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mixer [17]
3 years ago
5

47. A car travels 85 km in the first half hour of a trip. The car continues to travel for 2 more hours and travels 200 km. What

was the average speed of the car for the trip? a. 39 km/h b. 95 km/h c. 114 km/h d. 285 km/h
Physics
1 answer:
Otrada [13]3 years ago
3 0

Answer: 114 km/h

Explanation:

The formula for determining average speed is expressed as

Average speed = total distance/total time

The car travels 85 km in the first half hour of a trip. The car continues to travel for 2 more hours and travels 200 km. It means that the total distance that the car travels is

85 + 200 = 285 km

The total time spent by the car is

0.5 + 2 = 2.5 hours

Therefore,

Average speed = 285/2.5 = 114 km/h

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Using a rope that will snap if the tension in it exceeds 356 N, you need to lower a bundle of old roofing material weighing 478
klasskru [66]

Answer:

a) 2.5 m/s²

b) 6.12 m/s

Explanation:

Tension of rope = T = 356N

Weight of material = W = 478 N

Distance from the ground = s = 7.5 m

Acceleration due to gravity = g = 9.81 m/s²

Mass of material = m = 478/9.81 = 48.72

Final velocity before the bundle hits the ground = v

Initial velocity = u = 0

Acceleration experienced by the material when being lowered = a

a) W-T = ma

⇒478-356 = 48.72×a

\Rightarrow \frac{122}{48.72} = a

⇒a = 2.5 m/s²

∴ Acceleration achieved by the material is 2.5 m/s²

b) v²-u² = 2as

⇒v²-0 = 2×2.5×7.5

⇒v² = 37.5

⇒v = 6.12 m/s

∴ Velocity of the material before hitting the ground is 6.12 m/s

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The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9
kirill [66]

Answer:

Speed of the car 1 =V_1=8.98m/s

Speed of the car 2 =V_2=17.96m/s

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

KE =\frac{1}{2}mv^2

where,

m = mass of the body

v = velocity of the body

thus,

\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

2M_2V_1^2=0.5\times M_2V_2^2

or

2V_1^2=0.5\times V_2^2

or

4V_1^2= V_2^2

or

2V_1= V_2  .................(1)

also,

\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2

or

\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2

or

2(V_1+9.0)^2=(2V_1+9.0)^2

or

\sqrt{2}(V_1+9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)

or

(2V_1-\sqrt{2}V_1)=(12.72-9.0)

or

(0.404V_1)=(3.72)

or

V_1=8.98m/s

and, from equation (1)

V_2=2\times 8.98m/s = 17.96m/s

Hence,

Speed of car 1 =V_1=8.98m/s

Speed of car 2 =V_2=17.96m/s

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2 years ago
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