Magma forms when solid rock in the crust and upper mantle. True or False

Final answer:

The work required to move a satellite from an orbit of radius 2R to 3R around a planet is calculated using the gravitational potential energy formula and is found to be 3.897×1010 J.

Explanation:

To calculate the work required to move a satellite from one circular orbit to another around a planet, we must consider the gravitational potential energy differences in the two orbits.

The gravitational potential energy (U) of an object of mass m in orbit around a planet of mass M at a distance r is given by U = -GmM/r, where G is the gravitational constant (6.67×10-11 N m2/kg2).

For the initial orbit at radius 2R, the potential energy is U1 = -GmM/(2R), and for the final orbit at radius 3R, the potential energy is U2 = -GmM/(3R). The work done (W) in moving the satellite is the difference in gravitational potential energy, W = U2 - U1. Substituting the values, we get:

W = (-GmM/3R) - (-GmM/2R) = (GmM/6R)

Let's calculate the work required using the given values: G = 6.67×10-11 N m2/kg2, m = 571 kg, M = 7.76×1024 kg, R = 8.8×106 m.

W = (6.67×10-11 N m2/kg2 × 571 kg × 7.76×1024 kg) / (6 × 8.8×106 m)

W = 3.897×1010 J

Therefore, the work required to move the satellite from a circular orbit of radius 2R to one of radius 3R is 3.897×1010 J.

Answer:

The answer to your question is 11.2 m/s

Explanation:

Data

Initial speed (vo) = 6.0 m/s

Acceleration (a) = 3.0 m/s²

time = 15 s

Final speed = ?

Formula

                d = vot + [tex]\frac{1}{2} at^{2}[/tex]

                vf² = vo² + 2ad

Process

                d = (6)(15) + [tex]\frac{1}{2} (3)(15)^{2}[/tex]

                d = 90 + 337.5

                d = 427.5 m

                vf² = (6)² + 2(3)(15)

                vf² = 36 + 90

                vf² = 126

                vf = 11.2 m/s

Answer:

Increase in temperature will be [tex]0.389^{\circ}C[/tex]

Explanation:

We have given mass of the aluminium m = 0.095 kg

Height h = 55 m

Specific heat of aluminium c = 900 J/kg°C

We know that potential energy is given as

[tex]PE=mgh=0.095\times 9.8\times 55=51.205[/tex]

Now 65 % of potential energy [tex]=\frac{51.205\times 65}{100}=33.28[/tex]

Now this energy is used to increase the temperature

So [tex]mc\Delta T=33.28[/tex]

[tex]0.095\times 900\times \Delta T=33.28[/tex]

[tex]0.095\times 900\times \Delta T=33=0.389^{\circ}C[/tex]

Answer:

V = 100.8 V

Explanation:

given,

hum being can be electrocuted with current = 48 mA = 0.048 A

Resistance of the man = 2100 Ω

Fatal voltage  = ?

we know,

V = I R

V is the fatal voltage in Volts

R is the resistance provided by the human body

I is current

V = I R

V = 0.048 x 2100

V = 100.8 V

the voltage which can be considered as fatal is equal to  V = 100.8 V

Answer:

Explanation:

The static charges are generated due to excess or deficiency of electrons, because these are the smallest quanta of charge available at the molecular level which can get transferred with minimal energy requirement.

These charges are usually generated by friction between  the two surfaces leading to the transfer of electron from one to another.

When a plastic pipe is rubbed with a cloth then due to friction the surface of the cloth loses electron which gets stuck at the surface of the pipe making it negatively charged.

Final answer:

The moment of inertia of the wheel for an axis perpendicular to the wheel at its center is 0.964 kg * m^2.

Explanation:

To calculate the moment of inertia of the wheel, we can use the principle of conservation of energy. The initial gravitational potential energy of the mass is equal to the final rotational kinetic energy of the wheel. This can be represented by the equation:

mg * h = 1/2 * Iω^2

Where m is the mass, g is the acceleration due to gravity, h is the distance the mass has descended, I is the moment of inertia of the wheel, and ω is the angular velocity of the wheel. Rearranging the equation:

I = 2mg * h / ω^2

Substituting the given values:

I = 2 * 6.32 kg * 9.8 m/s^2 * 2.5 m / (4.0 m/s)^2

I = 0.964 kg * m^2

Therefore, the moment of inertia of the wheel for an axis perpendicular to the wheel at its center is 0.964 kg * m^2.

