The sun is the ultimate source for energy on the Earth.
a. True
b. False

Answers

Answer 1

Answer: So we want to know is it true or false that the Sun is the ultimate source of energy on Earth. The answer is true because fossil fuels are solar energy trapped in the Earth's surface. Water cycle is dependent on solar radiation and we use water to produce electricity. We use solar panels to produce electrical energy. And also, all plant life harvests solar energy.

Answer 2

Answer:

This statement is TRUE.

Related Questions

When a sound source is moving toward the observer, the observer will hear a higher pitch. This is due toa. a decrease in the amplitude of the sound wave. b. a decrease in the frequency of the sound wave. c. an increase in the frequency of the sound wave. d. an increase in the amplitude of the sound wave.
Which explains why light bends when it passes from air to water?a) It generates more energy. b) It is reflected. c) It generates heat. d) It changes speed.
What is true of an object pulled inward in an electric field?a) The object has a neutral charge. b) The object has a charge opposite that of the field. c) The object has a negative charge. d) The object has a charge the same as that of the field.
two vectors of same units have magnitude of 8 unit and 5 unit. what are the maximum and minimum magnitude of resultant that can be obtained with the two vectors?
The total amount of electrical energy used by a315-watt television during 30.0 minutes ofoperation is(1) 5.67 × 10^5 J (3) 1.05 × 10^1 J(2) 9.45 × 10^3 J (4) 1.75 × 10^1 J

What is refractive index?

Answers

The refractive index of a substance is the ratio . . .

(speed of light in vacuum) divided by (speed of light in the substance) .

So the speed of light in a substance is

(299,792,458 meters per sec) divided by (the refractive index of the substance).

At its closest point, Mercury is approximately 46 million kilometers from the sun. What is this distance in AU?

Answers

This problem can be solved using the following relation: 1 kilometer = 6.6846e-9 AU. Since we are already given the number of kilometers, we simply have to multiply it to its equivalent in AU to solve for the equivalent distance. This is done below:

46 000 000 * 6.6846e-9 = 0.3 AU

At what distance from a long, straight wire carrying a current of 8.7 A is the magnetic field due to the wire equal to the strength of Earth’s field, approximately 5.4 × 10−5 T? The permeabilty of free space is 1.25664 × 10−6 T · m/A. Answer in units of cm.

Answers

Answer:

r= 3.2 cm

Explanation:

Given that

I= 8.7 A

B= 5.4 x 10⁻⁵ T

μo=1.25664 x 10⁻⁶

We know that magnetic filed in wire at a distance r given as

$B=(\mu_oI)/(2\pi r)$

$r=(\mu_oI)/(2\pi B)$

By putting the values

$r=(1.25664* 10^(-6)* 8.7)/(2* \pi * 5.4* 10^(-5))\ m$

r=0.032 m

r= 3.2 cm

Final answer:

The distance from a long straight wire at which the magnetic field equals the strength of Earth’s field, given a current of 8.7 A and Earth's field of 5.4 × 10−5 T, can be calculated using the formula for the magnetic field around a current-carrying wire. Substituting the given values, the answer is approximately 37.22 cm.

Explanation:

To solve this physics problem, we will use the formula for the magnetic field produced by a current carrying long, straight wire. The formula is: B = μI / (2πr), where 'B' is the magnetic field strength, 'μ' is the permeability of free space, 'I' is the current, and 'r' is the radial distance away from the wire.

In this case, Earth’s magnetic field, 'B', is given as 5.4 × 10−5 T, the current, 'I', is given as 8.7 A, and the permeability of free space, 'μ', is given as 1.25664 × 10−6 T · m/A. We need to find 'r', the distance away from the wire, and we want this answer in centimeters.

So, rearrange the formula to solve for 'r': r = μI / (2πB).

Substitute our known values into the equation: r = (1.25664 × 10−6 T · m/A × 8.7 A) / (2π × 5.4 × 10^-5 T). After calculating, we need to convert from meters to centimeters by multiplying by 100. The final answer is approximately 37.22 cm.

