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Studentka2010 [4]
3 years ago
5

Which option below correctly compares the average annual dose of background radiation to the dose liked to an increased cancer r

isk?
A. The average annual dose of background radiation is 250 times larger than the dose linked to increased cancer risk.

B. The average annual dose of background radiation is 250 times smaller than the dose linked to increased cancer risk.

C. The average annual dose of background radiation is 100,000 times smaller than the dose linked to increased cancer risk.

D. The average annual dose of background radiation is 100,000 times larger than the dose linked to increased cancer risk.

Physics
1 answer:
julia-pushkina [17]3 years ago
6 0
The answer for this question would be choice "<span>B. The average annual dose of background radiation is 250 times smaller than the dose linked to increased cancer risk."

You only have to compare 4.0 x 10^-4 and 1.0 x 10^-1. And if you can observe carefully, when you try to multiply the average annual dose of background radiation by 250, you would get 0.1 which is equivalent to the amount of annual dose linked to increased cancer risk. Therefore, the answer is B.</span>
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500 nm

Explanation:

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10

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A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a
KatRina [158]

Answers:

a) 0.80 kg

b) 2.12 s

c) 1.093 m/s^{2}

Explanation:

We have the following data:

k=7 N/m is the spring constant

A=12.5 cm \frac{1 m}{100 cm}=0.125 m is the amplitude of oscillation

V=32 cm/s=0.32 m/s is the velocity of the block when x=\frac{A}{2}=0.0625 m

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

U_{o}+K_{o}=U_{f}+K_{f} (1)

Where:

U_{o}=k\frac{A^{2}}{2} is the initial potential energy

K_{o}=0  is the initial kinetic energy

U_{f}=k\frac{x^{2}}{2} is the final potential energy

K_{f}=\frac{1}{2} m V^{2} is the final kinetic energy

Then:

k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2} (2)

Isolating m:

m=\frac{k(A^{2}-x^{2})}{V^{2}} (3)

m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}} (4)

m=0.80 kg (5)

<h3>b) Period</h3>

The period T is given by:

T=2 \pi \sqrt{\frac{m}{k}} (6)

Substituting (5) in (6):

T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}} (7)

T=2.12 s (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration a_{max} is when the force is maximum F_{max}, as well :

F_{max}=m.a_{max}=k.x_{max} (9)

Being x_{max}=A

Hence:

m.a_{max}=kA (10)

Finding a_{max}:

a_{max}=\frac{kA}{m} (11)

a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg} (12)

Finally:

a_{max}=1.093 m/s^{2}

5 0
3 years ago
On a hot day, the deck of a small ship reaches a temperature of 48
AlekseyPX

The final temperature of the seawater-deck system is 990°C.

<h3>What is heat?</h3>

The increment in temperature adds up the thermal energy into the object. This energy is Heat energy.

The deck of a small ship reaches a temperature Ti= 48.17°C seawater on the deck to cool it down. During the cooling, heat Q =3,710,000 J are transferred to the seawater from the deck. Specific heat of seawater= 3,930 J/kg°C.

Suppose for 1 kg of sea water, the heat transferred from the system is given by

3,710,000 = 1 x 3,930 x (T - 48.17)

T = 990°C  to the nearest tenth.

The final temperature of the seawater-deck system is 990°C.

Learn more about heat.

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