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Murljashka [212]

2 years ago

8

A geological process Select one: A. is limited to acting on rocks B. shapes and changes the earth C. starts in the outer atmosph

ere of the earth D. causes no long-term change

1 answer:

navik [9.2K]2 years ago

5 0

It is B i believe thank and good bye

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Most of the elements (85%) in the periodic table are.

Rasek [7]

Most of the elements are metals

4 0

2 years ago

Read 2 more answers

A child goes down the slide,starting from rest. If the length of the slide is 2m and it takes the child 3 seconds to go down the

lapo4ka [179]

Answer:

$0.44 m/s^2$

Explanation:

We have the following data:

- distance covered by the child: d = 2 m (length of the slide)

- time taken to cover this distance: t = 3 s

- initial velocity of the child: 0 m/s (he starts from rest)

So we can find the acceleration by using the equation:

$d=ut+\frac{1}{2}at^2$

Where a is the acceleration.

Substituting the values and solving for a,

$a=\frac{2d}{t^2}=\frac{2(2)}{3^2}=0.44 m/s^2$

3 0

2 years ago

What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

Sedbober [7]

Answer:

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

Explanation:

We first identify the elements of this simple harmonic motion:

The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.

The angular frequency ω can be calculated with the formula:

$\omega =\sqrt{\frac{k}{m}}$

Where k is the spring constant and m is the mass of the particle.

Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

$x=A\cos(\omega t)$

Finally, the equation of the motion of the system is:

$x=(0.088m)\cos(\omega t)$

or

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

7 0

3 years ago

A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back

elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity= $9.8 m/s^2$

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates= $\mu mg=0.202\times 9.8\times 35$

Force by which box accelerates= $62.286 N$

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

$ma=\mu mg$

$a=\mu g=9.8\times 0.202=1.9796 m/s^2$

Hence, the truck can have maximum acceleration before the box slide= $1.9796 m/s^2$

3 0

3 years ago

if the time is taken to bring a ball to rest from a certain velocity 'v'. is reduced to half, what will be the change in the val

777dan777 [17]

The answer is C) rate of change of momentum. The answer is not initial or final momentum as the start and end points are not changing. On the other hand, the time it takes for the ball to change velocity is. This change relates to the change of momentum. Hope this helped :))

6 0

2 years ago