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dezoksy [38]
3 years ago
15

The diagram shows monochromatic light passing through two openings.

Physics
1 answer:
geniusboy [140]3 years ago
4 0

Answer: constructive interference in which waves strengthen each other

Explanation:

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Temperature is a measure of work.<br> True<br> False
Veseljchak [2.6K]
The answer is false.
5 0
3 years ago
A 50 kg laboratory worker is exposed to 30 mJ of neutron radiation with an RBE of 10. Part A What is the dose in mSv
lys-0071 [83]

Answer:

The dose is 6 mSV

Explanation:

The absorbed dose (in gray - Gy) is the amount of energy that ionizing radiation deposits per unit mass of tissue. That is,

Absorbed dose = Energy deposited / Mass

while Dose equivalent (DE) (in Seivert -Sv) is given by

DE = Absorbed dose × RBE (Relative biological effectiveness)

First, we will determine the Absorbed dose

From the question, Energy deposited = 30mJ and Mass = 50kg

From,

Absorbed dose = Energy deposited / Mass

Absorbed dose = 30mJ/50kg

Absorbed dose = 0.6 mGy

Now, for the Dose equivalent (DE)

DE = Absorbed dose × RBE

From the question, RBE = 10

Hence,

DE = 0.6mGy × 10

DE = 6 mSv

5 0
3 years ago
What is the external-internal pressure difference when the diver's lungs are at a depth of 6.1m (about 20ft)? assume that the di
Nesterboy [21]
Snorkel connects the diver's lungs with the air above the water. Because of this, the pressure inside the diver's lungs is the same as the atmospheric pressure.
The static pressure of a fluid is given with this equation:
P=\rho gh
The density of water is 1000kg/m^3.
P=1000\cdot 9.81 \cdot 6.1=59841$Pa
Standart atmospheric pressure is 101<span>325Pa. 
</span>The difference is:
\Delta P=101325-59841=41484 $Pa

8 0
3 years ago
Free body diagram for a 4kg
Svetradugi [14.3K]
999999999999999999999999999999999999999999999999999999
4 0
3 years ago
A 75 kg soccer player jumps vertically upwards and heads the 0.45 kg ball as it is descending vertically with a speed of 24 m/s.
Vikentia [17]

Answer:

31.66 m/s

Explanation:

mass of player, M = 75 kg

mass of ball, m = 0.45 kg

initial velocity of player, U = + 4 m/s

initial velocity of ball, u = - 24 m/s

Let the final speed of player is V and the ball is v.

use conservation of momentum

Momentum before collision = momentum after collision

75 x 4 - 0.45 x 24 = 75 x V + 0.45 x v

289.2 = 75 V + 0.45 v    .... (1)

As the collision is perfectly elastic, coefficient of restitution,e = 1

So, e=\frac{v_{1}-v_{2}}{u_{2}-u_{1}}

V - v = u - U

V - v = -24 - 4 = - 28

V = v - 28, put this value in equation (1), we get

289.2 = 75 (v - 28) + 045 v

289.2 = 75 v - 2100 + 0.45 v

2389.2 = 75.45 v

v = 31.66 m/s

Thus, the velocity of ball after collision is 31.66 m/.

7 0
3 years ago
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