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3 years ago

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Police investigators examining the scene of an accident involving two cars measure 79-meter long skidmarks of one of the cars wh

ich nearly came to a stop before colliding. The coefficient of kinetic friction between rubber and the pavement is about 0.82. Estimate the initial speed of that car assuming a level road.

1 answer:

EastWind [94]3 years ago

8 0

Answer:

$v_i = 36.65 m/s$

Explanation:

As we know that friction force on the car during it applied brakes is given as

$F_f = -\mu mg$

so the deceleration of the car is given as

$a = -\mu g$

now we can use kinematics to find the initial speed of the car

$v_f^2 - v_i^2 = 2 a d$

so we will have

$0 - v_i^2 = 2(-\mu g)(d)$

now plug in all the data in it

$v_i^2 = 2(0.82)(9.81)(79)$

$v_i = 36.65 m/s$

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Given the Earth's density as 5.5 g/cm3 and the Moon's density as 3.34 g/cm3, determine the Roche limit for the Moon orbiting the

Assoli18 [71]

Answer:

The Roche limit for the Moon orbiting the Earth is 2.86 times radius of Earth

Explanation:

The nearest distance between the planet and its satellite at where the planets gravitational pull does not torn apart the planets satellite is known as Roche limit.

The relation to determine Roche limit is:

$Roche\ limit=2.423\times R_{P}\times\sqrt[3]{\frac{D_{P} }{D_{M} } }$ ....(1)

Here $R_{P}$ is radius of planet and $D_{P}\ and\ D_{M}$ are density of planet and moon respectively.

According to the problem,

Density of Earth, $D_{P}$ = 5.5 g/cm³

Density of Moon, $D_{M}$ = 3.34 g/cm³

Consider $R_{E}$ be the radius of the Earth.

Substitute the suitable values in the equation (1).

$Roche\ limit=2.423\times R_{E}\times\sqrt[3]{\frac{5.5 }{3.34 } }$

$Roche\ limit= 2.86R_{P}$

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3 years ago

While no one knows why, astronomers do now know that the solar nebula has ___________?

iris [78.8K]

While no one knows why, astronomers do now know that the solar nebula has collapsed. Scientists believed that our Solar system was made when gas and dust cloud was being disturbed by an explosion of supernova. Due to pressure, this cloud collapsed.

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3 years ago

Read 2 more answers

A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta

Vika [28.1K]

Answer:

$t_t=4.131\ s$

Explanation:

Given:

height above the horizontal form where the ball is hit, $y=1\ m$

angle of projectile above the horizontal, $\theta=30^{\circ}$

initial speed of the projectile, $u=40\ m.s^{-1}$

<u>Firstly we find the </u><u>vertical component of the initial velocity</u><u>:</u>

$u_y=u.\sin\theta$

$u_y=40\times \sin30^{\circ}$

$u_y=20\ m.s^{-1}$

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

$v_y=0\ m.s^{-1}$

Using eq. of motion:

$v_y^2=u_y^2-2g.h$ (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

$0^2=20^2-2\times 9.8\times h$

$h=20.4082\ m$ (from the height where it is thrown)

<u>Now we find the time taken to ascend to this height:</u>

$v_y=u_y-g.t$

$0=20-9.8t$

$t=2.041\ s$

<u>Time taken to descent the total height:</u>

we've total height, $h'=h+y$ $=20.4082+1$

$h'=u_y'.t'+\frac{1}{2} g.t'^2$

during the course of descend its initial vertical velocity is zero because it is at the top height, so $u_y'=0\ m.s^{-1}$

$21.4082=0+4.9t'^2$

$t'=2.09\ s$

<u>Now the total time taken by the ball to hit the ground:</u>

$t_t=t'+t$

$t_t=2.09+2.041$

$t_t=4.131\ s$

3 0

3 years ago

a chemical reaction a temperature change may occur because of the breaking or formation of chemical bonds that release excess en

Hoochie [10]

Answer:

it cannot be considered a physical change because it occur from the chemical reaction.

Explanation:

because physical change is the change of the size,volume, density,flexibility. in other words its a change in the outer appearance or change in which we can see or feel without using a tool or a machine

hope it help just my opinion

5 0

4 years ago

A beam of 1.0 MHz ultrasound begins with an intensity of 1000 W/m². After traveling 12 cm through tissue with no significant ref

Ksenya-84 [330]

Answer:

Option C

Explanation:

Given:

- Depth of tissue d = 12 cm

- frequency of ultrasound f = 1 MHz

- Input Intensity I_i = 1000 W/m^2

- attenuation coefficient soft tissue a = 0.54

Find:

- Out-put intensity at the required depth

Solution

- The amount of attenuation in (dB) with the progression of depth is given by:

Attenuation = a*f*d

Attenuation = 0.54*12*1

Attenuation = 6.48 dB

- The relation with attenuation and ratio of input and output intensity is given by:

Attenuation = 10*log_10 (I_i / I_o)

6.48 dB = 10*log_10 (I_i / I_o)

I_i / I_o = 10^(0.648)

I_o = 1000 / 10^(0.648)

I_o = 225 W/m^2

- Hence the answer is option C: I_o = 250 W/m^2

4 0

4 years ago