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3 years ago

PLEASE HELP ASAP BEST ANSWER WILL BE MARKED BRAINLIEST!!!!!

Physics

1 answer:

umka2103 [35]3 years ago

3 0

Answer:

i think its The movement of large pieces of ice from one place to another.

Explanation:

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A uniform, 0.0300-kg rod of length 0.400 m rotates in a horizontal plane about a fixed axis through its center and perpendicular

AURORKA [14]

Answer:

A) ω₂ = 28 rev/min

B) ω_3 = 28 rev/min

Explanation:

A) Initial moment of inertia (I₁) is the rod's own ((1/12)ML²) plus that of the two rings (two lots of mr² if we treat each ring as a point mass).

Thus;

I₁ = ((1/12)ML²) + 2(mr²)

Where;

M is mass of rod = 0.03 kg

L is length of rod = 0.4m

m is mass of small rings = 0.02 kg

r is radius of small rings = 0.05 m

Thus;

I₁ = ((1/12) x 0.03 x 0.4²) + 2(0.02 x 0.05²)

I₁ = 0.0021 kg.m²

Now, when the rings slide and reach the ends of the rod i.e at r = L/2 = 0.4/2 = 0.2m from axis, the new Moment of inertia is;

I₂ = ((1/12) x 0.03 x 0.4²) + 2(0.02 x 0.2²)

I₂ = 0.0036 kg.m²

From conservation of angular momentum, we know that:

I₁ω₁ = I₂ω₂

We are given ω₁ = 48 rev/min.

Thus; plugging in the relevant values;

0.0021 x 48 = 0.0036ω₂

0.1008 = 0.0036ω₂

ω₂ = 0.1008/0.0036

ω₂ = 28 rev/min

b) For this, we have the same scenario as the case above where the ring just reaches the ends of the rods.

Thus,

I₂ω₂ = I_3•ω_3

So,0.0021 x 48 = 0.0036ω_3

ω_3 = 28 rev/min

7 0

3 years ago

A spherical raindrop 2.5 mm in diameter falls through a vertical distance of 3900 m. Take the cross-sectional area of a raindrop

Karolina [17]

Answer:

276.62 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (positive downward and negative upward)

Equation of motion

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 3900+0^2}\\\Rightarrow v=276.62\ m/s$

<u>Neglecting air drag</u> the velocity of the spherical drop would be 276.62 m/s

5 0

4 years ago

What is the force when an object exerts 15 pascal pressure on a square surface of side 5 meters

olga55 [171]

Answer:375N

Explanation:

pressure =15 pascal

Length of side =5 m

Area=length x length

Area=5 x 5

Area=25 m^2

Force=pressure x area

Force=15 x 25

Force=375N

3 0

4 years ago

A cart loaded with bricks has a total mass of 20.4 kg and is pulled at constant speed by a rope. The rope is inclined at 26.1 ◦

Readme [11.4K]

Answer:

Normal force, N = 154.5 N

Explanation:

Given that,

Total mass of the cart, m = 20.4 kg

Angle of inclination, $\theta=26.1^{\circ}$

Distance moved, d = 20.1 m

The coefficient of kinetic friction between ground and cart is 0.6

The value of acceleration due to gravity, $g=9.8\ m/s^2$

Let N is the normal force exerted on the cart by the floor. It is given by :

$N=mg-T\ sin\theta$

$T\ cos\theta=\mu N$

$T\ cos\theta=\mu (mg-T\ sin\theta)$

$T\ cos(26.1)=0.6 (20.4\times 9.8-T\ sin(26.1))$

T = 103.23 N

So, the normal force is :

$N=20.4\times 9.8-103.23\ sin(26.1)$

N = 154.5 N

So, the normal force exerted on the cart by the floor is 154.5 N. Hence, this is the required solution.

4 0

3 years ago

What is the acceleration of a softball if it hits the ground with a force 0.50 kg and hits the catchers glove with a force of 25

Rom4ik [11]

Acceleration of the ball is $50 m/s^2$

Explanation:

The acceleration of the ball can be found by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between the mass of the object and its acceleration:

$F=ma$

where

F is the net force

m is the mass

a is the acceleration

For the ball in this problem, we have

m = 0.50 kg (mass)

F = 25 N (force)

thereofre, the acceleration of the ball is

$a=\frac{F}{m}=\frac{25}{0.50}=50 m/s^2$

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0

3 years ago