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3 years ago

9

Mrs. Botts applies brakes on a car help it to decelerate at the rate of -0.80m/s2. What distance is required to stop the car whe

n it is moving 17 m/s.

1 answer:

lyudmila [28]3 years ago

8 0

Answer:

Explanation:

vf=0

vi = 17

a = -0.8

Δx = ?

vf^2 = vi^2 + 2 a Δx

Δx = (vf^2 - vi^2) / 2a

Δx = (0-17^2) / 2 (-0.8)

Δx = 180.625 m

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A music fan at a swimming pool is listening to a radio on a diving platform. The radio is playing a constant- frequency tone whe

joja [24]

Answer:

The Doppler Effect is given by the following relation;

$f' = \left (\dfrac{v + v_0}{v - v_s} \right) \times f$

Where;

f' = The frequency the observer hears

f = Actual frequency of the wave

v = The velocity of the sound wave

$v_o$ = The velocity of the observer

$v_s$ = The velocity of the source

Where the observer is stationary, we have;

(i) When the source is moving in the direction of the observer

$f' = \left (\dfrac{v }{v - v_s} \right) \times f$

(ii) When the source is receding from the observer, we have;

$f' = \left (\dfrac{v }{v + v_s} \right) \times f$

Therefore;

(a) A person left behind on the platform

For a person left behind on the platform, we have that the radio source is receding, therefore, we have;

$f' = \left (\dfrac{v }{v + v_s} \right) \times f$

(1) Given that (v + $v_s$ ) > v, therefore, v < (v + $v_s$ ), f' < f, the frequency heard by the person left on the platform, f', is smaller (lower) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively smaller

(b) A person down below floating on a rubber raft

For the the person down below on the rubber raft, the radio source is advancing

Therefore, the radio source is moving towards the person at rest down on the rubber raft, therefore, we have;

$f' = \left (\dfrac{v }{v - v_s} \right) \times f$

(1) Given that (v - $v_s$ ) < v, therefore, f' > f, the frequency heard by the person down below floating on the rubber raft, f', is greater (higher) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively greater (higher)

Explanation:

7 0

3 years ago

Which astronomer spent 20 years plotting the positions of the planets

klio [65]

That was Tycho Brahe, and I thought it was actually more years than that.

5 0

3 years ago

Read 2 more answers

1. Calculate the momentum of each car before the collision: SHOW YOUR WORK!

a_sh-v [17]

Answer:

Momentum of red car = 5kgm/s

Momentum of blue car = 0kgm/s

Explanation:

Momentum = mass × velocity

For the red car

Mass = 1kg

Velocity = 5m/s

Momentum of the red car = 1kg × 5m/s

Momentum of the red car = 5kgm/s

For the blue car.

Mass = 1kg

Velocity = 0m/s(shows that the blue car is stationery)

Momentum = 1kg ×0m/s

Momentum of the blue car = 0kgm/s

3 0

3 years ago

Question Part Points Submissions Used A pitcher throws a 0.200 kg ball so that its speed is 19.0 m/s and angle is 40.0° below th

postnew [5]

Answer:

The impulse is (10.88 i^ + 7.04 j^) N s

maximum force on the ball is (4.53 10 2 i^ + 2.93 102 j ^) N

Explanation:

In a problem of impulse and shocks we must use the impulse equation

I = dp = pf-p₀ (1)

p = m V

With we have vector quantities, let's decompose the velocities on the x and y axes

V₀ = -19 m / s

θ₀ = 40.0º

Vf = 46.0 m / s

θf = 30.0º

Note that since the positive direction of the x-axis is from the batter to the pitcher, the initial velocity is negative and the angle of 40º is measured from the axis so it is in the third quadrant

Vcx = Vo cos θ

Voy = Vo sin θ

Vox= -19 cos (40) = -14.6 m/s

Voy = -19 sin (40) = -12.2 m/s

Vfx = 46 cos 30 = 39.8 m/s

Vfy = 46 sin 30 = 23.0 m/s

a) We already have all the data, substitute and calculate the impulse for each axis

Ix = pfx -pfy

Ix = m ( vfx -Vox)

Ix = 0.200 ( 39.8 – (-14.6))

Ix = 10.88 N s

Iy = m (Vfy -Voy)

Iy = 0.200 ( 23.0- (-12.2))

Iy= 7.04 N s

In vector form it remains

I = (10.88 i^ + 7.04 j^) N s

b) As we have the value of the impulse in each axis we can use the expression that relates the impulse to the average force and your application time, so we must calculate the average force in each interval.

I = Fpro Δt

In the first interval

Fpro = (Fm + Fo) / 2

With the Fpro the average value of the force, Fm the maximum value and Fo the minimum value, which in this case is zero

Fpro = (Fm +0) / 2

In the second interval the force is constant

Fpro = Fm

In the third interval

Fpro = (0 + Fm) / 2

Let's replace and calculate

I = Fpro1 t1 +Fpro2 t2 +Fpro3 t3

I = Fm/2 4 10⁻³ + Fm 20 10⁻³+ Fm/2 4 10⁻³

I = Fm 24 10⁻³ N s

Fm = I / 24 10⁻³

Fm = (10.88 i^ + 7.04 j^) / 24 10⁻³

Fm = (4.53 10² i^ + 2.93 10² j ^) N

maximum force on the ball is (4.53 10 2 i^ + 2.93 102 j ^) N

3 0

3 years ago

A tiger paces back and forth along the front of its cage, which is 8 m wide. The tiger starts from the right side of the cage, p

Black_prince [1.1K]

Answer:

The total distance is 24 m.

The tiger's resultant displacement is 8 m.

Explanation:

Given that,

Width of tiger= 8 m

Tiger starts from the right side of the cage,

Paces to the left side

$\Delta x_{1}=+8\ m$

Then, back to the right side,

$\Delta x_{2}=-8\ m$

finally, back to the left

$\Delta x_{3}=+8\ m$

We need to calculate the total distance covered

Using formula of distance

$d= x_{1}+x_{2}+x_{3}$

Put the value into the formula

$d=8+8+8$

$d=24\ m$

We need to calculate the tiger's resultant displacement

Using formula of displacement

$D=\Delta x_{1}+\Delta x_{2}+\Delta x_{3}$

Put the value into the formula

$D=8+(-8)+8$

$D=8\ m$

Hence, The total distance is 24 m.

The tiger's resultant displacement is 8 m.

3 0

3 years ago