Final answer:

The weight of a spacecraft at 6450 km above Earth's surface is 1767.5 N and at 33700 km above Earth's surface is 182.35 N, calculated using Newton's law of universal gravitation and considering the increased distance from the Earth's center.

Explanation:

The weight of a spacecraft can be calculated using Newton's law of universal gravitation which states that every mass attracts every other mass with a force that is directly proportional to the product of their masses and inversely proportional to the distance squared between their centers, F = G * (m₁ * m₂) / r², where G is the gravitational constant, m₁ and m₂ are the masses involved, and r is the distance between the centers of the two masses. To find the weight of the spacecraft at a certain height, we need to use the spacecraft's mass and the new distance from the Earth's center, which includes both the Earth's radius and the altitude above the surface.

To answer the first part of the question, we calculate the weight at 6450 km above Earth's surface. Since the radius of the Earth is also 6450 km, the distance from the center of the Earth to the spacecraft is now 2 * 6450 km. Applying the law of gravitation, the gravitational force, and hence the weight, will be (6450 km / 2 * 6450 km)² = 1/4 of the original weight, which is 7070 N/4 = 1767.5 N.

For the second part of the question, at a height of 33700 km above the Earth's surface, the distance from the center is 6450 km + 33700 km = 40150 km. Repeating the calculation, the weight at this height will be (6450 km / 40150 km)² times the original weight, giving us a reduced weight of (7070 N * (1/6.23)²) ≈ 182.35 N.

Answer:0.25 kg-m/s

Explanation:

Given

mass of blob [tex]m=50 gm [/tex]

initial velocity [tex]u=-5 m/s\ \hat{i}[/tex]

time of collision [tex]t=20 ms[/tex]

we know Impulse is equal to change in momentum

initial momentum [tex]P_i=mu[/tex]

[tex]P_i=50\times 10^{-3}\times (-5)=-0.25 kg-m/s[/tex]

Final momentum [tex]P_f=50\times 10^{-3}v[/tex]

[tex]P_f=0[/tex] as final velocity is zero

Impulse [tex]J=P_f-P_i[/tex]

[tex]J=0-(-0.25)[/tex]

[tex]J=0.25 kg-m/s[/tex]

Answer:

a)P =14288.4 W

b)P = 19.16  horsepower

Explanation:

Given that

m= 108 kg

Initial velocity ,u= 0 m/s

Final velocity ,v= 19.6 m/s

t= 2.9 s

Lets take acceleration of Cheetah is a m/s²

We know that

v= u  + a t

19.6 = 0 + a x 2.9

a= 6.75 m/s²

Now force F

F= m a

F= 108 x 6.75 N

F= 729 N

Now the power P

P = F.v

P = 729 x 19.6 W

P =14288.4 W

We know that

1 W= 0.0013  horsepower

P = 19.16  horsepower

P =14288.4 W

Answer:

3.747 μm

Explanation:

To answer this question, the fundamental wave equation will be used. Light is an electromagnetic wave so we will use the speed of light for this electromagnetic wave.

v = fλ

299 792 458 m/s = 80,000 *10^9 * λ

λ = 3.747 *10^-6 = 3.747 μm

Answer:

the angular velocity increase by a factor of 2

Explanation:

using the law of the conservation of the angular momentum

[tex]L_i = L_f[/tex]

where [tex]L_i[/tex] is the inicial angular momentum and [tex]L_f[/tex] is the final angular momentum.

also, the angular momentum can be calculated by:

L = IW

where I is the inertia momentum and the W is the angular velocity.

so:

[tex]I_i W_i = I_fW_f[/tex]

we know that [tex]I_f = \frac{1}{2}I_i[/tex] then,

[tex]I_iW_i = \frac{1}{2}I_iW_f[/tex]

solving for [tex]W_f[/tex]:

[tex]W_f = 2W_i[/tex]

When a figure skater pulls her arms in while spinning, her moment of inertia decreases and her angular speed increases. In the provided case, with the moment of inertia decreasing by half, the angular speed will double.

