What is the speed of a wave if it is 8 meters long and has a frequency of 3 hz?

Avery likes to walk 30 feet in 10<br>seconds. What is her speed?

Sliva [168]

Answer:

If she is walking 30 feet every 10 seconds, that means she is walking 180 feet per minute. Multiply that by the 60 minutes in an hour, means she walks (180x60)= 10,800 feet an hour.

She walks 3 feet a second.

She walks 180 feet a minute.

She walks 10,800 feet an hour.

Explanation:

6 0

3 years ago

Tubby and his twin brother Libby have a combined mass of 200 kg and are zooming along in a 100 kg amusement park bumper car at 1

harkovskaia [24]

Answer: 14.1 m/s

Explanation:

We can solve this with the Conservation of Linear Momentum principle, which states the initial momentum $p_{i}$ (before the elastic collision) must be equal to the final momentum $p_{f}$ (after the elastic collision):

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=m_{1}V_{i} + m_{2}U_{i}$

$p_{f}=m_{1}V_{f} + m_{2}U_{f}$

Where:

$m_{1}=200 kg +100 kg=300 kg$ is the combined mass of Tubby and Libby with the car

$V_{i}=10 m/s$ is the velocity of Tubby and Libby with the car before the collision

$m_{2}=25 kg + 100 kg=125 kg$ is the combined mass of Flubby with its car

$U_{i}=0 m/s$ is the velocity of Flubby with the car before the collision

$V_{f}=4.12 m/s$ is the velocity of Tubby and Libby with the car after the collision

$U_{f}$ is the velocity of Flubby with the car after the collision

So, we have the following:

$m_{1}V_{i} + m_{2}U_{i}=m_{1}V_{f} + m_{2}U_{f}$ (2)

Finding $U_{f}$ :

$U_{f}=\frac{m_{1}(V_{i}-V_{f})}{m_{2}}$ (3)

$U_{f}=\frac{300 kg(10 m/s-4.12 m/s)}{125 kg}$ (4)

Finally:

$U_{f}=14.1 m/s$

8 0

3 years ago

In a vacuum, two particles have charges of q1 and q2, where q1 = +3.11 μC. They are separated by a distance of 0.241 m, and part

lesya [120]

Answer:

Charge of particle 2, $q_2=-7.13\ \mu C$

Explanation:

Given that,

Charge 1, $q_1=3.11\ \mu C=3.11\times 10^{-6}\ C$

The distance between charges, r = 0.241 m

Force experienced by particle 1, F = 3.44 N

We need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$q_2=\dfrac{Fr^2}{kq_1}$

$q_2=\dfrac{3.44\times (0.241)^2}{9\times 10^9\times 3.11\times 10^{-6}}$

$q_2=7.13\times 10^{-6}\ C$

or

$q_2=7.13\ \mu C$

So, the magnitude of electric charge 2 is $q_2=7.13\ \mu C$ . Since, the force is attractive then the magnitude of charge 2 must be negative.

5 0

3 years ago

Explain how force and surface area affect the applied pressure.

vlabodo [156]

Answer:

force and surface area are two factors affecting pressure on solids

more the force you apply, more will be the pressure

pressure and force are directly proportional meaning if Force is greater, pressure will also be greater

more the surface area of the solid less will be the pressure

surface area and pressure are inversely proportional meaning if surface are is big, pressure will be less, surface area small, pressure will be greater

6 0

3 years ago

Read 2 more answers

Three sharks are located at the vertices of an equilateral triangle whose side equals LL. They all start moving simultaneously w

salantis [7]

Answer:

They will never meet.

Explanation:

Given that three sharks are located at the vertices of an equilateral triangle whose side equals LL. They all start moving simultaneously with a constant speed vv, with the first shark heading constantly toward the second, the second toward the third, and the third toward the first. How soon will the sharks converge at one point ?

Since the three sharks are located at the vertices of an equilateral triangle whose side equals, and the first shark is moving toward the second, the second is moving toward the third, and the third shark is moving toward the first, the movement of the three shark will form a circular motion.

Since the three sharks are moving at a constant speed, the three sharks will never meet because the distances between them are also the same.

The three sharks will therefore continue to move till infinity.

6 0

3 years ago