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3 years ago

Which of these is an example of a chemical change?

1 answer:

5 0

A; A chemical change makes a substance that wasn't there before. Moldy cheese has a new substance, color change and an odor.

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3. A sprinter leaves the starting blocks with an acceleration of 4.5 m/s2. What is the

UkoKoshka [18]

Hi there! :)

$\large\boxed{v_{f} = 18 m/s}$

Use the following kinematic equation to solve for the final velocity:

$v_{f} = v_{i} + at$

In this instance, the runner started from rest, so the initial velocity is 0 m/s. We can rewrite the equation as:

$v_{f} = at$

Plug in the given acceleration and time:

$v_{f} = 4.5 * 4 = 18 m/s$

5 0

3 years ago

Scientists are working on a new technique to kill cancer cells by zapping them with ultrahigh-energy (in the range of 1012 W) pu

ASHA 777 [7]

Answer:

the best way to prevent cancer is to eat right

Explanation:

4 0

3 years ago

Suddenly a worker picks up the bag of gravel. Use energy conservation to find the speed of the bucket after it has descended 2.3

fiasKO [112]

Explanation:

A worker picks up the bag of gravel. We need to find the speed of the bucket after it has descended 2.30 m from rest. It is case of conservation of energy. So,

$\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh}$

h = 2.3 m

$v=\sqrt{2\times 9.8\times 2.3} \\\\v=6.71\ m/s$

So, the speed of the bucket after it has descended 2.30 m from rest is 6.71 m/s.

8 0

3 years ago

A man starts walking north at 3 ft/s from a point P. Five minutes later a woman starts walking south at 4 ft/s from a point 500

mrs_skeptik [129]

Answer:

ds/dt = 6.98 ft/s

Explanation:

Given:

- The speed of man due north Vm = 3 ft/s

- The speed of woman due south Vw = 4 ft/s

- Woman starts walking 5 mins later than man

Find:

At what rate are the people moving apart 15 min after the woman starts walking?

Solution:

- The total time for which the man is walking due north from P, is Tm:

Tm = 5 + 15 = 20 mins

- The total distance traveled by man in Tm mins is:

Dm = Tm*Vm

Dm = 20*60*3

Dm = 3,600 ft

- The total time for which the woman is walking due south from 500 ft due east from P, is Tw:

Tw = 15 = 15 mins

- The total distance traveled by man in Tw mins is:

Dw = Tw*Vw

Dw = 15*60*4

Dw = 3,600 ft

- The displacement between man and woman at any instance is (s) which can be related by pythagoras theorem as follows:

s^2 = (dm + dw)^2 + 500^2

Where, dm : Distance travelled by man at any time Tm

dw : Distance travelled by woman at any time Tw

- Differentiate s with respect to t:

2s*ds/dt = 2*(dm + dw)*(Vm + Vw)

s*ds/dt = (dm + dw)*(Vm + Vw)

ds/dt = [ (dm + dw)*(Vm + Vw) ] / s

- Evaluate the rate of separation of man and woman ds/dt by evaluating at instance Tm = 20 mins and Tw = 15 mins. We have:

ds/dt = [ (Dm + Dw)*(Vm + Vw) ] / sqrt ( (Dm + Dw)^2 + 500^2 )

- Plug in the values:

ds/dt = [ (3600 + 3600)*(3 + 4) ] / [sqrt ( (3600 + 3600)^2 + 500^2 )]

ds/dt = 6.98 ft/s

7 0

3 years ago

A disc of mass m slides with negligible friction along a flat surface with a velocity v. The disc strikes a wall head-on and bou

qwelly [4]

Answer:

-v/2

Explanation:

Given that:

a disc of mass m

Collides with the wall going through a sliding motion on on the plane smooth surface.

Upon rebounding from the wall its kinetic energy becomes one-fourth of the initial kinetic energy before collision.

<u>We know, kinetic energy is given as:</u>

$KE_i=\frac{1}{2}. m.v^2$

consider this to be the initial kinetic energy of the body.

<u>Now after collision:</u>

$KE_f=\frac{1}{4}\times KE_i$

$KE_f=\frac{1}{4} \times \frac{1}{2}\times m.v^2$

Considering that the mass of the body remains constant before and after collision.

$KE_f=\frac{1}{2}\times m.(\frac{v}{2})^2$

Therefore the velocity of the body after collision will become half of the initial velocity but its direction is also reversed which can be denoted by a negative sign.

3 0

3 years ago