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lina2011 [118]
3 years ago
10

If the energy in a mechanical wave increases or decreases, what is also going to increase or decrease? A wavelength B amplitude

с frequency d refraction​
Physics
1 answer:
lisov135 [29]3 years ago
8 0

Answer:

i think wave lengh not 100% sure tho

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A 4 kg object is moving at a speed of 5 m/sec. How much kinetic energy does the object have? A. 10 joules B. 20 joules C. 50 jou
sp2606 [1]

Answer:

A. 10 joules

Explanation:

Given parameters:

Mass of object  = 4kg

Speed  = 5m/s

Unknown:

Kinetic energy of the the object  = ?

Solution:

Kinetic energy is the energy due to a moving body.

So;

           mathematically;

    K.E  = \frac{1}{2} x m v²

m is the mass

v is the velocity

  K.E  =  \frac{1}{2}  x 4 x 5  = 10J

5 0
3 years ago
A weather forecaster uses a computational model on a Monday to predict the weather on Friday. Why might that forecast change? (1
grandymaker [24]

Answer:

(d) is your answer for your question

7 0
3 years ago
A positive charge is placed within the sphere without touching it. You grounded the sphere by touching it with your finger. Then
zhuklara [117]

Answer:

Negative

Explanation:

If a negatively charged object is used to charge a neutral object by induction, then the neutral object will acquire a positive charge. And if a positively charged object is used to charge a neutral object by induction, then the neutral object will acquire a negative charge:

We have a neutral sphere

1) Electrons are attracted to the positive charge

2) Electrons enter the sphere from ground, attrated to the + charge in the sphere

3) The sphere has an excess of e- having entered from the ground

4) Electrons redistribute uniformly

3 0
4 years ago
Use the worked example above to help you solve this problem. A car traveling at a constant speed of 27.9 m/s passes a trooper hi
Sidana [21]
H h h huh h jvjvhvuvuvuvyv
7 0
4 years ago
A grapefruit falls from a tree and hits the ground .75 seconds later. How far did the grapefruit drop? What was its speed?
Ivanshal [37]
1) The grapefruit is in free fall, so it moves by uniformly accelerated motion, with constant acceleration g=9.81 m/s^2. Calling h its height at t=0, the height at time t is given by
h(t)=h- \frac{1}{2}gt^2
We are told thatn when t=0.75 s the grapefruit hits the ground, so h(0.75 s)=0. If we substitute these data into the equation, we can find the initial height h of the grapefruit:
0=h- \frac{1}{2}gt^2
h= \frac{1}{2}gt^2= \frac{1}{2}(9.81 m/s^2)(0.75 s)^2=2.76 m

2) The speed of the grapefruit at time t is given by
v(t)=v_0 +gt
where v_0=0 is the initial speed of the grapefruit. Substituting t=0.75 s, we find the speed when the grapefruit hits the ground:
v(0.75 s)=gt=(9.81 m/s^2)(0.75 s)=7.36 m/s
3 0
4 years ago
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