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3 years ago

11

On a cold day (0C) you observe a person far away from you hit a piece of metal with a hammer. Then 0.75 seconds later you hear

the sound of the hammer hitting the metal. How far are you from the other person?

1 answer:

bearhunter [10]3 years ago

3 0

Answer:

248.48 meters

Explanation:

When there is low temperature than the normal room temperature, the speed of sound in air decreases because at lower temperatures the movement of air molecules through which the sound travels gets decreases.

Speed of sound at 0⁰ is found to be approximately 331.3 m/s.

The question says that sound was heard by the observer 0.75 seconds later after hammer hit the metal.

As we know, Distance traveled = Speed × Time taken

So, distance traveled by sound in 0.75 seconds = 331.3 × 0.75 = 248.48 m

So, the observer was standing 248.48 meters away from the other person hitting the hammer.

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The spring in the muzzle of a child's spring gun has a spring constant of 730 N/m. To shoot a ball from the gun, first the sprin

Korolek [52]

Answer:

a. V=11.84 m/s

b.x=0.052m

Explanation:

a).

Given

$K=730 N/m$ , $m=0.053kg$ , $h=1.90m$ .

$v_f^2=v_i^2+2*g*h$

$v_i^2=2*g*h=2*9.8m/s^2*1.9m$

$v_i=\sqrt{2*9.8m/s^2*1.9m}=\sqrt{37.24 m^2/s^2}$

$v_i=6.1 m/s$

$v_i=V*sin(31)$

$V=\frac{v_i}{sin(31)}=\frac{6.1m/s}{sin(31)}$

$V=11.84 m/s$

b).

$K_k=\frac{1}{2}*K*x^2$

No friction on the ball so:

$x^2=\frac{2*K_k}{K}$

$x=\sqrt{\frac{2*0.053kg*9.8m/s^2*1.9m}{730N/m}}$

$x=\sqrt{2.7x10^{-3}m^2}=0.052m$

5 0

3 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

We can use the following equation of motion to calculate how far the car has travel since braking to stop

$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

8 0

3 years ago

You and your friend are having a discussion about weight. He claims he weighs less on the 100th floor of a building that he does

balu736 [363]

No the acceleration of gravity on earth is still 9.8ms-2 unless he was on a different planet

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3 years ago

HELP PLEASE 15 POINTS !!!

Nookie1986 [14]

B(I)84.0*10^5J
B(ll)1400W

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3 years ago

A 3-kW resistance heater in a water heater runs for 3 hours to raise the water temperature to the desired level. Determine the a

Kipish [7]

Answer:

Energy, 9 kWh or 32400 kJ

Explanation:

Given that,

The power of heater, P = 3 kW

It runs for 3 hours to raise the water temperature to the desired level. We need to find the amount of electric energy used. We know that the electrical power of an object is given by total energy delivered per unit time. It is given by :

$P=\dfrac{E}{t}$

$E=P\times t$

$E=3\ kW\times 3\ h$

E = 9 kWh

Since, 1 kWh = 3600 kJ

E = 32400 kJ

So, the amount of electric energy used is 9 kWh or 32400 kJ. Hence, this is the required solution.

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