What is albedo? What on Earth has high albedo?

Answer:

If there is a system of magnets being held in place, there is potential energy. When you let go the potential energy makes to kinetic energy and the magnets move.

Magnets cause objects to have kinetic energy as a result of the magnetic

force present in it.

Magnets is a material which has a strong metallic field which attracts

ferromagnetic substances such as steel, iron etc.

When  the magnets are put near these substances , the magnets attract them and causes them to move towards its direction . The magnets causes the conversion of the potential energy into kinetic energy through the strong metallic field.

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Answer:

18 kg * m/s

Explanation:

The equation for momentum is:

momentum = mass * velocity

And although the problem already gave velocity, you must calculate the mass from the graviational force.

Fg = 3.4N = mg = 9.8m

mass = 3.4 N / 9.8 m/s^2 = 0.3469 kg

Momentum = 52m/s * 0.3469 kg - 18.041 kg * m/s = 18

Answer:

A longitudinal wave is a wave where the displacement of the medium is in the same direction than the propagation of the wave.

This means that as a P wave travels through the Earth, the relative motion between the p wave and the particles is near zero, as the motion of the particles is parallel to the motion of the wave.

An example of this would be the waves generated when you throw a rock in water, you can see how the water particles move along the waves in the water.

The relative motion between the p wave and the particles is almost negligible.

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                                    ...................

Answer:

B)

Explanation:

The answer is B)- The object's speed must be changing.

So will I get the Brainliest...

Answer:

the answer is b mark me as the brqinlist

Answer:

The frequency stays the same and the wavelength decreases.

Explanation:

When sound wave enters a new medium where sound travels faster, its frequency will remain same because it depends only on the source.

The relation between wavelength and speed is inverse, it means when the speed of sound increases, its wavelength will decrease.

So, the frequency stays the same and the wavelength decreases. Hence, the correct option is (d).

Answer:

1200cm²

Explanation:

Width= 30cm

Length= 40cm

Area of base(A) = Width×Length

= 30cm×40cm

= 1200cm²

Taking into account the definition of reactanguar prism and area of rectangle, the area of its base is 1200 cm².

A rectangular prism is a polyhedron whose surface is formed by two equal and parallel rectangles called bases and by four lateral faces that are also parallel and equal rectangles two by two.

So, the base being a rectangle, its area is calculated as the multiplication between the base and the height. In this case, these values ​​correspond to the width and length of the figure.

Then, in this case, you know:

Being:

Area of base (A) = Width×Length

Then, the are of base (A) is calculated as:

Area of base (A)= 30cm×40cm

Solving:

Area of base (A)= 1200cm²

Finally, the area of its base is 1200 cm².

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Answer:

Capital (G) is a universal gravitation law. and small (g) is acceleration of gravity of the each (9.8m/s^2)

Explanation:

According to the web

Answer:

Width of a box, [tex]l=6.41\times 10^{-15}\ m[/tex]

Explanation:

The ground level energy of a proton in a box is, E = 5 MeV

[tex]E =5\times 10^6\ eV\\\\=5\times 10^6\times 1.6\times 10^{-19}\\\\=8\times 10^{-13}\ J[/tex]

Energy in a box is given by :

[tex]E=\dfrac{n^2h^2}{8ml^2}[/tex]

For ground state, n = 1

m is mass of proton

h is Planck's constant

l is width of the box

[tex]l^2=\dfrac{n^2h^2}{8mE}\\\\l^2=\dfrac{1^2\times (6.63\times 10^{-34})^2}{8\times 1.67\times 10^{-27}\times 8\times 10^{-13}}\\\\l=\sqrt{\dfrac{1^{2}\times(6.63\times10^{-34})^{2}}{8\times1.67\times10^{-27}\times8\times10^{-13}}}\\\\l=6.41\times 10^{-15}\ m[/tex]

So, the width of the bx is [tex]6.41\times 10^{-15}\ m[/tex].

The width of the box, for the ground level energy with which the particles in a nucleus are bound, is 6.41×10⁻¹⁵ m.

The energy in a box can be calculated with the following formula.

[tex]E=\dfrac{n^2h^2}{8mL^2}[/tex]

Here, (E) is the energy at the nth state, (n) n is the quantum number, (h) is plank's constant and (L) is the width of the box.

The proton is in a box of width L. The width of the box be for the ground level energy to be 5.0 MeV, a typical value for the energy with which the particles in a nucleus are bound.

The energy of the box is at ground level. Then the value of nth state will be 1. It is known that the value of plank's constant is 6.63×10⁻³⁴ m²kg/s.

Put this values in the above formula as,

[tex]8\times10^{-13}=\dfrac{(1)^2(6.63\times10^{-34})^2}{8(1.67\times10^{27})(l)^2}\\l=6.41\times10^{-15}\rm\; m[/tex]

Thus, the width of the box, for the ground level energy with which the particles in a nucleus are bound, is 6.41×10⁻¹⁵ m.