When a figure skater is spinning with her arms outstretched, and she pulls her arms in close to her body, her moment of inertia decreases. According to the law of conservation of angular momentum, if the moment of inertia of a spinning object decreases, its angular speed must increase to keep the angular momentum constant. In this case, since the skater's moment of inertia decreases by half, her angular speed will correspondingly double, or increase by a factor of 2.

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Answer:

The time can catch before it smashes on the ground is [tex]t=5.73 s[/tex]

Explanation:

Using the force equation

[tex]F=m*a[/tex]

[tex]F_{net}=m*a[/tex]

So replacing and solving to find the acceleration

[tex]a = (m_1*g-m_2*g) / m_1+m_2[/tex]

Finding the factor

[tex]a = g *( m_1-m_2)/m_1+m_2[/tex]

[tex]a=9.8m/s^2 *( 4.0 kg- 3.0 kg) / (4.0 + 3.0) kg[/tex]

[tex]a=1.4 m/s^2[/tex]

Now replacing in Newtons law to find  the time before can catch so:

[tex]d= \frac{1}{2}*a*t^2[/tex]

[tex]t=\sqrt{\frac{2*d}{a}}=\sqrt{\frac{2* 23m}{1.4 m/s^2}}[/tex]

[tex]t=5.73 s[/tex]

Final answer:

Facilitated transport, also known as facilitated diffusion, is the process by which a molecule moves down its concentration gradient using transport proteins in the plasma membrane.

Explanation:

Facilitated transport, also known as facilitated diffusion, is the process by which a molecule moves down its concentration gradient using transport proteins in the plasma membrane. This process does not require the input of energy and allows substances to diffuse across the membrane more easily. For example, glucose is transported into cells using glucose transporters that utilize facilitated transport. This process is important for the movement of larger or charged molecules that cannot freely diffuse across the cell membrane.

Answer:

The speed in the smaller section is [tex]43.2\,\frac{m}{s}[/tex]

Explanation:

Assuming all the parts of the pipe are at the same height, we can use continuity equation for incompressible fluids:

[tex] \Delta Q=0 [/tex] (1)

With Q the flux of water that is [tex] Av[/tex] with A the cross section area and v the velocity, so by (1):

[tex] A_{2}v_{2}-A_{1}v_{1}=0 [/tex]

subscript 2 is for the smaller section and 1 for the larger section, solving for [tex] v_{2} [/tex]:

[tex]v_{2}=\frac{A_{1}v_{1}}{A_{2}} [/tex] (2)

The cross section areas of the pipe are:

[tex] A_{1}=\frac{\pi}{4}d_{1}^{2} [/tex]

[tex] A_{2}=\frac{\pi}{4}d_{2}^{2} [/tex]

but the problem states that the diameter decreases 86% so [tex] d_{2}=0.86d_{1} [/tex], using this on (2):

[tex] v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\approx1.35v_{1} [/tex]

[tex]v_{2}\approx(1.35)(32)\approx43.2\,\frac{m}{s} [/tex]

Answer:

The final velocities of all the six balls will be same.

Explanation:

According to law of conservation of energy:

Gain in K.E = Loss in potential energy

   ½ mv^2 = mgh  

Where “m” and “g” are constant. The interchange in energies will occur only with the change in velocity and height. Since, balls are thrown from the same hight with the same initial velocity so, their final velocities will also be same just before striking the ground.

The six balls will reach the ground at the same time, hence the final velocity of the balls will be the same.

During a downward motion of an object, the speed of the object increases as the object moves downwards and becomes maximum before the object hits the ground.

The equation for estimating the final velocity of the six balls is given as;

[tex]v_f = v_i + gt[/tex]

If air resistance is negligible, the six balls will reach the ground at the same time, hence the final velocity of the balls will be the same.

