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2 years ago

Can someone help me with this I don’t understand how to do this

Physics

1 answer:

olga nikolaevna [1]2 years ago

4 0

Answer:

You are doing this wrong. Make sure that if you have 2 rows and on the top it has pounds then on the second bottom row you also have pounds so that they cancel each other out.

Explanation:

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Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150

Sever21 [200]

Answer: $2.89\approx 2.9^{\circ}C/s$

Explanation:

Given

$Power\left ( P\right )=150 MW$

mass of core $\left ( m\right )=1.60\times 10^5 kg$

Average specific heat $\left ( C\right )=0.3349 KJ/kg^{\circ}C$

And rate of increase of temperature = $\frac{\mathrm{d}T}{\mathrm{d} t}$

Now

P= $mc\frac{\mathrm{d}T}{\mathrm{d} t}$

$150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}$

Thus $\frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}$

$\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s$

6 0

3 years ago

How can meteorologists use the jet stream to predict the weather ?

barxatty [35]

<span>Jet streams act as an invisible director of the atmosphere and are largely responsible for changes in the weather across the globe.
Hope this helps</span>

5 0

3 years ago

Superman does an exhibition run at a track meet. When he runs the 200 m

Ber [7]

Answer:

6.32s

Explanation:

Given parameters:

Length of track and distance covered = 200m

Acceleration = 10m/s²

Unknown:

Time taken to cover the track = ?

Solution:

To solve this problem, we apply one of the motion equations as shown below:

S = ut + $\frac{1}{2}$ at²

S is the distance covered

t is the time taken

a the acceleration

u is the initial velocity

The initial velocity of Superman is 0;

So;

S = $\frac{1}{2}$ at²

200 = $\frac{1}{2}$ x 10 x t²

200 = 5t²

t² = 40

t = 6.32s

5 0

2 years ago

When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the c

e-lub [12.9K]

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = $e^{kt}$

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = $e^{-0.01t}$

V = V₀ $e^{-0.01t}$

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀ $e^{-0.01t}$

0.1V₀ = V₀ $e^{-0.01t}$

0.1V₀/V₀ = $e^{-0.01t}$

0.1 = $e^{-0.01t}$

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

3 0

3 years ago

The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t

dem82 [27]

Answer:

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

now for the time period of moon around the earth we can say

$T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}$

here we know that

$T_1 = 0.08 year$

$r_1 = 0.0027 AU$

$M_e$ = mass of earth

Now if the same formula is used for revolution of Earth around the sun

$T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}$

here we know that

$r_2 = 1 AU$

$T_2 = 1 year$

$M_s$ = mass of Sun

now we have

$\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}$

$\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}$

$12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}$

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

4 0

3 years ago