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3 years ago

7

What is the speed of a bobsled whose distance-time graph indicates that it traveled 123m in 28s?

1 answer:

Irina18 [472]3 years ago

6 0

4.39 seconds per meter

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Three students attempt to define what it means for lines lll and mmm to be perpendicular. Can you match the teacher's comments t

Ber [7]

Incompletevquestion. However, I inferred from a general perspective about perpendicular lines.

<u>Explanation:</u>

Put simply, <u>perpendicular lines</u> are lines that are at right angles (90°) to each other. Thus, we could say based on this definition that for lines lll and mmm to be perpendicular they intersect and be at right angles (90°) to each other as <u>found on the attached image.</u>

3 0

3 years ago

The 6 kg block is then released and accelerates to the right, toward the 4 kg block. The surface is rough and the coefficient of

KatRina [158]

Answer:

v =3.41 m/s

Explanation:

given,

mass of block 1 = 6 Kg

mass of another block 2 = 4 Kg

coefficient of friction = 0.3

Assuming 6 Kg block is attached to the spring of spring constant 350 N/m

and distance between the two block is equal to 0.5 m

using formula

$U = \dfrac{1}{2}kx^2$

$U = \dfrac{1}{2}\times 350 \times 0.5^2$

U = 43.75 J

using conservation of energy

KE = U - f.d

where f is the frictional force acting

$\dfrac{1}{2}mv^2 = 43.75- \mu m g d$

$\dfrac{1}{2}\times 6 \times v^2 = 43.75- 0.3\times 6 \times 9.8 \times 0.5$

$v= \sqrt{11.643}$

v =3.41 m/s

4 0

4 years ago

Light travels through a vacuum at a speed of 3 x 10 m/s. What is the speed of

Andrew [12]

Answer:

v = c / n (n = 1 for air)

v = c / 1.33 = 3 * 10E8 m/s / 1.33 = 2.25 * 10E8 m/s

3 0

3 years ago

The only drawback to solar energy is that it

MA_775_DIABLO [31]

It is to highly priced for most of civilization.

3 0

4 years ago

If the earth were twice the distance from the sun that it is now, the gravitational force exerted on it by the sun would be: a)

devlian [24]

Answer:

a) 1/4 what it is now

Explanation:

As we know that force of gravitation between two planets at some distance "r" from each other is given as

$F_g = \frac{Gm_1m_2}{r^2}$

now since we know that if the distance between Earth and Sun is changed

So the force of gravity will be given as

$\frac{F_g'}{F_g} = \frac{r_1^2}{r_2^2}$

now we know that the distance between sun and earth is changed to twice the initial distance between them

so we have

$r_2 = 2r_1$

so new gravitational force between sun and earth is given as

$F_g' = \frac{r_1^2}{(2r_1)^2}F_g$

$F_g' = \frac{1}{4}F_g$

4 0

4 years ago