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Answer:

option C

Explanation:

Let mass of the bullet be m and velocity be v

mass of gun be M and bullet be V

now,

using conservation of momentum for gun 1

(M+m) V' = 2 mv + 3 MV

V' = 0

3 M V = - 2 mv

momentum of gun 1 =- 2 mv---------(1)

now for gun 2

(M+m) V' = mv + MV

V' = 0

M V = - mv

momentum of gun 1 = -mv-----------(2)

dividing equation (1) by (2)

[tex]\dfrac{P_m1}{P_m2} = \dfrac{- 2mv}{-mv}[/tex]

[tex]\dfrac{P_m1}{P_m2} = \dfrac{2}{1}[/tex]

the correct answer is option C

The correct option is Option C (2:1).The ratio of the momentum imparted to gun #1 to that imparted to gun #2 is 2:1. This conclusion follows from the principle of conservation of momentum. Thus, the correct answer is option c) 2:1.

Define the mass of the bullet from gun #2 as m. The bullet from gun #1 then has a mass of 2m since it is twice as heavy. Denote the velocity of both bullets as v:

So, the momentum imparted to gun #1 is 2m * v, and for gun #2 it is m * v.

To find the ratio of the momentum imparted to gun #1 to that imparted to gun #2, we calculate:

Therefore, the correct answer is option c) 2:1

Answer:

Catastrophic theories

Explanation:

The theory of catastrophe is a collection of methods used to analyze and describe the ways in which a system can experience sudden significant behavioral changes when one or more of the variables that govern it are continuously modified.

Georges Louis de Buffon suggested in 1745 the first destructive theory — that a comet pulled material from the Sun to form the planets.

Therefore the answer is -

Catastrophic theories

Answer:

Sa

Explanation:

Spiral Galaxies  -

It is a disk shaped galaxies which have spiral structure , is refereed to as spiral galaxies .

According to Hubble , these galaxies are classified as Sa , Sb , Sc .

Where ,

Sa - have the structure , which is bulged from the central portion , along with a tightly wrapped spiral structure .

Sb - have a lesser bulge and the spiral is looser .

Sc - It has very weak bulge with the open spiral structure .

Hence , from the question ,

The given information is about the Sa .

Answer:

gdsz

Explanation:

dsgzz cxvzdgctfgdsvftgdsftrdsfdtsardtgasfd5t6sgftsfdrstfdtgsv6cr5vsd5rw5

Answer:

t = 8.5 s

Explanation:

Kinematic equation of the movement of the tennis ball that is thrown upwards :

y = y₀ + v₀*t -½ g*t²   Equation (1)

Where :  

y : position of the ball as a function of time

y₀ : Initial position of the ball

t: time  

g: acceleration due to gravity in m/s²

Known data  

g = 32 ft/s²

y₀ = 680 ft

v₀ = 56 ft/s

Calculation of the time it takes for the ball to thit the ground

We replace data en the equation (1)

y = y₀ + v₀*t -½ g*t²  

0 = 680+(56)*t -½( 32) *t²

16*t²-(56)*t- 680 = 0  equation (2)

solving equation (2) quadratic:

t₁ = 8.5 s

t₁ = -5 s

Time cannot be negative so the time it takes for the ball to hit the ground  is t = 8.5 s

The ball will accelerate at a rate of [tex]3.11 m/s^2[/tex]

Explanation:

We can describe the motion of the ball by using Newton's second law, which states that the net force exerted on an object is equal to the product between the mass of the object and its acceleration:

[tex]F=ma[/tex]

where

F is the net force

m is the mass

a is the acceleration

In this problem,

F = 14 N is the force exerted on the ball

m = 4.5 kg is the mass of the ball

Solving the equation, we find its acceleration:

[tex]a=\frac{F}{m}=\frac{14}{4.5}=3.11 m/s^2[/tex]

So, the ball will accelerate at a rate of [tex]3.11 m/s^2[/tex].

