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3 years ago

15

A 1.4kg ball is swung in a vertical circle with a radius of 0.8m. If the frequency is 1.5Hz, find the tension in the rope at the

bottom position.

1 answer:

Artyom0805 [142]3 years ago

4 0

Answer:

43

Explanation:

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How long does it take for earth to complete one full rotation?

Gwar [14]

Answer:

23 hours 56 minutes 4.091 seconds

Explanation:

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2 years ago

A wooden box is pushed with 100 N towards North

anastassius [24]

0 N. The net force is found through thinking of the quantities as scalar. If you want to think of north as positive, think of south as negative. Then add them. 100 + (-100) = 0N

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3 years ago

Two factors on which the frequency of the sound generated by a wind instrument depends...

fenix001 [56]

The other factor that determines the frequency produced by a wind instrument is the speed at which the sound waves travel. Although typically about 343 m/s (at room temperature), the speed of sound does vary slightly with the temperature of the air, increasing as the temperature rises and decreasing as the air becomes colder.

i got this answer from g00gle

4 0

3 years ago

To lift a load of 100 kg a distance of 1 m an effort of 25 kg must be applied over an inclined plane of length 4 m. What must be

andreyandreev [35.5K]

Work = force * distance.
We must produce twice as much energy as we are lifting the weight twice as high.
But we are not increasing the force so we must increase the length of the ramp ( distance ) instead.
The new length will be twice as great as the previous length.
So 8 metres is required.

25 kg * 8 m = work = 100 kg * 2 m

7 0

3 years ago

A person with lymphoma receives a dose of 45 Gy in the form of γ radiation during a course of radiotherapy. Most of this dose is

amid [387]

Answer:

Explanation:

a) Energy absorbed by cancerous tissue E = R x M

R is the radiation dosage

M is the mass of the lymphatic

Energy absorbed by cancerous tissue

$=45\times18\times10^-^3J\\\\=810\times10^-^3J$

b) If only 1% of the total energy Et is absorbed over 20min session for 5 weeks = E

And the time of the period of the course is t = 20 x 5 = 100 min

E = 0.01Et

Total energy of gamma ray beam

Et = E x 100

= 810 x 10⁻³ x 100

= 8100J

Power of gamma ray beam is P

$P=\frac{Et}{t} \\\\=\frac{8100J}{100\times60} \\\\=0.0135W$

c) The total activity

$A=\frac{-dN}{dt} = dN decays/unit\ time$

The Number of decays dN = Es Total energy emitted at sourse per seconds / energy emitted per day

$dN=\frac{ Es}{ 0.03Mev}$

The Energy source Es per seconds = 0.05Ps

P= 0,0135W

The Power emitted at source

$Ps = \frac{0.0135}{0.05} \\\\=0.27W$

The Energy emitted per decay = 0.03MeV

$=0.03\times10^6\times1.602\times10^-^1^9J$

$A=\frac{-dN}{dt} \\\\=\frac{0.27}{0.03\times10^6\times1.602\times10^-^1^9} \\\\=\frac{0.27}{0.04806\times10^-^1^3} \\\\=5.618\times10^1^3/sec$

8 0

4 years ago