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3 years ago

Calculate the kinetic energy of a particle with a

1 answer:

6 0

K=1/2×mv²
m=3.34×10^-27
V=2.89×10^5
k=1/2 × 3.34×10^-27 × (2.89×10^5)²= 1/2 ×3.34×10^-27 × 8.3521×10^10=
1/2 × 27.9 × 10^-17= 13.95 × 10^-17 joule

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Consider the case of the car starting at rest and accelerating forward. A. Since the air inside the car is not leaking out, it m

Alchen [17]

Answer:

The force originates from the tires. Tires accelerate the car. The car interior in turn accelerates the air inside the car by providing a pushing force. Thus it is the car read windows, seats and side windows that push the air forward causing it to accelerate.

Explanation:

The air inside the car is being forced forward by the car interior. This includes the rear windows, seats, side windows etc.

This accelerating force originates from the tires, and makes the car interior accelerate with the car. The air inside (since it has no direct connection to the tires) is then pushed forward by the windows, seats etc.

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3 years ago

If a wheel rotates 5 times in 90 seconds, what is the period and frequency

Sati [7]

Answer:

i think it should be 18

Explanation:

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3 years ago

If 27 J of work are needed to stretch a spring from 15 cm to 21 cm and 45 J are needed to stretch it from 21 cm to 27 cm, what i

kupik [55]

Answer:

9 cm.

Explanation:

The energy used for stretch the spring from $15 cm$ to $21 cm$ will be , $E_{1}=27J$

The energy used for stretch the spring from $21 cm$ to $27 cm$ will be , $E_{2}=45J$

using the energy of spring formula ,we find that

$27 = \frac{1}{2}K((21-L^{2})-(15-L^{2}))$

$45 = \frac{1}{2}K((27-L^{2})-(21-L^{2}))$

Dividing both the equation will get,

$\frac{3}{5}=\frac{(21-L)^{2}-(15-L)^{2}}{(27-L)^{2}-(21-L)^{2}}\\5((21-L)^{2}-(15-L)^{2})=3((27-L)^{2}-(21-L)^{2})\\3(729 - 54L + L^{2}- 441 + 42L - L^{2} ) = 5(441 - 42L + L^{2} - 225 + 30L - L^{2} )\\3(288 - 12L) = 5(216 - 12L)\\24L = 216\\L = 9 cm$

Therefore, the natural length of the spring is, 9 cm.

4 0

3 years ago

A mass m = 75 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.2 m and finally

dalvyx [7]

Answer:

The velocity is 19.39 m/s

Solution:

As per the question:

Mass, m = 75 kg

Radius, R = 19.2 m

Now,

When the mass is at the top position in the loop, then the necessary centrifugal force is to keep the mass on the path is provided by the gravitational force acting downwards.

$F_{C} = F_{G}$

$\frac{mv^{2}}{R} = mg$

where

v = velocity

g = acceleration due to gravity

$v = \sqrt{2gR} = \sqrt{2\times 9.8\times 19.2} = 19.39\ m/s$

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3 years ago

A 1500 W electric heater is plugged into the outlet of a 120 V circuit that has a 20 A circuit breaker. You plug an electric hai

Mila [183]

Answer:

Explanation:

Current drawn by electric heater = power/volt =1500/120 = 12.5 A.

current drawn by hair drier at 600 watt = 600/120 =5 A

current drawn by hair drier at 900 watt = 900/120 = 7.5 A.

Total current drawn by heater and hair drier used at 900 watt

= 12.5 + 7.5 = 20 A

Breaking current =20 A

So fuse will trip at this point .

3 0

3 years ago