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Answer:

Mass attached to the spring is 41.95 kg

Explanation:

We have given time period of the spring T = 2.1 sec

Let the mass attached is m

And spring constant is k

We know that time period is given by

[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]

[tex]2.1=2\pi \sqrt{\frac{m}{k}}[/tex]---------eqn 1

Now if the mass is increased by 68.10 kg then time period become 3.4 sec

So [tex]3.4=2\pi \sqrt{\frac{m+68.10}{k}}[/tex]------eqn 2

Now dividing eqn 1 by eqn 2

[tex]\frac{2.1}{3.4}=\sqrt{\frac{m}{m+68.10}}[/tex]

[tex]0.381=\frac{m}{m+68.10}[/tex]

[tex]m=41.95 kg[/tex]

So mass attached to the spring is 41.95 kg

Final answer:

To find the value of mass m, use the formula for the period of a mass-spring system.

Explanation:

In order to find the value of mass m, we can use the formula for the period of a mass-spring system:

T = 2π√(m/k)

Where T is the period, m is the mass, and k is the spring constant.

For the initial system with period 2.1 seconds, we have:

2.1 = 2π√(m/k)

For the system with mass increased by 6.810×10^1 kg and period 3.4 seconds, we have:

3.4 = 2π√((m + 6.810×10^1)/k)

Using these two equations, we can solve for the value of m.

Answer:

the value of acceleration due to gravity in moon is 1.6m/[tex]s^{2}[/tex] along downward direction

Explanation:

Here, the acceleration is constant and it is equal to acceleration due to gravity in moon. Therefore the question depicts a situation of uniformly accelerated motion in a straight line. So, let us refresh the three equations of uniformly accelerated straight line motion.

v = u + at

[tex]s = ut + \frac{1}{2}at^{2}[/tex]

[tex]v^{2} = u^{2} +2as[/tex]

where,

u = initial velocity

v = final velocity

s = displacement

a = acceleration

t = time

Since we are dealing with vectors (velocity, acceleration and displacement), we have to take their directions in to account. So we must adopt a coordinate system according to our convenience. Here, we are taking point of throwing as origin, vertically upward direction as positive y axis and vertically downward direction as negative y axis.

t = 1s

u = 0 (since the hammer is dropped)

v = -1.6m/s (since its direction is downward)

a = ?

The only equation that connects all the above quantities is

v = u + at

therefore,

a = [tex]\frac{v - u}{t}[/tex]

substituting the values

a = [tex]\frac{-1.6 - 0}{1}[/tex]

a = -1.6m/[tex]s^{2}[/tex]

Thus, the value of acceleration due to gravity in moon is 1.6m/[tex]s^{2}[/tex]. The negative sign indicates that it is along downward direction.

Answer:

a

Explanation:

a sound cannot be soft

Answer:

Quite or loud

Explanation:

Answer:

Qtot = 6C * deltaV

Explanation:

you can find the total capacitance from adding 1C+2C+3C=6C. and the total voltage is 1V. Capacitance = charge/voltage--> C = Q / V--> 6C = Q / deltaV. this makes Qtot = 6C* deltaV

The Total charge for the equivalent circuit is =  [tex]Q_{tot}[/tex] = 6c * ΔV

  Although your question is incomplete I found the missing part online and used it to resolve the question

Given data :

Total capacitance ( C ) = 6C ( 1 + 2 + 3 )

voltage = 1 V

Three capacitors having values of ; 1 C, 2 C,  3 C

Determine the total charge ( Qtot )

Applying the formula ; Q = CV ----  ( 1 )

 where; Q =  charge

              C = capacitance

              change in V = ΔV

∴ [tex]Q_{tot}[/tex] = 6c * ΔV

Hence the total charge Qtot for the equivalent capacitor =  6c * ΔV

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Answer:

a) 6.04 m/s

b) 11.02 m/s

Explanation:

a) Let the father mass be M, and his speed be V. His son mass is m = 3M/5. Since his kinetic energy initially is half of after he increases his speed by 2.5m/s

[tex]E_2 = 2E_1[/tex]

[tex]\frac{M(V+2.5)^2}{2} = 2\frac{MV^2}{2}[/tex]

