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Alexxx [7]
2 years ago
12

Can someone help me for this question?​

Physics
1 answer:
LUCKY_DIMON [66]2 years ago
4 0

Answer:

A, 0.050 Hz

Explanation:

1) Frequency = speed divided by wavelength

time is 2* 60 = 120 seconds

distance = 6 wave lengths

speed = distance divided by time

speed = 6 wave lengths divided by 120

Hope this helps!

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Two 100 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Bat -10 m/s when th
tatiyna

Answer:

b) -10 m/s

Explanation:

In perfectly elastic head on collisions of identical masses, the velocities are exchanged with one another.

3 0
2 years ago
A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 c
Mashcka [7]

Answer with Explanation:

We are given that

Diameter of pipe,d_1=0.573 m

v_1=13.5 m/s

v_2=5.83 m/s

Volume flow rate of the petroleum along the pipe=Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})

Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s

By equation of continuity

A_1v_1=A_2v_2

\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2

d^2_2=\frac{v_1}{v_2}d^2_1

d_2=\sqrt{\frac{v_1}{v_2}}d_1

d_2=0.573\sqrt{\frac{13.5}{5.83}}

d_2=0.87 m

d_2=0.87\time 100=87 cm

1 m=100 cm

4 0
3 years ago
A car traveling at 50 m/s comes to a stop in 5 seconds. What is the<br> acceleration of this car?
Alexandra [31]

Answer:10

Explanation:

You have to do speed divided by time so your answers 10

4 0
2 years ago
The ms = 0.5645 and mk=0.1113 for a 30 N box on level ground. What level force is needed to start the box moving
nevsk [136]

Answer:

16.935 N

Explanation:

In order to make the box start moving, the level force applied on the box (F) must be greater than the force of static friction that keeps the box at rest, which is equal to

F_f = \mu_s (mg)

where

\mu_s = 0.5645 is the coefficient of static friction

(mg) = 30 N is the weight of the box

Therefore, the condition for F must be:

F \geq F_f\\F \geq \mu_k (mg)=(0.5645)(30 N)=16.935 N

So, the applied force must be greater than this value.

3 0
3 years ago
Two equal forces are applied to a door. The first force is applied at the midpoint of the door, the second force is applied at t
Orlov [11]

Answer:

D) the second at the doorknob

Explanation:

The torque exerted by a force is given by:

\tau = Fdsin \theta

where

F is the magnitude of the force

d is the distance between the point of application of the force and the centre of rotation

\theta is the angle between the direction of the force and d

In this problem, we have:

- Two forces of equal magnitude F

- Both forces are perpendicular to the door, so \theta=90^{\circ}, sin \theta=1

- The first force is exerted at the midpoint of the door, while the 2nd force is applied at the doorknob. This means that d is the larger for the 2nd force

--> therefore, the 2nd force exerts a greater torque

4 0
2 years ago
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