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Artist 52 [7]
3 years ago
13

You are driving home from school steadily at 91 km/h for 160 km . It then begins to rain and you slow to 64 km/h instantly. You

arrive home after driving 4.0 hours.How far is your hometown from school?
Physics
1 answer:
Korolek [52]3 years ago
3 0

Total distance = (average speed) x (total time)

What do we know, and what else can we figure out ?

-- Total time = 4 hours.

-- First piece of the trip:

. . . distance . . . 160 km

. . . speed . . . 91 km/h.  

. . . time . . . (160 km) / (91 km/hr) = 1.758 hours

-- Second piece of the trip:

. . . speed . . . 64 km/h

. . . time . . . (4 - 1.758) = 2.242 hours

. . . distance (64 km) x (2.242 hr) = 143.47 km

What we're trying to find is the total distance.  Looking back over our work for the two pieces of the trip, we see that we have the distance for each piece now, so all we have to do is addummup.

Total distance = (160 km) + (143.47 km)

<em>Total distance = 303.47 km</em>

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A billiard ball of mass 0.28 kg hits a second, identical ball at a speed of 7.2 m/s and comes to rest as the second ball flies o
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Answer:

(a) -8064 N

(b) 8064 N

Explanation:

(a)

From Newton’s law of motion, Force, F=ma where m is mass and a is acceleration.

Since acceleration is the rate of change of velocity per unit time, then where v is velocity and the subscripts f and I denote final and initial

For the first ball, the mass is 0.28 Kg, final velocity is zero since it finally comes to rest, t is 0.00025 s and initial velocity is given as 7.2 s. Substituting these values we obtain

F=0.28\times \frac {0-7.2}{0.00025}=-8064 N

(b)

For the second ball, the mass is also 0.28 Kg but its initial velocity is taken as zero, the final velocity of the second ball will be equal to the initial velocity of the second ball, that is 7.2 m/s and the time is also same, 0.00025 s. By substitution

F=0.28\times \frac {7.2-0}{0.00025}=8064 N

Here, we prove that action and reaction are equal and opposite

3 0
3 years ago
A car accelerates uniformly from rest at a speed of 1.67 ft s^2 over a distance of 5 yards.What is the acceleration of a car?
Nina [5.8K]

Answer:

a= 17.69 m/s^2

Explanation:

Step one:

given data

A car accelerates uniformly from rest to 23 m/s

u= 0m/s

v= 23m/s

distance= 30m

Step two:

We know that

acceleration= velocity/time

also,

velocity= distance/time

23= 30/t

t= 30/23

t= 1.30 seconds

hence

acceleration= 23/1.30

accelaration= 17.69 m/s^2

5 0
2 years ago
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While John is traveling along a straight interstate
lora16 [44]

Answi dont know

Explanation:

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7 0
3 years ago
Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges
Nostrana [21]

Answer:

r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

E_1 = E_2

Let the position where net field is zero will lie at distance "r" from q1

\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}

now we will have

\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}

now square root both sides

\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}

now we have

\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1

so we have

r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})

8 0
3 years ago
Why are there shadows?
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Shadows are formed when an opaque object or an object that doesn't allow light to pass through is in the way or infront of etc. a source of light.
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