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3 years ago

At a race car driving event, a staff member notices that the skid marks left by the race car are

1 answer:

5 0

A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

A car experiences a deceleration (a) of -41.62 m/s² and comes to a stop (final velocity = v = 0 m/s) after 10.99 m (s).

We can calculate the initial velocity of the car (u) using the following kinematic equation.

$v^{2} = u^{2} + 2as\\\\u = \sqrt[]{v^{2}-2as} = \sqrt[]{(0m/s)^{2}-2(-42.61m/s^{2} )(10.99m)} = 30.60m/s$

A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

Learn more: brainly.com/question/14851168

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Over millions of years weathered rock soil dead plant and animal remains are pressed and cemented together under the ground form

Eduardwww [97]

Answer:

true

Explanation:

The statement being made is completely true. This layer of rock is called a Sedimentary Rock level and is slowly formed over millions of years with minerals and organic remains from the bottom of the Oceans that may no longer be covered in water anymore. Since it is made up of all these minerals and remains, it is studied widely by Geologists and Archeologists to better understand the Earth's past.

5 0

3 years ago

Star A and Star B have different apparent brightnesses but identical luminosities. Star A is 10 light years away from earth and

STALIN [3.7K]

intensity of a star is inversely depends on the square of the distance from the star

we can say it is given as

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

here we know that

$\frac{I_1}{I_2} = 36 times$

also we know that

$r_1 = 10 Ly$

now we will have

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

$36 = \frac{r_2^2}{10^2}$

$r_2 = 60 Ly$

so other star is at distance 60 Light years

8 0

4 years ago

An electron is initially at rest in a uniform electric field having a strength of 1.85 × 106 V/m. It is then released and accele

kirza4 [7]

Answer:

$W = 462.5 keV$

Explanation:

As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron

So here we have to find the work done by electric field on moving electron

So we have

$F = qE$

$F = (1.6 \times 10^{-19})(1.85 \times 10^6)$

$F = 2.96 \times 10^{-13} N$

now the distance moved by the electron is given as

$d = 0.25 m$

so we have

$W = F.d$

$W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)$

$W = 7.4 \times 10^{-14} J$

now we have to convert it into keV units

so we have

$1 keV = 1.6 \times 10^{-16} J$

$W = 462.5 keV$

5 0

3 years ago

Problem 5 You are playing a game where you drop a coin into a water tank and try to land it on a target. You often find this gam

Dvinal [7]

Answer:

Option D: 1.5in in front of the target

Explanation:

The object distance is $y= 6in$ .

Because the surface is flat, the radius of curvature is infinity .

The incident index is $n_i=\frac{4}{3}$ and the transmitted index is $n_t= 1$ .

The single interface equation is $\frac{n_i}{y}+\frac{n_t}{y^i}=\frac{n_t-n_i}{r}$

Substituting the quantities given in the problem,

$\frac{\frac{4}{3}}{6in}+\frac{1}{y^i}= 0$

The image distance is then $y^i=-\frac{18}{4}in =-4.5in$

Therefore, the coin falls $1.5in$ in front of the target

6 0

3 years ago

You throw a baseball directly upward at time ????=0 at an initial speed of 14.9 m/s. What is the maximum height the ball reaches

xeze [42]

Answer:

Maximum height, h = 11.32 meters

Explanation:

It is given that,

The baseball is thrown directly upward at time, t = 0

Initial speed of the baseball, u = 14.9 m/s

Ignoring the resistance in this case and using a = g = 9.8 m/s²

We have to find the maximum height the ball reaches above where it leaves your hand. Let the maximum height is h. Using third equation of motion as :

$v^2-u^2=2ah$

At maximum height, v = 0

and a = -g = -9.8 m/s²

$h=\dfrac{v^2-u^2}{2a}$

$h=\dfrac{0-(14.9\ m/s)^2}{2\times -9.8\ m/s^2}$

h = 11.32 meters

Hence, the maximum height of the baseball is 11.32 meters.

8 0

3 years ago