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Answer:

The value is  [tex]\frac{\Delta S }{ L} = - 0.0721 \ J / km[/tex]

Explanation:

From the question we are told that

   The  force is  [tex]F = 22.7 \ N[/tex]

    The value of room temperature is [tex]T = 298 \ K[/tex]

Generally the rate at which its entropy changes as it stretches is mathematically represented as

         [tex]\frac{\Delta S }{ L} = - \frac{F}{T}[/tex]

=>      [tex]\frac{\Delta S }{ L} = - \frac{21.5}{ 298 }[/tex]

=>      [tex]\frac{\Delta S }{ L} = - 0.0721 \ J / km[/tex]

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]

m is the mass

v is the velocity

From the question we have

[tex]k = \frac{1}{2} \times 74 \times {11}^{2} \\ = 37 \times 121[/tex]

We have the final answer as

Hope this helps you

Answer:

Approximately [tex]3.99\times 10^{4}\; \rm J[/tex] (assuming that the melting point of ice is [tex]0\; \rm ^\circ C[/tex].)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

[tex]\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}[/tex]

The energy required comes in three parts:

The following equation gives the amount of energy [tex]Q[/tex] required to raise the temperature of a sample of mass [tex]m[/tex] and specific heat capacity [tex]c[/tex] by [tex]\Delta T[/tex]:

[tex]Q = c \cdot m \cdot \Delta T[/tex],

where

For the first part of energy input, [tex]c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}[/tex] whereas [tex]m = 0.100\; \rm kg[/tex]. Calculate the change in the temperature:

[tex]\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}[/tex].

Calculate the energy required to achieve that temperature change:

[tex]\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}[/tex].

Similarly, for the third part of energy input, [tex]c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}[/tex] whereas [tex]m = 0.100\; \rm kg[/tex]. Calculate the change in the temperature:

[tex]\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}[/tex].

Calculate the energy required to achieve that temperature change:

[tex]\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}[/tex].

The second part of energy input requires a different equation. The energy [tex]Q[/tex] required to melt a sample of mass [tex]m[/tex] and latent heat of fusion [tex]L_\text{f}[/tex] is:

[tex]Q = m \cdot L_\text{f}[/tex].

Apply this equation to find the size of the second part of energy input:

[tex]\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}[/tex].

Find the sum of these three parts of energy:

[tex]\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}[/tex].

Answer: the required mass is 1.7628 × 10²⁵ μ

Explanation:

Given that;

energy released in the fission of one ²³⁵U₉₂ is 206.6 MeV

power p = 28.7 Mega Watts = 28.7 × 10⁶ W = 28.7 × 10⁶/ 1.6× 10⁻¹³ = 17.9375 × 10¹⁹ MeV/s

now fission need per second will be;

⇒ power / energy released i  fission

= 17.9375 × 10¹⁹ MeV /  206.6 MeV = 8.68 × 10¹⁷ per second

now fission need per day will be;

⇒ ( 8.68 × 10¹⁷ × 360 × 24 ) = 7.5 × 10²² per day

hence mass of ²³⁵U₉₂ that is consumed in one day in this reactor will be;

⇒ (235 × 7.5 × 10²²)μ

= 1.7628 × 10²⁵ μ

Therefore the required mass is 1.7628 × 10²⁵ μ

Complete Question

Suppose a wireless charging system of an electric toothbrush. The charger has Solenoid 1 with 800 turns and the toothbrush has Solenoid 2 with 150 turns. The current in the charger is 220 mA and the corresponding average magnetic flux through Solenoid 2 is 0.056 Wb. What is the mutual inductance of the pair of the solenoids?

Answer:

The value is [tex]M = 38.2 \ H[/tex]

Explanation:

From the question we are told that

     The number of turns of solenoid 1 is [tex]N_1 = 800[/tex]

     The  number of turns for solenoid 2 is  [tex]N_2 = 150 \ turns[/tex]

     The current in the charger is  [tex]I = 220 \ mA = 220 *10^{-3} \ A[/tex]

      The  magnetic flux through solenoid 2 is  [tex]\phi = 0.056 \ Wb[/tex]

Generally mutual inductance is mathematically represented as

        [tex]M = \frac{N_2 * \phi }{ I }[/tex]

=>     [tex]M = \frac{150 * 0.056 }{ 220 *10^{-3} }[/tex]

=>    [tex]M = 38.2 \ H[/tex]

You need to know the coefficient of static friction between a wooden object and a wooden surface. I'll denote it with µ. If you're given a specific value you should obviously use that.

