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emmainna [20.7K]
3 years ago
7

A wave created by a certain source travels from medium 1 into another medium 2. It is noticed that its velocity is faster in med

ium 2 than in medium 1. Three students are discussing what happens to the wave as it moves into medium 2. student 1: frequency increases. student 2: wavelength increases. student 3: both change. which student is right?
Physics
1 answer:
poizon [28]3 years ago
3 0
After thorough researching, if a wave created by a certain source travels from medium 1 into another medium 2. It is noticed that its velocity is faster in medium 2 than in medium 1. The student that discussed rightly what happens to the wave as it moves into the medium is the student 1.
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When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam
Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s

b) the change in the combined kinetic energy of the two-car system during this collision

\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))

substitute the value in the equation above

=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J

Hence, the change in combine kinetic energy is -2534.78J

8 0
3 years ago
As part
umka21 [38]

Answer:

Part a)

a = 3.68 m/s^2

Part b)

a = 11.8 m/s^2

Explanation:

Part a)

For force conditions of two blocks we will have

m_1g - T = m_1 a

T - m_2g = m_2 a

now from above equations we have

(m_1 - m_2) g = (m_1 + m_2) a

a = \frac{m_1 - m_2}{m_1 + m_2} g

now we know that

m_1 = \frac{908}{9.8} = 92.65 kg

m_2 = \frac{412}{9.8} = 42 kg

now from above equation we have

a = \frac{92.65 - 42}{92.65 + 42}(9.8)

a = 3.68 m/s^2

Part b)

When heavier block is removed and F = 908 N is applied at the end of the string then we have

F - mg = ma

908 - 412 = 42 a

a = 11.8 m/s^2

8 0
3 years ago
in a softball game, a batter hits the ball at the velocity of 27m/s and angle of 40 shown below. What is the maximum range of th
Nata [24]

Answer:

R = 73.25 m

Explanation:

We have,

Initial speed of the ball is 27 m/s

It is projected at an angle of 40 degrees

The maximum range of the ball is given by :

R=\dfrac{u^2\sin2\theta}{g}

Plugging all the values we get :

R=\dfrac{(27)^2\sin2(40)}{9.8}\\R=73.25\ m

So, the maximum range of the ball is 73.25 m

8 0
3 years ago
Two technicians are discussing the testing of a catalytic converter. Technician A says that a vacuum gauge can be used and obser
Andrej [43]

Answer:

Answer is C. Both technicians A and B.

Refer below.

Explanation:

Two technicians are discussing the testing of a catalytic converter. Technician A says that a vacuum gauge can be used and observed to see if the vacuum drops with the engine at 2500 RPM for 30 seconds. Technician B says that a pressure gauge can be used to check for backpressure. The following technician is correct:

Both technicians A and B

7 0
3 years ago
Read 2 more answers
A force in the negative direction of an x-axis is applied for 17ms to a 0.18 kg ball initially moving at 16.0 m/s in the positiv
Marizza181 [45]
Impulse = Ft=mΔv => Δv = Ft/m = 4.28/0.18 = 23.78 m/s

But,
Δv = v1-v2, where v1 = initial velocity = 16 m/s, v2 = final velocity

Therefore,
v1 - v2 = 23.78 => v2 = v1 - 23.78 => v2 = 16 - 23.78 = -7.78 m/s

The velocity of ball after the force is 7.78 m/s in the direction of the force.

5 0
3 years ago
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