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2 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]2 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 student measured the volume of three rocks. The diagram above shows the

victus00 [196]

Answer:

option B is right

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2 years ago

Because of surface tension, it is possible, with care, to support an object heavier than water on the water surface. The maximum

boyakko [2]

Answer:

$\pi 1 = \frac{h}{l}$

$\pi 2 =$ б / $l^2gp$

$\frac{h}{l} =$ Ф ( б / $l^2gp$ )

Explanation:

<u>Develop a suitable set of dimensionless parameters for this problem</u>

The set of dimensionless parameters for this problem is :

$\pi 1 = \frac{h}{l}$

$\pi 2 =$ б / $l^2gp$

$\frac{h}{l} =$ Ф ( б / $l^2gp$ )

and they are using the pi theorem, MLT systems

attached below is a detailed solution

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2 years ago

On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s

dsp73

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

= 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2

= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

4 0

2 years ago

regrine falcons frequently grab prey birds from the air. Sometimes they strike at high enough speeds that the force of the impac

solmaris [256]

Answers:

a) 30 m/s

b) 480 N

Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

<h3>a) Final speed</h3>

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum $p_{i}$ before the collision must be equal to the final momentum $p_{f}$ after the collision:

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=MV_{i}+mU_{i}$

$p_{f}=(M+m) V$

Where:

$M=480 g \frac{1 kg}{1000 g}=0.48 kg$ the mas of the peregrine falcon

$V_{i}=45 m/s$ the initial speed of the falcon

$m=240 g \frac{1 kg}{1000 g}=0.24 kg$ is the mass of the pigeon

$U_{i}=0 m/s$ the initial speed of the pigeon (at rest)

$V$ the final speed of the system falcon-pigeon

Then:

$MV_{i}+mU_{i}=(M+m) V$ (2)

Finding $V$ :

$V=\frac{MV_{i}}{M+m}$ (3)

$V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg}$ (4)

$V=30 m/s$ (5) This is the final speed

<h3>b) Force on the pigeon</h3>

In this part we will use the following equation:

$F=\frac{\Delta p}{\Delta t}$ (6)

Where:

$F$ is the force exerted on the pigeon

$\Delta t=0.015 s$ is the time

$\Delta p$ is the pigeon's change in momentum

Then:

$\Delta p=p_{f}-p_{i}=mV-mU_{i}$ (7)

$\Delta p=mV$ (8) Since $U_{i}=0$

Substituting (8) in (6):

$F=\frac{mV}{\Delta t}$ (9)

$F=\frac{(0.24 kg)(30 m/s)}{0.015 s}$ (10)

Finally:

$F=480 N$

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3 years ago

For two objects with masses 100 kg and 500 kg to be 20 meters apart , find the gravity between them

Vikki [24]

Answer:

8,3375*10^-9 N

Explanation:

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2 years ago