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2 years ago

In which scenario does radiation occur?

Physics

2 answers:

Airida [17]2 years ago

7 0

Suntanning outside or in tanning booths are both examples.

marin [14]2 years ago

7 0

Answer:drying towels at the beach

Explanation:

Just took the test

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A boat sails south with the help of a wind blowing in the direction S36°E with magnitude 300 lb. Find the work done by the wind

goldfiish [28.3K]

Answer:

The work done by the wind as the boat moves 130 ft is (rounded) W= 31,550 ft-lb.

Explanation:

F= 300 lb < -54º

Fsouth= 300 lb * cos(36º)

Fsouth= 242.7 lb

d= 130 ft

W= F*d

W= 31551 ft-lb

6 0

2 years ago

Calculate a pendulum's frequency of oscillation (in Hz) if the pendulum completes one cycle in 0.5 s.

Marina86 [1]

Time taken to complete one oscillation for a pendulum is Time Period, T = 0.5 s
Frequency of the pendulum oscillation = 1 / Time Period => f = 1 / T = 1 / 0.5
Frequency f = 2 Hz

3 0

3 years ago

Suppose that a steel bridge, 1000 m long, were built without any expansion joints. Suppose that only one end of the bridge was h

Stels [109]

Answer:

The difference in the length of the bridge is 0.42 m.

Explanation:

Given that,

Length = 1000 m

Winter temperature = 0°C

Summer temperature = 40°C

Coefficient of thermal expansion $\alpha= 10.5\times10^{-6}\ K^{-1}$

We need to calculate the difference in the length of the bridge

Using formula of the difference in the length

$\Delta L=L\alpha\Delta T$

Where, $\Delta T$ = temperature difference

$\alpha$ =Coefficient of thermal expansion

L= length

Put the value into the formula

$\Delta L=1000\times10.5\times10^{-6}(40^{\circ}-0^{\circ})$

$\Delta L=0.42\ m$

Hence, The difference in the length of the bridge is 0.42 m.

5 0

3 years ago

Two boats - Boat A and Boat B - are anchored a distance of 24 meters apart. The incoming water waves force the boats to oscillat

ozzi

Answer:

wavelength = 24 m

Period = 10 s

f = 0.1 Hz

Amplitude = 4 m

Explanation:

Wavelength:

Since the boats are at crest and trough, respectively at the same time. Hence, the horizontal distance between them is the wavelength of the wave:

wavelength = 24 m

Period:

The period is given as:

$Period = \frac{time}{no.\ of\ cycles} \\\\Period = \frac{10\ s}{1}\\\\$

Period = 10 s

Frequency:

The frequency is given as:

$f = \frac{1}{time\ period}\\\\f = \frac{1}{10\ s}\\\\$

f = 0.1 Hz

Amplitude:

Amplitude will be half the distance between extreme points, that is, crest and trough:

Amplitude = 8 m/2

Amplitude = 4 m

5 0

3 years ago

two negative charges that are both -3.0 C push each other apart with a force of 19.2 N how far apart are the charges

Hoochie [10]

The electrostatic force between two charges q1 and q2 is given by
$F=k_e \frac{q_1 q_2}{r^2}$
where $k_e =8.99 \cdot 10^9 N m^2 C^{-2}$ is the Coulomb's constant and r is the distance between the two charges.

If we use F=19.2 N and q1=q2=-3.0 C, we can find the value of r, the distance between the two charges by re-arranging the previous formula:
$r= \sqrt{k_e \frac{q_1 q_2}{F} }= \sqrt{ 8.99 \cdot 10^9 N m^2 C^{-2} \frac{(-3.0C)^2}{19.2 N} } =6.49 \cdot 10^4 m=64.9 km$

5 0

3 years ago