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Amiraneli [1.4K]

2 years ago

7

an object at rest will remain at rest and an object in motion stays in straight-line motion unless acted upon by a_____ or unbal

anced force

1 answer:

LekaFEV [45]2 years ago

5 0

Equal is the answer your looking for

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6.6 kg block initially at rest is pulled to theright along a horizontal, frictionless surfaceby a constant, horizontal force of

neonofarm [45]

Answer:

v = 3.04 m/s

Explanation:

given,

mass of the block, M = 6.6 Kg

horizontal force, F = 12.2 N

distance, L = 2.5 m

initial speed = 0 m/s

speed of the block,v = ?

we now

Work done is equal to change in Kinetic energy.

Work done = Force x displacement

W = Δ K E

Δ K E = Force x displacement

$\dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2= F .s$

$\dfrac{1}{2}\times 6.6 \times v^2 - 0= 12.2\times 2.5$

3.3 v² = 30.5

v² = 9.242

v = 3.04 m/s

speed of the block is equal to 3.04 m/s

4 0

2 years ago

Long periods of unusually low precipitation are called?

DiKsa [7]

I think it is called a drought but i would look it up

6 0

3 years ago

T/F all compounds are molecules but NOT all molecules are compounds.

Solnce55 [7]

True because molecules don't have to be compounds but compounds have to be molecules

7 0

2 years ago

What is the moment of inertia of a 2.0 kg, 20-cm-diameter disk for rotation about an axis (a) through the center, and (b) throug

FinnZ [79.3K]

Answer:

(a) I=0.01 kg.m²

(b) I=0.03 kg.m²

Explanation:

Given data

Mass of disk M=2.0 kg

Diameter of disk d=20 cm=0.20 m

To Find

(a) Moment of inertia through the center of disk

(b) Moment of inertia through the edge of disk

Solution

For (a) Moment of inertia through the center of disk

Using the equation of moment of Inertia

$I=\frac{1}{2}MR^{2}\\ I=\frac{1}{2}(2.0kg)(0.20m/2)^{2}\\ I=0.01 kg m^{2}$

For (b) Moment of inertia through the edge of disk

We can apply parallel axis theorem for calculating moment of inertia

$I=(1/2)MR^{2}+MD\\ Here\\D=R\\I=(1/2)(2.0kg)(0.20m/2)^{2}+(2.0kg)(0.20m/2)^{2}\\ I=0.03kgm^{2}$

8 0

3 years ago

A ball is dropped off the roof of a tall building. If the ball reaches the ground in 8 seconds, how tall is the building, in met

Ghella [55]

Let <em>b</em> be the height of the building, and <em>y</em> the height of the ball at time <em>t</em>, given by

<em>y</em> = <em>b</em> - 1/2 <em>gt</em>²

where <em>g</em> = 9.8 m/s² is the magnitude of the acceleration due to gravity.

It takes the ball 8 s to reach the ground, at which point <em>y</em> = 0, so that

0 = <em>b</em> - 1/2 (9.8 m/s²) (8 s)²

<em>b</em> = 1/2 (9.8 m/s²) (8 s)²

<em>b</em> = 313.6 m

5 0

3 years ago