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Likurg_2 [28]
3 years ago
15

Technician A says that valve springs should be kept with the valve at the time of disassembly and tested for squareness and prop

er spring force. Technician B says that the exhaust valve is about 50 percent of the size of the intake valve. Who is right
Physics
2 answers:
otez555 [7]3 years ago
8 0

Answer:

Technician A is right

Explanation:

Technician B is wrong because exhaust gasses are usually under more pressure and hence are less dense, it therefore means that the exhaust valve are usually smaller in size than the intake valve. In order to achieve efficient functioning, the exhaust value is expected to be about 75 to 80 percent of the intake valve. Hence Technician B is wrong to assume that the exhaust valve is about 50% the size of the intake valve. On the other hand, it is correct practice to always leave the valve spring coupled with the valve during disassembly and it should undergo testing for accurate spring force and squareness.

3241004551 [841]3 years ago
6 0

Answer:

Technician A is absolutely correct that the valve spring should be kept with the valve at the time of disassembly and tested for squareness and proper spring force.

Technician B is not correct because the size of exhaust valve is about 75-80% of the intake valve for proper functioning.

Explanation:

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An object that is slowing down in a positive direction must have
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Answer:

Positive velocity and negative acceleration

Explanation:

An object moving in the positive direction has a positive velocity.

An object that's slowing down while moving in the positive direction has a negative acceleration.

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3 years ago
This experiment is to see if water flows faster out of a smaller can or a larger can.
ddd [48]

Answer:rh

Explanation:fjdj

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2 years ago
Consider two objects (Object 1 and Object 2) moving in the same direction on a frictionless surface. Object 1 moves with speed v
d1i1m1o1n [39]

1) A) Object 1 has the greater momentum

The magnitude of the momentum of an object is given by:

p=mv

where

m is the mass of the object

v is its speed

Object 1 has a mass of m_1 = 2m and a speed of v_1 = v, so its momentum is

p_1 = m_1 v_1 = (2m)(v)=2mv

Object 2 has a mass of m_2 = m and a speed of v_2 = \sqrt{2} v, so its momentum is

p_2 = m_2 v_2 = (m)(\sqrt{2} v)=\sqrt{2}mv

So we see that p_1 > p_2, so object 1 has the greater momentum.

2) The objects have the same kinetic energy.

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

Object 1 has a mass of m_1 = 2m and a speed of v_1 = v, so its kinetic energy is

K_1 = \frac{1}{2}m_1 v_1^2 = \frac{1}{2}(2m)(v)^2=mv^2

Object 2 has a mass of m_2 = m and a speed of v_2 = \sqrt{2} v, so its kinetic energy is

K_2 = \frac{1}{2}m_2 v_2^2 = \frac{1}{2}(m)(\sqrt{2} v)^2=mv^2

So we see that K_1 =K_2, so the objects have same kinetic energy

5 0
3 years ago
How does the law of conservation of matter apply to chemical equations? A. The sum of the coefficients on each side of the equat
irinina [24]
The answer is letter D.
5 0
3 years ago
Read 2 more answers
Resonances of the ear canal lead to increased sensitivity of hearing, as we’ve seen. Dogs have a much longer ear canal—5.2 cm—th
krek1111 [17]

Answer:

B. 1700 Hz, 5100 Hz

Explanation:

Parameters given:

Length of ear canal = 5.2cm = 0.052 m

Speed of sound in warm air = 350 m/s

The ear canal is analogous to a tube that has one open end and one closed end. The frequency of standing wave modes in such a tube is given as:

f(m) = m * (v/4L)

Where m is an odd integer;

v = velocity

L = length of the tube

Hence, the two lowest frequencies at which a dog will have increased sensitivity are f(1) and f(3).

f(1) = 1 * [350/(4*0.052)]

f(1) = 1682.69 Hz

Approximately, f(1) = 1700 Hz

f(3) = 3 * [350/(4*0.052)]

f(3) = 5048 Hz

Approximately, f(3) = 5100 Hz

7 0
3 years ago
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