All categories

3 years ago

Most of the energy on Earth comes from which of the following?

1 answer:

3 0

<span>The earth gets most of its energy from the sun. The answer is letter B. This is because most of the living organisms on earth need sunlight to perform activities that would balance the ecosystem. Sunlight is the reason for the climate system on earth, the production of food through cellular respiration and many more.</span>

You might be interested in

Cylindrical beaker of height 0.100 mm and negligible weight is filled to the brim with a fluid of density rhorhorho = 890 kg/m3k

Nesterboy [21]

Incomplete part of the question

A ball of density ρb = 5000 kg/m3 and volume V = 60.0 cm3 is then submerged in the fluid, so that some of the fluid spills over the side of the beaker. The ball is held in place by a stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration due to gravity is g = 9.81 m/s2 . What is the weight Wb of the ball? Express your answer numerically in newtons.

What is the reading W2 of the scale when the ball is held in this submerged position? Assume that none of the water that spills over stays on the scale. Calculate your answer from the quantities given in the problem and express it numerically in newtons.

What is the force Fr applied to the ball by the rod? Take upward forces to be positive (e.g., if the force on the ball is downward, your answer should be negative). Express your answer numerically in newtons.

The rod is now shortened and attached to the bottom of the beaker. The beaker is again filled with fluid, the ball is submerged and attached to the rod, and the beaker with fluid and submerged ball is placed on the scale.

What weight W3 does the scale now show?

Answer:

(a) 2.94 N

(b) 1 N

(d) 3.42 N

Explanation:

(a)

From the definition of density, it's mass per unit volume hence mass is a product of density and volume. To get weight, we multiply mass by acceleration due to gravity

The weight of the ball is $W=\rho g V$

Where $\rho$ is the density, V is volume and g is acceleration due to gravity

Substituting density for 5000 Kg/m3 and g for 9.8 m/s2 and v for 0.00006 m3 then

$W= 5000 kg/m^{3} * 9.8 m/s^{2} * 0.00006 m^{3}=2.94 N$

(b)

Because the ball is being held up mostly by the rod, the fluid pressure on the bottom of the cylinder is just the same as before.

The scale does not "know" the ball is there at all.

That's why it still reads 1 N.

Therefore, the reading is 1 N

(c)

The buoyant force of the fluid on the ball is equal to the weight of the displaced fluid, namely,

$890 kg/m^{3} * 9.8 m/s^{2} * 0.00006 m^{3} = 0.52 N$

so the force needed for the rod to hold up the ball is 2.94 N - 0.52 N = 2.42 N.

(d)

Now the scale "feels" the weight of the ball,

so the scale reads the weight of the ball

PLUS the weight of the original fluid

MINUS the weight of the fluid that was displaced

= 2.94 N + 1.00 N - 0.52 N = 3.42 N

6 0

3 years ago

At a sports car rally, a car starting from rest accelerates uniformly at a rate of 5 m/s/s over a straight-line distance of 291

Bogdan [553]

Answer:

10.8 s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Acceleration (a) = 5 m/s/s

Distance travelled (s) = 291 m

Time (t) taken =?

We can calculate the time taken for the car to cover the distance as follow:

s = ut + ½at²

291 = 0 × t + ½ × 5 × t²

291 = 0 + 2.5 × t²

291 = 2.5 × t²

Divide both side by 2.5

t² = 291 / 2.5

t² = 116.4

Take the square root of both side

t = √116.4

t = 10.8 s

Thus, it will take the car 10.8 s to cover the distance.

8 0

3 years ago

A charge of −20 µC is distributed uniformly over the surface of a spherical conductor of radius 11.0 cm. Determine the electric

Alex73 [517]

Answer:

(a) $-6.76\times 10^{12}\ N/C$

(b) $-1.352\times 10^{13}\ N/C$

(c) $-7.2\times 10^{11}\ N/C$

Explanation:

(a)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 5 cm = 0.05 m

Coulomb's constant (k) = $9\times 10^{9}\ Nm^2/C^2$

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.05\\\\E_{in}=-1.352\times 10^{14}\times 0.05\\\\E_{in}=-6.76\times 10^{12}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

(b)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 10 cm = 0.10 m

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.10\\\\E_{in}=-1.352\times 10^{14}\times 0.10\\\\E_{in}=-1.352\times 10^{13}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$