag = v/u = Hi/H

Hi = image height = ?

H = object height = 3.50 cm

45.09/24.8 = Hi/3.50

Hi = (45.09*3.50)/24.8

Hi = 6.36 cm

The image height is 6.36 cm

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2 years ago

A jet can travel the 6000 km distance between washington,

Andrei [34K]

d = r*t is the basic distance equation
d = 6000 km
t with the tail wind = 6 hr
r with the tail wind = speed of the plane + wind speed = s + w
t with the head wind = 7.5 hr
r with the head wind = speed of the plane - wind speed = s-w
(s+w)*6 = 6000
(s-w)*7.5 = 6000
s + w = 1000
s - w = 800
 2s = 1800
s = 900 km/h
s + w = 1000
w = 100
Check the anwer by calculating the return trip.
(900-100) * 7.5 = 800 * 7.5
800 * 7.5 = 6000 km
Answer: The rate of the jet in still air is 900 km/h. The rate of the wind is 100 km/hr.

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2 years ago