Learn more about Magnetic Field Calculation here:

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A small object with mass m = 0.0900 kg moves along the +x-axis. The only force on the object is a conservative force that has the potential-energy function U(x)=−αx2+βx3, where α=2.00J/m2 and β=0.300J/m3. The object is released from rest at small x. When the object is at x = 4.00 m, what are its (a) speed and (b) acceleration (magnitude and direction)? (c) What is the maximum value of x reached by the object during its motion?

Answers

The conservation of mechanical energy allows to find the results for the questions of the motion of mass in a conservative force are;

a) the velocity at x = 4 m is: v = 16.9 m / s

b) The acceleration at x = 4m is: a = 17.8 m / s²

c) The maximum elongations is: x = 6.67 m

Given parameters

The potential energy U = - α x² + β x³ with α= 2 j/m² and β= 0.3 j/m³

The mass of the body m = 0.0900 kg

The distance x = 4.00 m

To find

a) The speed

b) The acceleration

c) The maximum value of the distance

The conservation of mechanical energy is one of the most important concepts in physics, stable that if there is not friction the mechanical energy remains constant at all points.

Em = K + U

Where Em mechanical energy, K the kinetical energy ang U the potential energy.

a) Let's find the velocity using the conservation of mechanical energy

Starting point. Where the mass is released.

Em₀ = U (0)

Final point. When for a distance of x = 4 m here we have potential and kinetic energy.

$Em_f$ = K + U (4)

They indicate that the only force is conservative, therefore mechanicalenergy is conserved

Em₀ = $Em_f$

0 = ½ m v² + U (4)

½ m v² = -U (4)

v² = 2 / m (αx² - β x³)

Let's calculate

v² = $(2)/(0.09)$ (2 4² - 0.3 4³)

v = $√(284.44)$

v = 16.9 m / s

b) Acceleration is requested at this point.

We use that potential energy and force are related

F = $- ( dU)/(dx)$

We carry out the derivatives

F = 2αx - 3βx²

Let's calculate

F = 2 2 4 - 3 0.3 4²

F = 1.6 N

Now we use Newton's second law that relates the net force with the product of the mass and the acceleration of the body.

F = ma

a = $(F)/(m)$

a = $(1.6)/(0.09)$

a = 17.8 m / s²

c) At maximum displacement.

Let's use conservation of mechanical energy

Starting point. Where x = 0 is released

Emo = U (0) = 0

Final point. Point of maximum elongation, kinetic energy is zero

$Em_f$ = U (xmax)

Energy is conserved

Em₀ = $Em_f$

$0 = U(x_(max))$

-αx² + βx³ = 0

x² (-α + βx) = 0

the solutions of this equation is:

x = 0

-α + βx = 0

x = $(\alpha)/(\beta )$

Let's calculate

x = $(2)/(0.3)$

x = 6.67 m

In conclusion using the conservation of mechanical energy we can find the results for the questions of the motion of mass in a conservative force are;

a) the velocity at x = 4 m is: v = 16.9 m / s

b) The acceleration at x = 4m is: a = 17.8 m / s²

c) The maximum elongations is: x = 6.67 m

Learn more here: brainly.com/question/2615468

Answer:

a) v= 284.44 $(m)/(s^(2))$

b) a=17.78 $(m)/(s^(2) )$

c) x=6.67m

Explanation:

a).

$U(x)=-\alpha *x^(2) +\beta*x^(3)\n\alpha=2(J)/(m^(2))\n\beta=0.3(J)/(m^(2))\nU(0)+V(0)=U(4)+V(4)\nU(0)=-\alpha *0^(2) +\beta*0^(3)=0\nU(0)=0\nV(0)=0\n0=U(4)+V(4)\nU(4)=-\alpha *4^(2) +\beta*4^(3)\nU(4)=-2*4^(2)+0.3*(4^(3))\nU(4)=-12.8 J\n0=-12.8J+V(4)\n12.8=(1)/(2)*m*(v_(4))^(2) \n v_(4)^(2) =(2*12.8J)/(0.09kg)\n v_(4)^(2)=284.44$

$v_(4)=√(284.44)\nv_(4)=16.8 (m)/(s)$

b).