By Newton's second law, the horizontal and vertical net forces are

• net horizontal:

∑ F = p - f = 0

• net vertical:

∑ F = n - w = 0

where

p = magnitude of the pushing force

f = mag. of friction

n = mag. of the normal force

w = weight of the crate

The second equation gives

n = w = (5 kg) (9.8 m/s²) = 49 N

Friction is proportional to the normal force by a factor of µ, so

f = µ (49 N) = 49µ N

To overcome static friction, the push has to exceed this in magnitude, so that

p > 49µ N

For instance, if p = 0.25, then p would need to greater than 12.25 N. (This example isn't particularly helpful, though, since both possibly correct options are larger than 12.25 N...)

Answer:

F = 385.56 N

Explanation:

Given that,

Mass, m = 37.8 kg

He applies a horizontal force and cross the wooden floor.

We need to find the force that must be applied to move a dog with a constant speed of 1 m/s.

The coefficient of kinetic friction between the dog in the floor is 1.02.

The net force acting on it is given by :

F = f- μmg

f is force due to constant speed, f = 0 (since, a = 0)

F = μmg

= 1.02 × 37.8 × 10

= 385.56 N

Hence, the required force is 385.56 N.

Explanation:

Fnet = 12.0N - 5.0N = 7.0N

a = Fnet/m = 7.0N/6.5kg = 1.07m/s².

Answer:

The density of the block is 1113.216 kilograms per cubic meter.

The weight of the block is 43669.237 newtons.

Explanation:

According to the Archimedes' Principle, the drag force experimented by the metal block is equal to the weight of the volume of water displace by the block. Besides, the block has a weight that cannot be neglected and experiments a tension from the cable. Given that the metal block is suspended, then we could consider that block is at rest.

From Newton's Laws of Motion we obtain the following equation of equilibrium:

[tex]\Sigma F = T-\rho_{m}\cdot V_{m}\cdot g + \rho_{w}\cdot V_{m}\cdot g = 0[/tex] (1)

Where:

[tex]T[/tex] - Tension on the cable, measured in newtons.

[tex]\rho_{w}[/tex], [tex]\rho_{m}[/tex] - Densities of salt water and the metal block, measured in kilograms per cubic meter.

[tex]V_{m}[/tex] - Volume of the metal block, measured in cubic meters.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

If we know that [tex]T = 42600\,N[/tex], [tex]\rho_{w} = 1030\,\frac{kg}{m^{3}}[/tex], [tex]V_{m} = 4\,m^{3}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the density of the block is:

[tex]T+\rho_{w}\cdot V_{m}\cdot g = \rho_{m}\cdot V_{m}\cdot g[/tex]

[tex]\rho_{m} = \frac{T+\rho_{w}\cdot V_{m}\cdot g}{V_{m}\cdot g}[/tex]

[tex]\rho_{m} = \frac{42600\,N+\left(1030\,\frac{kg}{m^{3}} \right)\cdot (4\,m^{3})\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{(4\,m^{3})\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]\rho_{m} = 1113.216\,\frac{kg}{m^{3}}[/tex]

The density of the block is 1113.216 kilograms per cubic meter.

Lastly, the weight of the block ([tex]W[/tex]), measured in newtons:

[tex]W = \rho_{m}\cdot V_{m}\cdot g[/tex] (2)

[tex]W = \left(1113.216\,\frac{kg}{m^{3}}\right)\cdot (4\,m^{3}) \cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]W = 43669.237\,N[/tex]

The weight of the block is 43669.237 newtons.

Answer:

The speed of the box at the end of the 6 s interval is 7.5 m/s.

Explanation:

Given;

magnitude of the force, f = 15 N

mass of the box, m = 12 kg

initial velocity of the box, u = 0

time of the box motion, t = 6s

The applied force on the box is given by Newton's second law of motion;

f = ma

where;

a is the acceleration of the box

a = f / m

a = (15) / (12)

a = 1.25 m/s²

The final velocity of the box is calculated as;

v = u + at

v = 0 + (1.25 x 6)

v = 7.5 m/s.

Therefore, the speed of the box at the end of the 6 s interval is 7.5 m/s.

Answer:

Explanation:

Nucleus

Answer:

in a plane mirror distance of object from mirror is equal to distance of image

so it will be 24 ÷ 2 = 12 cm

Answer:

f = 0 N

Explanation:

It is given that,

A horse pulls a wagon with 2000N of at constant velocity.

We need to find the friction acting on the wagon.

As the wagon is pulled with a constant velocity. If its velocity is constant, it means its acceleration is 0. As a result friction force is 0.

Hence, the correct option is (a).

The required average speed of bicycle rider is 12 km/h.

Given data:

The distance travelled by the bicycle rider is, d = 15 km.

Time taken to cover the distance is, t = 1.25 h.

