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Sati [7]
4 years ago
12

Why are you teachers regarded as professionals​

Physics
1 answer:
Murljashka [212]4 years ago
3 0

Answer:

coz teaching is their profession.

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A truck moving at 13.3 m/s hits a concrete wall. As a result of the collision, a 6-kg wrench moves forwards and strikes the wall
Naddik [55]

Answer:

Explanation:

The velocity of the wrench must be equal to the velocity of the truck . So momentum of the wrench before it hits the wall

= mv = 6 x 13.3 = 79.8 kg m /s

If resisting force of wall be F , impulse on the wrench = F x time

= F x .07

Impulse = change in momentum of the wrench = mv - 0 = mv = 79.8 kgm/s

So F x .07 = 79.8

F = 1140 N .

8 0
3 years ago
When checking for noncondensables inside a recovery cylinder why should the technician allow the temperature of the cylinder to
77julia77 [94]
Before taking a pressure reading, it is necessary for the technician to first allow the temperature of the cylinder to stabilize to room temperature because a comparison with a temperature-pressure chart is only valid and true when both temperature and pressure of the refrigerant are stable. 
3 0
3 years ago
20- A gram of distilled water at 4° C:
Usimov [2.4K]

Answer:

D. will decrease slightly in volume when heated to 6° C

Explanation:

6 0
3 years ago
A stationary 500 kg tank fires a 20 kg miegile at 100 m/s. What is the velocity of the tank after the missile is fired? Assume t
dedylja [7]

Answer:

v₁ = 4 [m/s].

Explanation:

This problem can be solved by using the principle of conservation of linear momentum. Where momentum is preserved before and after the missile is fired.

P=m*v

where:

P = linear momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

(m_{1}*v_{1})=(m_{2}*v_{2})

where:

m₁ = mass of the tank = 500 [kg]

v₁ = velocity of the tank after firing the missile [m/s]

m₂ = mass of the missile = 20 [kg]

v₂ = velocity of the missile after firing = 100 [m/s]

(500*v_{1})=(20*100)\\v_{1}=2000/500\\v_{1}=4[m/s]

8 0
3 years ago
A student produces a power of p = 0.87 kw while pushing a block of mass m = 75 kg on an inclined surface making an angle of θ =
Paul [167]

When block is pushed upwards along the inclined plane

the net force applied on the block will be given as

F_{net} = mg sin\theta + \mu_k mg cos\theta

here we know that

m = 75 kg

\theta = 8.5 degree

\mu_k = 0.16

now plug in all values into this

F_{net} = 75\times 9.8 sin8.5 + 0.16 \times 75\times 9.8 cos8.5

F = 225 N

now for finding the power is given as

P = Fv

0.87 \times 10^3 = 225 \time v

v = \frac{870}{225} = 3.87 m/s

6 0
3 years ago
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