$F_(x)=(dU)/(dt)\nF_(x)=2*\alpha*x-3*\beta *x^(2) \nF_(4)=2*2*4-3*0.3*(4)^(2)\nF_(4)=1.6N\nF_(4)=m*a\na=(F_(4))/(m)=(1.6N)/(0.09kg)\na=17.7 (m)/(s^(2) )$

c).

$F_(x)=m*ax\nax=(F_(x))/(m) \nax=(4x-0.9x^(2))/(0.09kg)\n(dVx)/(dt)= (4x-0.9x^(2))/(0.09kg)\n\int\limits^x_x {(1)/(0.09)*(4x-0.9x^(2)) } \, dx\n (Vx^(2) )/(2)=22.2x^(2) -3.3x^(3)$

Value x,0

$Vx^(2) =44.4x^(2) -6.6x^(3)\n Vx=\sqrt{44.4x^(2) -6.6x^(3)}$

Inside the square root is the value of maximum value of x

$44.4x^(2) -6.6x^(3)=0\nx^(2)(44.4-6.6x)=0\n x=0$ but that value is not real so:

$44.4-6.6x=0\n6.6x=44.4\nx=(44.4)/(6.6)\n x=6.67m$

Explain three differences between a science and a pseudoscience.

Answers

1) science does not accept personal story's as evidence, pseudoscience relies on these story's as evidence.
2) science argues from scientific knowledge, pseudoscience argues from ignorance
3) science progresses, pseudoscience does not progress
and 4) (just in case) science holds pier review, pseudoscience does not

You throw a rock out of a 7th story window. you time that it takes 3.7 seconds to hit the ground, and measure that it hit the ground 115 m from the base of the building? How fast must you have thrown the rock?
How high up must the 7th floor be?
If you had thrown it at the same speed but at the 28th floor, how far from the base of the building would it land?

Answers

1) 31.1 m/s

The rock has been thrown straight out of the window: its motion on the horizontal direction is simply a uniform motion, with constant speed $v$ , because no forces act in the horizontal direction. The speed in a uniform motion is given by

$v=(S)/(t)$

where S is the distance traveled and t the time taken.

In this case, the distance by the rock before hitting the ground is $S=115 m$ and the time taken is $t=3.7 s$ , so the initial speed is given by

$v=(115 m)/(3.7 s)=31.1 m/s$

2) 67.1 m

In this part of the problem we are only interested in the vertical motion of the rock. The vertical motion is a uniformly accelerated motion, with constant acceleration $a=9.8 m/s^2$ (acceleration of gravity) towards the ground. In a uniformly accelerated motion, the distance traveled by the object is given by

$S=(1)/(2)at^2$

where t is the time. Substituting a=9.8 m/s^2 and t=3.7 s, we can find S, the vertical distance covered by the rock, which corresponds to the height of the 7th floor:

$S=(1)/(2)(9.8 m/s^2)(3.7 s)^2=67.1 m$

3) 230.1 m

The height of the 7th floor is 67.1 m. So we can assume that the height of each floor is

$h=(67.1 m)/(7)=9.6 m$

And so, the height of the 28th floor is

$h=28\cdot 9.6 m=268.8 m$

We can find the total time of the fall in this case by using the same formula of the previous part:

$S=(1)/(2)at^2$

In this case, S=268.8 m, so we can re-arrange the formula to find t

$t=\sqrt{(2S)/(g)}=\sqrt{(2(268.8 m))/(9.8 m/s^2)}=7.4 s$

And now we can consider the motion of the rock on the horizontal direction: we know that the rock travels at a constant speed of v=31.1 m/s, so the distance traveled is

$S=vt=(31.1 m/s)(7.4 s)=230.1 m$