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3 years ago

In order to convert sunlight energy into usable energy the cell needs

1 answer:

6 0

Cells do not use the energy from oxidation reactions as soon as it is released. Instead, they convert it into small, energy-rich molecules such as ATP and nicotinamide adenine dinucleotide (NADH), which can be used throughout the cell to power metabolism and construct new cellular components.

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An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.275 T. If the kinetic energy of the electr

xxMikexx [17]

Answer:

Radius, $r=2.14\times 10^{-5}\ m$

Explanation:

It is given that,

Magnetic field, B = 0.275 T

Kinetic energy of the electron, $E=4.9\times 10^{-19}\ J$

Kinetic energy is given by :

$E=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{2E}{m}}$

$v=\sqrt{\dfrac{2\times 4.9\times 10^{-19}}{9.1\times 10^{-31}}}$

v = 1037749.04 m/s

The centripetal force is balanced by the magnetic force as :

$qvB\ sin90=\dfrac{mv^2}{r}$

$r=\dfrac{mv}{qB}$

$r=\dfrac{9.1\times 10^{-31}\times 1037749.04}{1.6\times 10^{-19}\times 0.275 }$

$r=2.14\times 10^{-5}\ m$

So, the radius of the circular path is $2.14\times 10^{-5}\ m$ . Hence, this is the required solution.

3 0

3 years ago

Read 2 more answers

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

Pleaseeee Please help, I will love you forever and ever

matrenka [14]

Answer:

The answer to your question is

Explanation:

Data

mass = 0.5kg

T1 = 35

T2 = ?

Q = - 6.3 x 10⁴ J = - 63000 J

Cp = 4184 J / kg°C

Formula

Q = mCp(T2 - T1)

T2 = T1 + Q/mCp

Substitution

T2 = 35 - 63000/(0.5 x 4184)

T2 = 35 - 63000/2092

T2 = 35 - 30.1

T2 = 4.9 °C

6 0

3 years ago

The acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

VladimirAG [237]

We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"

it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

From the question we are told

the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

Generally the equation for the Force is mathematically given as

F=\frac{F}{dx}

Therefore

F=-kdx

k=600Nm^{-1}

now

K.E=0.5x ds^2

K.E=600*(-0.1^2)

K.E=3J

Therefore

the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

For more information on this visit

brainly.com/question/23379286

6 0

3 years ago

A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h

Tasya [4]

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

$v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2$

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

$v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft$

The minimum stopping distance is 141.78 ft.

4 0

3 years ago