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pogonyaev
3 years ago
6

Connie flew from Phoenix to flagstaff, a distance of 180 miles at a constant speed of 180 miles per hour. She then returned to t

he same airport at a cj stand speed of 90. Miles per hour. What was connies average speed in miles/ hour
Physics
1 answer:
julsineya [31]3 years ago
5 0

Answer:

Average Velocity = 120 mi/h

Explanation:

First we need to calculate the time taken by Connie to complete first half from Phoenix to flagstaff. So, we use:

s₁ = v₁t₁

where,

s₁ = distance from Phoenix to flagstaff = 180 miles

v₁ = speed of travel from Phoenix to flagstaff = 180 mi/h

t₁ = time taken to travel from Phoenix to flagstaff = ?

Therefore,

180 miles = (180 m/h)t₁

t₁ = 1 h

Now, we calculate the time for the return trip

s₂ = v₂t₂

where,

s₂ = distance from flagstaff to Phoenix = 180 miles

v₂ = speed of travel from flagstaff to Phoenix = 90 mi/h

t₂ = time taken to travel from flagstaff to Phoenix = ?

Therefore,

180 miles = (90 m/h)t₂

t₂ = 2 h

Now, we find total time:

t = t₁ + t₂

t = 1 h + 2 h

t = 3 h

and the total distance was:

s = s₁ +s₂

s = 180 mi + 180 mi

s = 360 mi

So, the average velocity will be:

Average Velocity = s/t

Average Velocity = 360 mi/3 h

<u>Average Velocity = 120 mi/h</u>

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If a 20 g cannonball is shot from a 5 kg cannon with a velocity of 100
balu736 [363]

Strange as it may seem, the statement in the question appears to be <em>TRUE</em>.  

-- Before the shot, neither the cannon nor the ball is moving, so their combined momentum is zero.  

-- Since momentum is conserved, we know immediately that their combined momentum AFTER the shot also has to be zero.

-- (20g is rather puny for a "cannonball" ... about the same weight as four nickels. But we'll take your word for it and just do the Math and the Physics.)

-- Momentum = (mass) x (velocity)

After the shot, the momentum of the cannonball is

(0.02 kg) x (100 m/s ==> that way)

Momentum of the ball = 2 kg-m/s ==> that way.

-- In order for both of them to add up to zero, the momentum of the cannon must be (2 kg-m/s this way <==) .

Momentum of cannon = (5 kg) x (V m/s this way <==)

2 kg-m/s this way <== = (5 kg) x (V m/s this way <==)

Divide each side by (5 kg):

V m/s  =  (2/5) m/s this way <==

Speed of recoil of the cannon = <em>-- 0.4 m/s</em>

3 0
3 years ago
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0
3 years ago
A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. the hoist weighs 430 n. the ropes,
ICE Princess25 [194]
Refer to the diagram shown below.

For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂                (1)

For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392        (2)

Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N

Answer:
T₂ = 339.06 N
T₃ = 276.57 N

7 0
2 years ago
A 90 kg astronaut Travis is stranded in space at a point 12 m from his spaceship. In order to get back to his ship, Travis throw
insens350 [35]

Answer:

Explanation:

This is a recoil problem, which is just another application of the Law of Momentum Conservation. The equation for us is:

[m_av_a+m_ev_e]_b=[m_av_a+m_ev_e]_a which, in words, is

The momentum of the astronaut plus the momentum of the piece of equipment before the equipment is thrown has to be equal to the momentum of all that same stuff after the equipment is thrown. Filling in:

[(90.0)(0)+(.50)(0)]_b=[(90.0)(v)+(.50)(-4.0)]_a

Obviously, on the left side of the equation, nothing is moving so the whole left side equals 0. Doing the math on the right and paying specific attention to the sig fig's here (notice, I added a 0 after the 4 in the velocity value so our sig fig's are 2 instead of just 1. 1 is useless in most applications).

0 = 90.0v - 2.0 and

2.0 = 90.0v so

v = .022 m/s This is the rate at which he is moving TOWARDS the ship (negative was moving away from the ship, as indicated by the - in the problem). Now we can use the d = rt equation to find out how long this process will take him if he wants to reach his ship before he dies.

12 = .022t and

t = 550 seconds, which is the same thing as 9.2 minutes

7 0
3 years ago
Which statement describes the horizontal acceleration of a projectile? It is –9.8 m/s2. It is 9.8 m/s2. It is constant. It is ze
lina2011 [118]
In a projectile, the horizontal acceleration is zero. The velocity remains constant at all times. However, the <u>vertical acceleration</u> is -9.81m/s^2.

Hope this helps!
6 0
3 years ago
Read 2 more answers
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