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3 years ago

PLEASE HELPP ASAP Which of the following statements correctly compares the speed of waves?

Physics

1 answer:

Dvinal [7]3 years ago

7 0

Answer: A

Explanation: Since particles in solids are more tightly bound together than the particles in gases, mechanical waves can propagate more quickly through solids. ... Since electromagnetic waves do not require a medium to pass through, they are faster in matter that has fewer particles.

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If the half-life of a 2.0 gram sample of a radionuclide is 15 hours, then the half-life of a 1.0 gram sample of the same radionu

Lerok [7]

7.5 hours or 450 minutes. 15/2=7.5

5 0

3 years ago

The mass of this astronaut on the moon is 84kg. If the astronaut is on the moon for 3 weeks then returns home, his mass will be

saul85 [17]

Answer:

84kg

Explanation:

If the mass of an astronaut in on the moon is 84kg on the moon. After the 3weeks in returning home, his weight will not change. The mass of the astronaut will remain the same which is 84kg. This is so because the acceleration due to gravity experienced by the body remains the same both in the earth and in moon.

7 0

3 years ago

What is the energy of the 30 kg skateboarder at 2 m off the ground traveling at 3 m/s?

zubka84 [21]

1) The kinetic energy of an object is given by:

where m is the object's mass and v its speed.

By using this equation, we find the initial kinetic energy of the skateboarder:

and the final kinetic energy as well:

So, her change in kinetic energy is

2) The work-energy theorem states that the work done to increase the speed of an object is equal to the variation of kinetic energy of the object:

Therefore, the work done by the skateboarder is

8 0

3 years ago

A superball with a mass m = 61.6 g is dropped from a height h = [02]____________________ m. It hits the floor and then rebounds

aleksklad [387]

<em>There is not enough data to solve the problem, but I'm assuming the initial height as h = 10 m for the question to have a valid answer and the student can have a reference to solve their own problem</em>

Answer:

(a) $\Delta P=1.67 \ kg.m/s$

(b) $\Delta P=0.86\ m/s$

Explanation:

<u>Change of Momentum</u>

The momentum of a given particle of mass m traveling at a speed v is given by

$P=m.v$

When this particle changes its speed to a value v', the new momentum is

$P'=m.v'$

The change of momentum is

$\Delta P=m.v'-m.v$

$\Delta P=m.(v'-v)$

Defining upward as the positive direction, we'll compute the change of momentum in two separate cases.

(a) The initial height of the superball of m=61.6 gr = 0.0616 Kg is set to h= 10 m. This information leads us to have the initial potential energy of the ball just after it's dropped to the floor:

$U=m.g.h=0.0616\cdot 9.8\cdot 10 =6.0368\ J$

This potential energy is transformed into kinetic energy just before the collision occurs, thus

$\displaystyle \frac{1}{2}mv^2=6.0368$

Solving for v

$\displaystyle v=\sqrt{\frac{6.0368\cdot 2}{0.0616}}$

$v=-14\ m/s$

This is the speed of the ball just before the collision with the floor. It's negative because it goes downward. Now we'll compute the speed it has after the collision. We'll use the new height and proceed similarly as above. The new height is

$h'=88.5\% (10)=8.85\ m$

The potential energy reached by the ball at its rebound is

$U'=m.g.h'=0.0616\cdot 9.8\cdot 8.85 =5.342568\ J$

Thus the speed after the collision is

$\displaystyle v'=\sqrt{\frac{5.342568\cdot 2}{0.0616}}$

$v'=13.17\ m/s$

The change of momentum is

$\Delta P=0.0616\cdot (13.17+14)$

$\Delta P=1.67 \ kg.m/s$

(b) If the putty sticks to the floor, then v'=0

$\Delta P=0.0616\cdot (0+14)$

$\Delta P=0.86\ m/s$

3 0

4 years ago

A supersonic aircraft flies at 3 km altitude at a speed of 1000 m/s on a standard day. How long after passing directly above a g

Gre4nikov [31]

Answer:

<h3>After 3seconds</h3>

Explanation:

A supersonic aircraft flies at 3 km altitude at a speed of 1000 m/s on a standard day. How long after passing directly above a ground observer is the sound of the aircraft heard by the ground observer

Using the formula for calculating speed expressed as;

Speed = Distance/Time

Given;

Distance = 3km = 3000m

Speed = 1000m/s

Required

How long after passing directly above a ground observer is the sound of the aircraft heard by the ground observer (Time)

From the formula;

Time = Distance/speed

Time = 3000/1000

Time = 3seconds

Hence the sound of the aircraft is heard after 3 seconds

7 0

3 years ago