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3 years ago

10

Plz help will mark the brainliest

2 answers:

Vanyuwa [196]3 years ago

7 0

No
MARK ME BRAINLEST

fenix001 [56]3 years ago

3 0

Answer:

yesssssssssssssssssssssssssssssssssssss

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Which force acts as an object to move it from rest or a constant straight line motion

sleet_krkn [62]

Answer:

The force is Inertia

Explanation:

The force that acts on an object to move it from rest or a constant straight line motion is known as Inertia.

In physics the above statement is governed by Newton's first law of motion which is also known as Law of Inertia.

This law states that, an object that is at rest will remain at rest and an object that is moving will continue to move in a straight line with constant speed, if and only if the net force acting on the object is zero.

This implies that, A stationary object will remain motionless if no force acts on it while a object with constant velocity will continue moving with constant velocity until a force acts on it (neglecting resistance from air and friction).

8 0

3 years ago

Read 2 more answers

skater spins over a point at a speed of 3.0 rotations per second then the momentum of inertia is 0.60 kg.M2, what is its angular

laiz [17]

Answer:

$L=11.3\ kg-m^2/s$

Explanation:

Given that,

Angular speed of a skater, $\omega=3\ rot/s=18.84\ rad/s$

The moment of inertia of the skater, I = 0.6 kg-m²

We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :

$L=I\omega$

Substitute all the values,

$L=0.6\times 18.84\\\\L=11.3\ kg-m^2/s$

So, its angular momentum is equal to $11.3\ kg-m^2/s$ .

8 0

3 years ago

If mass of both the objects are doubled

Fofino [41]

Answer:

it should be four times

4 0

2 years ago

A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What will the horizontal sp

gulaghasi [49]

Answer:

C

Explanation:

horizintal speed stays same

only vertical speed changes

so H speed will stay 30 m/s

6 0

2 years ago

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º. What is the coefficient of

Alex_Xolod [135]

Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

To Find :

The coefficient of static friction between the box and the plane.

Solution :

Vertical component of force :

$mg\ sin\ \theta = 120\times 10 \times sin\ 47^\circ{}=877.62 \ N$

Horizontal component of force(Normal reaction) :

$mg\ cos\ \theta = 120\times 10 \times cos\ 47^\circ{}=818.40 \ N$

Since, box is on the verge of slipping :

$mg\ sin\ \theta= \mu(mg \ cos\ \theta)\\\\\mu = tan \ \theta\\\\\mu = tan\ 47^o\\\\\mu = 1.07$

Therefore, the coefficient of static friction between the box and the plane is 1.07.

Hence, this is the required solution.

7 0

3 years ago