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-Dominant- [34]
2 years ago
15

Salem and Vernonville are 160 miles apart. A car leaves Salem traveling towards​ Vernonville, and another car leaves Vernonville

at the same​ time, traveling towards Salem. The car leaving Salem averages 10 miles per hour more than the​ other, and they meet after 1 hour and 36 minutes. What are the average speeds of the​ cars?

Physics
1 answer:
Paul [167]2 years ago
8 0

Answer:

The speed of car leaves  Vernonville = 45 miles per hour

The speed of car leaves  Salem = 55 miles per hour

Explanation:

Given that distance between Salem and Vernonville = 160 miles

Lest take speed of car which leaves from Vernonville = V miles per hour

So the speed of car which leaves from Salem = V +10 miles per hour

These two car meet at time = 1 hr 36 min = 1.6 hr

Distance travel by car leaves from Salem = x

   x= (V+10) x 1.6    --------1

Distance travel by car leaves from Vernonville = 160 - x

  160 -  x= V x 1.6      ----------2

From equation 1 and 2

160 - (V+10) x 1.6  = V x 1.6

V= 45 miles per hour

So

The speed of car leaves  Vernonville = 45 miles per hour

The speed of car leaves  Salem = 55 miles per hour

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jek_recluse [69]

Answer:

a)v_{f}=0.88m/s

Explanation:

To solve this problem we use the Momentum's conservation Law, before and after the girl catch the ball:

\\ p_{1}=p_{2}\\m_{ball}*v_{o.ball}+m_{girl}*v_{o.girl} = m_{ball}*v_{f.ball} + m_{girl}*v_{f.girl}        (1)

At the beginning the girl is  stationary:

v_{o.girl}=0m/s       (2)

If the girl catch the ball, both have the same speed:

v_{f.girl}=v_{f.ball}=v_{f}       (3)

We replace (2) and (3) in (1):

m_{ball}*v_{o.ball} = (m_{ball}+m_{girl})*v_{f} \\

We can now solve the equation for v_{f}:

v_{f}=\frac{m_{ball}*v_{o.ball}}{(m_{ball}+m_{girl})}=\frac{15*4.7}{15+65}=0.88m/s

4 0
2 years ago
In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal
SpyIntel [72]

Answer:

A) F_g = 26284.48 N

B) v = 7404.18 m/s

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Explanation:

We are given;

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Mass of the earth; M = 6 x 10²⁴ Kg

Earth circular orbit radius; R = 7.3 x 10⁶ m

A) Formula for the gravitational force is;

F_g = GmM/r²

Where G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Plugging in the relevant values, we have;

F_g = (6.67 × 10^(-11) × 3500 × 6 x 10²⁴)/(7.3 x 10⁶)²

F_g = 26284.48 N

B) From the momentum principle, we have that the gravitational force is equal to the centripetal force.

Thus;

GmM/r² = mv²/r

Making v th subject, we have;

v = √(GM/r)

Plugging in the relevant values;

v = √(6.67 × 10^(-11) × 6 x 10²⁴)/(7.3 x 10⁶))

v = 7404.18 m/s

C) From the energy principle, the minimum amount of work is given by;

E = (GmM/r) - ½mv²

Plugging in the relevant values;

E = [(6.67 × 10^(-11) × 3500 × 6 × 10²⁴)/(7.3 x 10⁶)] - (½ × 3500 × 7404.18)

E = 19.19 × 10^(10) J

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2 years ago
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Answer:

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let initial mass= m

final mass = m-m(4/5) = m/5

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V-0 = 1500 ln (1/0.2)

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therefore, its exhaust speed v= 2413.5 m/s

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PLEASE HELP!!!!!! WILL GIVE BRAINLIEST TO WHOEVER ANSWERS WITH THE RIGHT ANSWER !!!!!!!! 
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It would be B and D your welcome


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Starting from the angular velocity, we can calculate the tangential velocity of the stone:
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