[tex]V^2 + 5V + 6.25 = 2V^2[/tex]

[tex]V^2 - 5V - 6.25 = 0[/tex]

[tex]V \approx 6.04m/s[/tex]

b) The son kinetic energy initially is:

[tex]E_s = 2E_1 = 2\frac{MV^2}{2} = MV^2 = M*6.04^2 = 36.43M J[/tex]

We can solve for the son speed by the following formula

[tex]E_s = \frac{mv^2}{2}[/tex]

[tex]v^2 = \frac{2E_s}{m} = \frac{2*36.43M}{3M/5} = \frac{10*36.43}{3} = 121.4m/s[/tex]

[tex]v = \sqrt{121.4} = 11.02 m/s[/tex]

Answer:

The enthalpy for dissolution is - 305.558 J/g

Solution:

Mass of the ionic compound, m = 10.00 g

Mass of water, m' = 75.0 g

Initial temperature, T = [tex]23.2^{\circ}C[/tex]

Final Temperature, T' = [tex]31.8^{\circ}C[/tex]

Now,

To calculate the change in enthalpy:

We know that the specific heat of water is 4.18 [tex]J/g^{\circ}C[/tex]

Total mass of the solution, M = m + m' = 10.00 + 75.0 = 85.0 g

Temperature, difference, [tex]\Delta T = T' - T = 31.8 - 23.2 = 8.6^{\circ}C[/tex]

Thus

The heat absorbed by the solution is given by:

[tex]Q = MC_{w}\Delta T = 85.0\times 4.18\times 8.6 = 3055.58\ J[/tex]

Enthalpy, [tex]\Delta H = -\frac{Q}{m} = - \frac{3055.58}{10} = - 305.558\ J/g[/tex]

Answer:

Coefficient of static friction = 0.37

Explanation:

At the point the the quoll slides, quoll attains its maximum velocity.

So Ne = (mv^2)/r ....equa 1

And N =mg....equ 2

Where N vertical force of qoull acting on the surface, e = coefficient of friction, m=mass, g=9.8m/s^2, r =radius =1.6m, v= max velocity of quill = 2.4m/s

Sub equ 2 into equ 1

Mge= (mv^2)/r ...equa3

Simplfy equ3

e = v^2/(gr)...equ 4

Sub figures above

e = 5.76/(9.8*1.6)

e = 0.37

Answer:

Work function of the metal, [tex]W_o=3.38\times 10^{-17}\ J[/tex]

Explanation:

We are given that  

Frequency of the electromagnetic radiation,  [tex]f=5.16\times 10^{16}[/tex] Hz

The maximum kinetic energy of the emitted electron, [tex]K=4.04\times 10^{-19}\ J[/tex]

We need to find the work function of the metal.

We know that the maximum kinetic energy of ejected electron

[tex]K=h\nu-w_o[/tex]

Where h=Plank's constant=[tex]6.63\times 10^{-34} J.s[/tex]

[tex]\nu[/tex] =Frequency of light source

[tex]w_o[/tex]=Work function

Substitute the values in the given formula  

Then, the work function of the metal is given by :

[tex]W_o=h\nu -K[/tex]

[tex]W_o=6.63\times 10^{-34}\times 5.16\times 10^{16}-4.04\times 10^{-19}[/tex]

[tex]W_o=3.38\times 10^{-17}\ J[/tex]

So, the work function of the metal is [tex]3.38\times 10^{-17}\ J[/tex]. Hence, this is the required solution.

Answer:

[tex]P_c=P_b>P_a[/tex]

Explanation:

E = Energy

T = Time

Power is given by the equation

[tex]P=\frac{E}{T}[/tex]

For first case

[tex]P_a=\frac{2}{2}\\\Rightarrow P_a=1\ W[/tex]

For second case

[tex]P_b=\frac{10}{5}\\\Rightarrow P_b=2\ J[/tex]

For third case

[tex]P_c=\frac{2\times 10^{-3}}{1\times 10^{-3}}\\\Rightarrow P_b=2\ J[/tex]

The rank of power would be [tex]P_c=P_b>P_a[/tex]