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Veronika [31]
3 years ago
14

Rahul runs a distance of 12

Physics
1 answer:
Radda [10]3 years ago
4 0

Answer:

Rita runs faster

Explanation:

To obtain the answer to the question given, we determine the speed of Rita. This can be obtained as follow:

Distance = 1.6 km

Time = 15 mins

Speed =?

Next, we shall convert 15 mins to hour. This can be obtained as follow:

60 mins = 1 h

Therefore,

15 mins = 15 mins × 1 h / 60 mins

15 mins = 0.25 h

Thus, 15 mins is equivalent to 0.25 h.

Finally, we shall determine speed of Rita as follow:

Distance = 1.6 km

Time = 0.25 h

Speed =?

Speed = distance / time

Speed = 1.6 / 0.25

Speed = 6.4 Km/h

Thus Rita's speed is 6.4 Km/h

Comparing the speed of Rahul and Rita

Rahul's speed = 4.5 km/h

Rita's speed = 6.4 Km/h

We can see that Rita's speed is greater than Rahul's speed. This simply indicates that Rita runs faster than Rahul.

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A 1.1-kg object is suspended from a vertical spring whose spring constant is 120 N/m. (a) Find the amount by which the spring is
andriy [413]

Answer:

e = 0.0898m

v = 2.07m/s

Explanation:

a) According to Hooke's law

F = ke

e is the extension

k is the spring constant

Since F = mg

mg = ke

e = mg/k

Substitute the given value

e = 1.1(9.8)/120

e = 10.78/120

e = 0.0898m

Hence it is stretched by 0.0898m from its unstrained length

2) Total Energy = PE+KE+Elastic potential

Total Energy = mgh +1/2mv²+1/2ke²

Substitute the given value

5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²

Solve for v

5.0 = 2.156+0.55v²+0.48338

5.0-2.156-0.48338= 0.55v²

2.36 =0.55v²

v² = 2.36/0.55

v² = 4.29

v ,= √4.29

v = 2.07m/s

Hence the required velocity is 9.28m/s

4 0
3 years ago
An ideal gas is enclosed in a piston, and 1600 J of work is done on the gas. As this happens, the internal energy of the gas inc
Phantasy [73]

Answer:

- 1100 J heat flows out

Explanation:

dW = - 1600 J (as work is done on the gas)

dU = 500 J

dQ = ?

According to the first law of thermodynamics

dQ = dU + dW

dQ = 500 - 1600

dQ = - 1100 J

As heat is negative so it flows out.

3 0
2 years ago
If I can travel 20m in 18 seconds how far can I go in 10 minutes?
kifflom [539]

Answer:

36 s

Explanation:

20 m = 18 s

10 m = ?

20 × 18 = 360÷ 10

= 36 sec

5 0
2 years ago
A pendulum is made by attaching a sphere to the end of a string of negligible mass and it oscillates when released from an angle
Rudik [331]

Answer:

Angular momentum conservation and kinetic energy. Torsional ... motion-observation of what a given object does in relation to other objects. Frames of ... shows that the rectangular and spherical polar coordinates are related as follows: ... 2mo, which are connected by a string over a pulley of negligible mass and prevented.

Explanation:

8 0
3 years ago
A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1
Archy [21]

Answer:

a)  y= 3.5 10³ m, b)   t = 64 s

Explanation:

a) For this exercise we use the vertical launch kinematics equation

Stage 1

          y₁ = y₀ + v₀ t + ½ a t²

          y₁ = 0 + 0 + ½ a₁ t²

Let's calculate

         y₁ = ½ 16 10²

         y₁ = 800 m

At the end of this stage it has a speed

        v₁ = vo + a₁ t₁

        v₁ = 0 + 16 10

        v₁ = 160 m / s

Stage 2

        y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

        y₂ = 800 + 150 5 + ½ 11 5²

        y₂ = 1092.5 m

Speed ​​is

        v₂ = v₁ + a₂ t

        v₂ = 160 + 11 5

        v₂ = 215 m / s

The rocket continues to follow until the speed reaches zero (v₃ = 0)

         v₃² = v₂² - 2 g y₃

         0 = v₂² - 2g y₃

         y₃ = v₂² / 2g

         y₃ = 215²/2 9.8

         y₃ = 2358.4 m

The total height is

          y = y₃ + y₂

          y = 2358.4 + 1092.5

          y = 3450.9 m

           y= 3.5 10³ m

b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

          v₃ = v₂ - g t₃

          v₃ = 0

          t₃ = v₂ / g

          t₃ = 215 / 9.8

          t₃ = 21.94 s

The time to climb is

          t_{s} = t₁ + t₂ + t₃

          t_{s} = 10+ 5+ 21.94

          t_{s} = 36.94 s

The time to descend from the maximum height is

          y = v₀ t - ½ g t²

When it starts to slow down it's zero

         y = - ½ g t_{b}²

         t_{b}  = √-2y / g

         

        t_{b} = √(- 2 (-3450.9) /9.8)

        t_{b} = 26.54 s

Flight time is the rise time plus the descent date

        t = t_{s} + t_{b}

        t = 36.94 + 26.54

        t =63.84 s

        t = 64 s

3 0
2 years ago
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