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3 years ago

What is the speed of a sound wave that takes 0.5 s to travel 750 m?

Physics

1 answer:

kvasek [131]3 years ago

4 0

Answer:

C. 1500.

Explanation:

750 / .5 = 1500.

Hope this helps & best of luck!

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A car with an initial velocity of 5.0 is acelerated at 3.0for 4.0seconds what os the velocity of tje car ofter the 4.0seconds

sasho [114]

Vf =Vi+at
Vf= 5 + 3*4
Vf=5+12
Vf=17m/sec

6 0

3 years ago

A pirate ship shoots a cannon towards an oncoming ship. The cannon ball has a mass of 25 kg and the ship has a mass of 2500 kg.

mars1129 [50]

Answer:

25 m/s in the opposite direction with the ship recoil velocity.

Explanation:

Assume the ship recoil velocity and velocity of the cannon ball aligns. By the law of momentum conservation, the momentum is conserved before and after the shooting. Before the shooting, the total momentum is 0 due to system is at rest. Therefore, the total momentum after the shooting must also be 0:

$m_sv_s + m_bv_b = 0$

where $m_s = 2500 kg, m_b = 25 kg$ are masses of the ship and ball respectively. $v_s = 0.25 m/s, v_b$ are the velocities of the ship and ball respectively, after the shooting.

$2500*0.25 + 25*v_b = 0$

$25v_b = -2500*0.25$

$v_b = -2500*0.25/25 = -25 m/s$

So the cannon ball has a velocity of 25 m/s in the opposite direction with the ship recoil velocity.

3 0

3 years ago

Please I need your help

frez [133]

1.14 km = Distance
2.30 m/s = Speed
5.12 cm/s2 = Speed
6.150 mph = Distance
8.3.2 sec = Speed
9.25 ft = Distance

5 0

3 years ago

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the horizontal. The negligent

tensa zangetsu [6.8K]

Answer:

(a): The car's relative position to the base of the cliff is x= 32.52m.

(b): The lenght of the car in the ir is tfall= 1.78 sec.

Explanation:

Vo= 0

V= ?

d= 50m

h= 30m

a= 4 m/s²

t= √(2*d/a)

t= 5 sec

V= a*t

V= 20 m/s

Vx= V * cos(24º)

Vx= 18.27 m/s

Vy= V* sin(24º)

Vy= 8.13 m/s

h= Vy*t + g*t²/2

clearing t:

tfall= 1.78 sec (b)

x= Vx * tfall

x= 32.52 m (a)

4 0

3 years ago

A lady bug is sitting on the bottom of a can while you twirl it overhead on a string that is 65.0

MA_775_DIABLO [31]

The linear speed of the ladybug is 4.1 m/s

Explanation:

First of all, we need to find the angular speed of the lady bug. This is given by:

$\omega=\frac{2\pi}{T}$

where

T is the period of revolution

The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:

T = 1 s

Therefore, the angular speed is

$\omega=\frac{2\pi}{1 s}=6.28 rad/s$