Here, we need to use the simple relation between the speed, time and distance. The distance covered covered by any object per unit of time is known as average speed of object. And the mathematical expression for the average speed is given as,

[tex]v = \dfrac{d}{t}[/tex]

Here, v is the average speed.

Solving as,

[tex]v = \dfrac{15}{1.25}\\\\v = 12\;\rm km/h[/tex]

Thus, we can conclude that the required average speed of bicycle rider is 12 km/h.

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Answer:

The value is [tex]R_f = \frac{4}{5} R[/tex]

Explanation:

From the question we are told that

   The  initial velocity of the  proton is [tex]v_o[/tex]

    At a distance R from the nucleus the velocity is  [tex]v_1 = \frac{1}{2} v_o[/tex]

    The  velocity considered is  [tex]v_2 = \frac{1}{4} v_o[/tex]

Generally considering from initial position to a position of  distance R  from the nucleus

 Generally from the law of energy conservation we have that  

       [tex]\Delta K = \Delta P[/tex]

Here [tex]\Delta K[/tex] is the change in kinetic energy from initial position to a  position of  distance R  from the nucleus , this is mathematically represented as

      [tex]\Delta K = K__{R}} - K_i[/tex]

=>    [tex]\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2[/tex]

=>    [tex]\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2[/tex]

=>    [tex]\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2[/tex]

And  [tex]\Delta P[/tex] is the change in electric potential energy  from initial position to a  position of  distance R  from the nucleus , this is mathematically represented as

          [tex]\Delta P = P_f - P_i[/tex]

Here  [tex]P_i[/tex] is zero because the electric potential energy at the initial stage is  zero  so

             [tex]\Delta P = k * \frac{q_1 * q_2 }{R} - 0[/tex]

So

           [tex]\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0[/tex]

=>        [tex]\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}[/tex]

=>        [tex]- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )[/tex]

Generally considering from initial position to a position of  distance [tex]R_f[/tex]  from the nucleus

Here [tex]R_f[/tex] represented the distance of the proton from the nucleus where the velocity is  [tex]\frac{1}{4} v_o[/tex]

     Generally from the law of energy conservation we have that  

       [tex]\Delta K_f = \Delta P_f[/tex]

Here [tex]\Delta K[/tex] is the change in kinetic energy from initial position to a  position of  distance R  from the nucleus  , this is mathematically represented as

      [tex]\Delta K_f = K_f - K_i[/tex]

=>    [tex]\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2[/tex]

=>    [tex]\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2[/tex]

=>    [tex]\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2[/tex]

And  [tex]\Delta P[/tex] is the change in electric potential energy  from initial position to a  position of  distance [tex]R_f[/tex]  from the nucleus , this is mathematically represented as

          [tex]\Delta P_f = P_f - P_i[/tex]

Here  [tex]P_i[/tex] is zero because the electric potential energy at the initial stage is  zero  so

             [tex]\Delta P_f = k * \frac{q_1 * q_2 }{R_f } - 0[/tex]      

So

          [tex]\frac{1}{2} * m * \frac{1}{8} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R_f }[/tex]

=>        [tex]\frac{1}{2} * m *v_o^2 [-\frac{15}{16} ] = k * \frac{q_1 * q_2 }{R_f }[/tex]

=>        [tex]- \frac{15}{32} * m *v_o^2 = k * \frac{q_1 * q_2 }{R_f } ---(2)[/tex]

Divide equation 2  by equation 1

              [tex]\frac{- \frac{15}{32} * m *v_o^2 }{- \frac{3}{8} * m *v_0^2 } } = \frac{k * \frac{q_1 * q_2 }{R_f } }{k * \frac{q_1 * q_2 }{R } }}[/tex]

=>           [tex]-\frac{15}{32 } * -\frac{8}{3} = \frac{R}{R_f}[/tex]

=>           [tex]\frac{5}{4} = \frac{R}{R_f}[/tex]

=>             [tex]R_f = \frac{4}{5} R[/tex]

Explanation:

When the Javelin is at rest on the ground, Potential Energy and Kinetic Energy are zero.  immediately the Athlete picks the javelin up from the ground, there is an increase in the Kinetic Energy this increase continues until the javelin comes to a halt. Potential Energy also increases.

As the Athlete throws javelin, there is a decrease in the Potential Energy, the Kinetic Energy increases simultaneously until the javelin hits the ground.

after which Potential Energy  and Kinetic Energy becomes zero.

Answer: the answer is B. 2.0

Explanation:

Work done by lifting is 3,300 Newton.

Given that;

Mass of thing = 15 kg

Height lifted = 22 m

Find:

Work done by lifting

Computation:

Work done = mgh

Work done by lifting = (15)(10)(22)

Work done by lifting = 3,300 Newton

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Answer:

use hot water

Explanation:

hot water helps dissolve things faster