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3 years ago

15

If a car is moving on a straight line with a velocity of 40 m/s and it changes its velocity to 60 m/s in 4 seconds, calculate it

s acceleration.

1 answer:

Bogdan [553]3 years ago

8 0

Answer:

5m/s²

Explanation:

Initial Velocity= 40m/s

Final Velocity=60m/s

Time = 4s

a=(v-u)/t

a=60-40/4

a=20/4

a=5m/s²

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How do scientists know the continents were once closer than they are today? Explain your answer in 2-3 complete sentences.

WITCHER [35]

Answer:

Continental drift describes one of the earliest ways geologists thought continents moved over time. Today, the theory of continental drift has been replaced by the science of plate tectonics.

The theory of continental drift is most associated with the scientist Alfred Wegener. In the early 20th century, Wegener published a paper explaining his theory that the continental landmasses were “drifting” across the Earth, sometimes plowing through oceans and into each other. He called this movement continental drift.

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3 years ago

If the mass of a material is 87 grams and the volume of the material is 14 cm3, what would the density of the material be?

tresset_1 [31]

Answer:

6.214g/cm³

Explanation:

The question is on density of a material

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3 years ago

A mass is oscillating with amplitude A at the end of a spring.

Dmitry_Shevchenko [17]

A) $x=\pm \frac{A}{2\sqrt{2}}$

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

$E=\frac{1}{2}kA^2$ (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

$U=\frac{1}{3}K$

Using (2) we can rewrite this as

$U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}$

And using (1), we find

$U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2$

Substituting $U=\frac{1}{2}kx^2$ into the last equation, we find the value of x:

$\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}$

B) $x=\pm \frac{3}{\sqrt{10}}A$

In this case, the kinetic energy is 1/10 of the total energy:

$K=\frac{1}{10}E$

Since we have

$K=E-U$

we can write

$E-U=\frac{1}{10}E\\U=\frac{9}{10}E$

And so we find:

$\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A$

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3 years ago

How much work does it take to move a 50 μC charge<br> against a 12 V potential difference?

lukranit [14]

<span>work =V*Q =12*50*10^-6

The total work done will be equal to

work = V.Q

which means

w= 12 . 50.10^-6
Hence,
w= 0.0006 J</span>

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3 years ago

Help ASAP ITS DUE IN 30 MIN

Katena32 [7]

can't read it, need larger picture

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4 years ago