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hjlf
2 years ago
11

It takes 15 min to drive 6.0 mi in a straight line to the local hospital. It takes 10 min to go the last 3.0 mi, 2.0 min to go t

he last mile, and only 30 s (0.50 min) to go the last 0.50 mi. What is the average speed of the trip? Take the positive xx ‑direction to be from the starting point toward the hospital.
Physics
1 answer:
Gala2k [10]2 years ago
6 0

Answer:

36.87 km/h

Explanation:

Convert all the units in SI system

1 mile = 1609.34 m

d1 = 6 mi = 9656.04 m

t1 = 15 min = 15 x 60 = 900 s

d2 = 3 mi = 4828.02 m

t2 = 10 min = 10 x 60 = 600 s

d3 = 1 mi = 1609.34 m

t3 = 2 min = 2 x 60 = 120 s

d4 = 0.5 mi = 804.67 m

t4 = 0.5 min = 0.5 x 60 = 30 s

Total distance, d = d1 + d2 + d3 + d4

d = 9656.04 + 4828.02 +  1609.34 + 804.67 = 16898.07 m = 16.898 km

total time, t = t1 + t2 + t3 + t4

t = 900 + 600 + 120 + 30 = 1650 s = 0.4583 h

The ratio of the total distance covered to the total time taken is called average speed.

Average speed = 16.898 / 0.4583 = 36.87 km/h

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Answer:

B

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a steel sphere and brass ring have diameter 25cm and 24.9cm at 15°C.If the sphere and the ring are heated together.what is the t
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Answer:

Explanation:

Due to heat energy , metal expands . Formula for linear expansion is as follows .

L = l ( 1 + α Δt )

where L is expanded length , l is original length , α is coefficient of linear expansion and Δt is increase in temperature .

To pass the sphere through the ring , the diameter of both ring and sphere should be same after heating . Let after increase of temperature Δt , their diameter becomes same as L  . The linear coefficient of brass and steel are

20 x 10⁻⁶ and 12 x 10⁻⁶ respectively .

For steel sphere ,

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For brass ring

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1.004( 1 + 12 x 10⁻⁶ Δt ) =  ( 1 + 20 x 10⁻⁶ Δt )

1.004 + 12.0482 x 10⁻⁶ Δt  =   1 + 20 x 10⁻⁶ Δt

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6 0
3 years ago
Suppose you are on a cart, initially at rest, which rides on a frictionless horizontal track. You throw a ball at a vertical sur
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Answer:

F_c t_ c = -F_b t_b

And the forces are equal but in the opposite direction. So then we can write by general rule:

m_c \Delta V_{c} = -m_b \Delta V_b

Or equivalently:

m_c \Delta V_{c} +m_b \Delta V_b =0

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m_c represent the mass of the car

m_b represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

Explanation:

If we assume that we have the situation in the figure attached.

For this case we assume that the momentum changes are equal in magnitude and opposite in direction, so then we satisfy this:

F_c t_ c = -F_b t_b

And the forces are equal but in the opposite direction. So then we can write by general rule:

m_c \Delta V_{c} = -m_b \Delta V_b

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m_c \Delta V_{c} +m_b \Delta V_b =0

Where: V_c represent the speed of the car and V_b the speed of the ball

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So then the correct option for this case is :

A.Yes, and it moves to the right.

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Answer:

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Here's what I found:

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