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hjlf
3 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]3 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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A car is parked on a steep incline, making an angle of 37.0° below the horizontal and overlooking the ocean, when its brakes fai
patriot [66]

Answer:

a) The speed of the car when it reaches the edge of the cliff is 19.4 m/s

b) The time it takes the car to reach the edge is 4.79 s

c) The velocity of the car when it lands in the ocean is 31.0 m/s at 60.2º below the horizontal

d) The total time interval the car is in motion is 6.34 s

e) The car lands 24 m from the base of the cliff.

Explanation:

Please, see the figure for a description of the situation.

a) The equation for the position of an accelerated object moving in a straight line is as follows:

x =x0 + v0 * t + 1/2 a * t²

where:

x = position of the car at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the car starts from rest and the origin of the reference system is located where the car starts moving, v0 and x0 = 0. Then, the position of the car will be:

x = 1/2 a * t²

With the data we have, we can calculate the time it takes the car to reach the edge and with that time we can calculate the velocity at that point.

46.5 m = 1/2 * 4.05 m/s² * t²

2* 46.5 m / 4.05 m/s² = t²

<u>t = 4.79 s </u>

The equation for velocity is as follows:

v = v0  + a* t

Where:

v = velocity

v0 =  initial velocity

a = acceleration

t = time

For the car, the velocity will be

v = a * t

at the edge, the velocity will be:

v = 4.05 m/s² * 4.79 s = <u>19.4 m/s</u>

b) The time interval was calculated above, using the equation of  the position:

x = 1/2 a * t²

46.5 m = 1/2 * 4.05 m/s² * t²

2* 46.5 m / 4.05 m/s² = t²

t = 4.79 s

c) When the car falls, the position and velocity of the car are given by the following vectors:

r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)

v =(v0x, v0y + g * t)

Where:

r = position vector

x0 = initial horizontal position

v0x = initial horizontal velocity

t = time

y0 = initial vertical position

v0y = initial vertical velocity

g = acceleration due to gravity

v = velocity vector

First, let´s calculate the initial vertical and horizontal velocities (v0x and v0y). For this part of the problem let´s place the center of the reference system where the car starts falling.

Seeing the figure, notice that the vectors v0x and v0y form a right triangle with the vector v0. Then, using trigonometry, we can calculate the magnitude of each velocity:

cos -37.0º = v0x / v0

(the angle is negative because it was measured clockwise and is below the horizontal)

(Note that now v0 is the velocity the car has when it reaches the edge. it was calculated in a) and is 19,4 m/s)

v0x = v0 * cos -37.0 = 19.4 m/s * cos -37.0º = 15.5 m/s

sin 37.0º = v0y/v0

v0y = v0 * sin -37.0 = 19.4 m/s * sin -37.0 = - 11. 7 m/s

Now that we have v0y, we can calculate the time it takes the car to land in the ocean, using the y-component of the vector "r final" (see figure):

y = y0 + v0y * t + 1/2 * g * t²

Notice in the figure that the y-component of the vector "r final" is -30 m, then:

-30 m = y0 + v0y * t + 1/2 * g * t²

According to our reference system, y0 = 0:

-30 m = v0y * t + 1/2 g * t²

-30 m = -11.7 m/s * t - 1/2 * 9.8 m/s² * t²

0 = 30 m - 11.7 m/s * t - 4.9 m/s² * t²

Solving this quadratic equation:

<u>t = 1.55 s</u> ( the other value was discarded because it was negative).

Now that we have the time, we can calculate the value of the y-component of the velocity vector when the car lands:

vy = v0y + g * t

vy = - 11. 7 m/s - 9.8 m/s² * 1.55s = -26.9 m/s

The x-component of the velocity vector is constant, then, vx = v0x = 15.5 m/s (calculated above).

The velocity vector when the car lands is:

v = (15.5 m/s, -26.9 m/s)

We have to express it in magnitude and direction, so let´s find the magnitude:

|v| = \sqrt{(15.5 m/s)^{2} + (-26.9 m/s)^{2}} = 31.0m/s

To find the direction, let´s use trigonometry again:

sin α = vy / v

sin α = 26.9 m/s / 31.0 m/s

α = 60.2º

(notice that the angle is measured below the horizontal, then it has to be negative).

Then, the vector velocity expressed in terms of its magnitude and direction is:

vy = v * sin -60.2º

vx = v * cos -60.2º

v = (31.0 m/s cos -60.2º, 31.0 m/s sin -60.2º)

<u>The velocity is 31.0 m/s at 60.2º below the horizontal</u>

d) The total time the car is in motion is the sum of the falling and rolling time. This times where calculated above.

total time = falling time + rolling time

total time = 1,55 s + 4.79 s = <u>6.34 s</u>

e) Using the equation for the position vector, we have to find "r final 1" (see figure):

r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)

Notice that the y-component is 0 ( figure)

we have already calculated the falling time and the v0x. The initial position x0 is 0. Then.

r final 1 = ( v0x * t, 0)

r final 1 = (15.5 m/s * 1.55 s, 0)

r final 1 = (24.0 m, 0)

<u>The car lands 24 m from the base of the cliff.</u>

PHEW!, it was a very complete problem :)

5 0
3 years ago
A 5.0 V battery contains 775 C of charge. How much electricity energy can it produce?
KatRina [158]

Answer:

3875J

Explanation:

Energy is defined as the power × time

And it's defined as

Power = IV - I- current and V- voltage

Now quantity of electricity; Q = I × t

Where I is current and t is time

Now Energy = I ×V×t = V× I×t = V× Q;

where Q is quantity of electricity 775C and V is 5.0volt

Hence 775 × (5) =3875J

7 0
3 years ago
What happens when a substance undergoes a physical change
zysi [14]

Physical changes are those which do not affect the chemical properties of a substance. For example, when a paper is cut, the two pieces obtained are still paper. The chemical properties have not changed. Only, physically when one piece is cut to two. A chemical change occurs when the chemical properties change and a new substance with different chemical properties is obtained. For example, when the paper is burnt, it turns to ashes. Hence, this is a chemical change.

When a substance undergoes a physical change, the physical properties of the substance changes like the state (solid to liquid during melting), shape, size, volume etc. Many physical changes are reversible whereas, in general many chemical changes are irreversible.

5 0
4 years ago
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Which two phrases best describe cells?
guapka [62]
<h3><u>Answer;</u></h3>
  • Make up tissues
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<h3><u>Explanation;</u></h3>
  • <em><u>Cells are the smallest part of an organism that can carry out all of life functions. They are the basic functional unit of all living organisms.</u></em>
  • According to cell theory all living things are composed of cells, cells are the basic units of structure and function in living things, and new cells are produced from existing cells
  • <em><u>A group of cells make up tissues. Tissues are a group of cells that are specialized to carry out specific functions in plants and animal bodies. </u></em>

<em><u /></em>

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What are the four components to a workout program?
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Answer:

Aerobic Fitness. Aerobic fitness improves overall health and well-being. ...

Muscular Fitness. Strength training improves your muscle and bone health and helps with weight loss. ...

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Stability and Balance

Explanation:

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