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3 years ago

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A car of mass M traveling with velocity v strikes a car of mass M that is at rest. The two cars’ bodies mesh in the collision. T

he loss of the kinetic energy the moving car undergo in the collision is
a) a quarter of the initial kinetic energy.
b) half of the initial kinetic energy.
c) all the initial kinetic energy.
d) zero.

1 answer:

krok68 [10]3 years ago

4 0

Answer:

the correct answer is B

Explanation:

Let's propose the solution of the problem, for this we form a system formed by the two cars, so that the forces during the collision are internal, the momentum is conserved

instantly starts. Before the crash

p₀ = M v +0

final instant. After the crash

m_f = (M + M) v_f

the moment is preserved

p₀ = p_f

M v = 2 M v_f

v_f = v / 2

let's look for kinetic energy

before the crash

K₀ = ½ M v²

after the crash

K_f = ½ 2M (v_f)²

K_f = ½ 2M (v/2)²

K_f = (½ M v²) ½

K_f = K₀ / 2

therefore the correct answer is B

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Lewiston and Vernonville are 208 miles apart. A car leaves Lewiston traveling towards Vernonville, and another car leaves Verno

iragen [17]

Answer:

Average speed of the car A = 70 miles per hour

Average speed of the car B = 60 miles per hour

Explanation:

Average speed of the car A is $v_{A} =\frac{x_{A} }{t_{A} }$ (Equation A) and Average speed of the car B is $v_{B} =\frac{x_{B} }{t_{B} }$ (Equation B), where $x_{A}$ and $x_{B}$ are the distances and $t_{A}$ and $t_{B}$ are the times at which are travelling the cars A and B respectively.

We have to convert the time to the correct units:

1 hour and 36 minutes = 96 minutes

$96 minutes . \frac{1 hour}{60 minutes} = 1.6 h$

From the diagram (Please see the attachment), we can see that at the time they meet, we have:

$v_{A} = \frac{208-x}{1.6h} + 10\frac{miles}{h}$ (Equation C)

$v_{B} = \frac{208-x}{1.6h}$ (Equation D)

From Equation A and C, we have:

$\frac{208-x}{1.6}+10 = \frac{x}{1.6}$

208-x+16 = x

208 + 16 = 2x

$x = \frac{224}{2}$

x = 112 miles

Replacing x in Equation A:

$v_{A} = \frac{112miles}{1.6h}$

$v_{A} = 70 miles per hour$

Replacing x in Equation B:

$v_{B} = \frac{208miles-112miles}{1.6h}$

$v_{B} = \frac{96miles}{1.6h}$

$v_{B} = 60 miles per hour$

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3 years ago

A baby is dropped from a 30 m cliff on Earth. Acceleration due to gravity on Earth is about 9.8 m/s^2. Calculate the following q

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Answer:Dropping is not a good predictor of when labor will begin. In first-time mothers, dropping usually occurs 2 to 4 weeks before delivery, but it can happen earlier. In women who have already had children, the baby may not drop until labor begins.

Explanation:

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3 years ago

The visible color of light with the highest frequency is? question 31 options:

chubhunter [2.5K]

The visible color of light with the highest frequency is blue.

<h3>What is visible light?</h3>

The visible light spectrum is the segment of the electromagnetic spectrum that the human eye can view.

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Visible color of light can be arranged in order of increasing frequency as follows;

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indigo
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Thus, the visible color of light with the highest frequency is blue.

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1 year ago

A small cork with an excess charge of +6.0µC is placed 0.12 m from another cork, which carries a charge of -4.3µC.

Volgvan

A) 16.1 N

The magnitude of the electric force between the corks is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

$q_1 = 6.0 \mu C=6.0 \cdot 10^{-6} C$ is the magnitude of the charge on the first cork

$q_2 = 4.3 \mu C = 4.3 \cdot 10^{-6}C$ is the magnitude of the charge of the second cork

r = 0.12 m is the separation between the two corks

Substituting numbers into the formula, we find

$F=(9\cdot 10^9 N m^2 C^{-2} )\frac{(6.0\cdot 10^{-6}C)(4.3\cdot 10^{-6} C)}{(0.12 m)^2}=16.1 N$

B) Attractive

According to Coulomb's law, the direction of the electric force between two charged objects depends on the sign of the charge of the two objects.

In particular, we have:

- if the two objects have charges with same sign (e.g. positive-positive or negative-negative), the force is repulsive

- if the two objects have charges with opposite sign (e.g. positive-negative), the force is attractive

In this problem, we have

Cork 1 has a positive charge

Cork 2 has a negative charge

So, the force between them is attractive.

C) $2.69\cdot 10^{13}$

The net charge of the negative cork is

$q_2 = -4.3 \cdot 10^{-6}C$

We know that the charge of a single electron is

$e=-1.6\cdot 10^{-19}C$

The net charge on the negative cork is due to the presence of N excess electrons, so we can write

$q_2 = Ne$

and solving for N, we find the number of excess electrons:

$N=\frac{q_2}{e}=\frac{-4.3\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=2.69\cdot 10^{13}$

D) $3.75\cdot 10^{13}$

The net charge on the positive cork is

$q_1 = +6.0\cdot 10^{-6}C$

We know that the charge of a single electron is

$e=-1.6\cdot 10^{-19}C$

The net charge on the positive cork is due to the "absence" of N excess electrons, so we can write

$q_1 = -Ne$

and solving for N, we find the number of electrons lost by the cork:

$N=-\frac{q_1}{e}=-\frac{+6.0\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=3.75\cdot 10^{13}$

6 0

3 years ago

Dave wishes to get produce from the store, which is a distance D- 2.2 km east of his house. Dave drives his bike to the store at

Neporo4naja [7]

Answer:

a) $t_1=338.4615\ s$

b) $t=1386.0805\ s$

c) $t=23.1013\ min$

d) $d=4400\ m$

e) Since Dave starts from house and finally returns to the house so displacement is zero.

Explanation:

Given:

distance between the house and the store, $s=2.2\ km=2200\ m$

speed of driving from house to store, $v_1=-6.5\ m.s^{-1}$

speed of driving back from store to house, $v_2=2.1\ m.s^{-1}$

Since the store is located towards east from his house and the velocity in this direction is taken negative and contrary to this the velocity in the west direction is taken positive.

a)

time taken in reaching the store:

$t_1=\frac{s}{v_1}$

$t_1=\frac{-2200}{-6.5}$

$t_1=338.4615\ s$

b)

Now the time taken in returning form the store:

$t_2=\frac{s}{v_2}$

$t_2=\frac{2200}{2.1}$

$t_2=1047.6190\ s$

therefore total time taken by the trip:

$t=t_1+t_2$

$t=338.4615+1047.6190$

$t=1386.0805\ s$

c)

the time taken by the trip in minutes:

$t=\frac{1386.0805}{60}$

$t=23.1013\ min$

d)

distance travelled in the whole trip:

$d=2\times s$

$d=2\times 2200$

$d=4400\ m$

e)

Since Dave starts from house and finally returns to the house so displacement is zero.

3 0

3 years ago