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3 years ago

A person paddles down river at an average speed of 4km

Physics

2 answers:

adoni [48]3 years ago

7 0

Nice paddling. Thanks for sharing.
Do you have some question to ask ?

almond37 [142]3 years ago

7 0

Is that a question or are you just telling us

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A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten

babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

$P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2$

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0

3 years ago

A 4.0 Ω resistor has a current of 3.0 A in it for 5.0 min. How many electrons pass 3. through the resistor during this time inte

musickatia [10]

Answer:

Number of electrons, $n=5.62\times 10^{21}$

Explanation:

It is given that,

Resistance, R = 4 ohms

Current, I = 3 A

Time, t = 5 min = 300 s

We need to find the number of electrons pass through the resistor during this time interval. Let the number of electron is n.

i.e. q = n e ...............(1)

And current, $I=\dfrac{q}{t}$

$I\times t=n\times e$

$n=\dfrac{It}{e}$

e is the charge of an electron

$n=\dfrac{3\ A\times 300\ s}{1.6\times 10^{-19}}$

$n=5.62\times 10^{21}$

So, the number of electrons pass through the resistor is $5.62\times 10^{21}$ . Hence, this is the required solution.

6 0

3 years ago

What enables the human eye to "see" light in the infrared range of the electromagnetic spectrum?

Phantasy [73]

A thermogram<span> enables the human eye to "see" light in the infrared range of the electromagnetic spectrum.</span>

6 0

3 years ago

A squirrel on a table clothesline is running 3.4 m./s. How far does the squirrel go in 2.5 minutes

statuscvo [17]

2.5 that would be the corect answer hope this helps

5 0

3 years ago

2. One mole of a monatomic ideal gas undergoes a reversible expansion at constant pressure, during which the entropy of the gas

Anna35 [415]

Answer:

The initial and final temperatures of the gas is 300 K and 600 K.

Explanation:

Given that,

Entropy of the gas = 14.41 J/K

Absorb gas = 6236 J

We know that,

$ds=\dfrac{dQ}{dt}$

At constant pressure,

$dQ=C_{p}dt$

$\Delta s=\int_{T_{1}}^{T_{2}}{\dfrac{C_{p}dT}{T}}$

$\Delta s=C_{p}ln\dfrac{T_{2}}{T_{1}}$

Put the value into the formula

$14.41=2.5\times8.3144(ln\dfrac{T_{2}}{T_{1}})$

$\dfrac{14.41}{2.5\times8.3144}=ln\dfrac{T_{2}}{T_{1}}$

$0.693=ln\dfrac{T_{2}}{T_{1}}$

$ln2=ln\dfrac{T_{2}}{T_{1}}$

$T_{2}=2T_{1}$ ...(I)

We need to calculate the initial and final temperatures of the gas

Using formula of energy

$\Delta Q=C_{p}\Delta T$

Put the value into the formula

$6236=2.5\times8.3144(T_{2}-T_{1})$

$6236=20.786(T_{2}-T_{1})$

$T_{2}-T_{1}=\dfrac{6236}{20.786}$

$T_{2}-T_{1}=300$

Put the value of T₂

$2T_{1}-T_{1}=300$

$T_{1}=300\ K$

Put the value of T₁ in equation (I)

$T_{2}=2\times300$

$T_{2}=600\ K$

Hence, The initial and final temperatures of the gas is 300 K and 600 K.

8 0

3 years ago