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vampirchik [111]

3 years ago

13

The driver of a 1,000 kg car travelling at a speed of 16.7 m/s applies the car's brakes when he sees a red light. the car's brak

es provide a frictional force of 8,000 n. determine the stopping distance of the car.

2 answers:

Dmitry [639]3 years ago

6 0

Energy conservation :
<span>kinetic energy Ek = braking work W </span>
<span>mV^2 = 2Fb*x </span>
<span>x = mV^2/2Fb = 1000*16.7^2/16.000 = 17.43 meters</span>

Sauron [17]3 years ago

5 0

Answer:

1.04 meters

Explanation:

Thinking process:

Gathering the data

mass = 1 000 kg

speed, u = 16.7 m/s

Frictional force = 8 000 N

distance, s = ?

final velocity = 0 (car stops)

We know that the final velocity is calculated as: $v^{2} = u^{2} + 2as$

But, a = negative (declaration)

And F = ma

$8 000 = 1000 (a)\\ a = -8 m^{-2}$

substituting, 0 = (16.7) - 2 (8) (s)

-16.7 = -16s

s = 1.04 m

Stopping distance = 1.04 meters

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A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total

Lynna [10]

Explanation:

Given Data

Total mass=93.5 kg

Rock mass=0.310 kg

Initially wagon speed=0.540 m/s

rock speed=16.5 m/s

To Find

The speed of the wagon

Solution

As the wagon rolls, momentum is given as

P=mv

where

m is mass

v is speed

put the values

P=93.5kg × 0.540 m/s

P =50.49 kg×m/s

Now we have to find the momentum of rock

momentum of rock = mv

momentum of rock = (0.310kg)×(16.5 m/s)

momentum of rock =5.115 kg×m/s

From the conservation of momentum we can find the wagons momentum So

wagon momentum=50.49 -5.115 = 45.375 kg×m/s

Speed of wagon = wagon momentum/(total mass-rock mass)

Speed of wagon=45.375/(93.5-0.310)

Speed of wagon= 0.487 m/s

Throwing rock backward,

momentum of wagon = 50.49+5.115 = 55.605 kg×m/s

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3 years ago

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xxTIMURxx [149]

Answer:

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Explanation:

Given parameters:

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Unknown:

Final velocity = ?

Solution:

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v² = u² + 2gh

v is the final velocity

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g is the acceleration due to gravity

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v² = 0² + (2 x 9.8 x 8.18)

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3 years ago

Explain where you observe reflection, refraction, and absorption of light in your everyday activities

Studentka2010 [4]

Reflection: you look in the mirror.
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5 0

3 years ago

A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa

Nataly_w [17]

Answer:

$V_p = 267.258 m/s$

$V_n = 38.375 m/s$

Explanation:

using the law of the conservation of the linear momentum:

$P_i = P_f$

where $P_i$ is the inicial momemtum and $P_f$ is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

$P_i = m(270 m/s)$

$P_f = mV_P + M_nV_n$

where m is the mass of the proton and $V_p$ is the velocity of the proton after the collision, $M_n$ is the mass of the nucleus and $V_n$ is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = m $V_p$ + 14m $V_n$

then, m is cancelated and we have:

270 = $V_p$ + $14V_n$

This is a elastic collision, so the kinetic energy K is conservated. Then:

$K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2$

and

Kf = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

then,

$\frac{1}{2}m(270)^2$ = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

here we can cancel the m and get:

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

now, we have two equations and two incognites:

270 = $V_p$ + $14V_n$ (eq. 1)

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

in the second equation, we have:

36450 = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$ (eq. 2)

from this last equation we solve for $V_n$ as:

$V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

and replace in the other equation as:

270 = $V_p +$ 14 $\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

so,

$V_p = -267.258 m/s$

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

$V_n = \frac{270+267.258}{14}$

$V_n = 38.375 m/s$

3 0

3 years ago