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4 years ago

Identify solar, wind, nuclear, and tidal power as nonrenewable or renewable energy

Physics

1 answer:

Nostrana [21]4 years ago

5 0

Answer:

Renewable energy: Tidal, solar and wind

Non renewable energy: nuclear power

Explanation:

Renewable energy is any energy in which its source is always replenished or renewed.

While non renewable energy sources cannot be replenished or renewed immediately because of its limited availability of resources.

Most renewable energy sources are natural sources.

Solar, tidal and wind energy are renewable energy known as clean energy. While nuclear energy is a non renewable energy in which its source of energy is fuel ( uranium e.t.c)

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The planet Mars has a mass of 6.1 × 1023 kg and radius of 3.4 × 106 m. What is the acceleration of an object in free fall near t

Oksanka [162]

The acceleration due to gravity of Mars is $3.5\ m / s^{2}$

<u>Explanation:</u>

As per universal law of gravity, the gravitational force is directly proportional to the product of masses and inversely proportional to the square of the distance between them. But in the present case, the gravity need to be determined between Mars and the object on Mars. Since the mass of Mars is greater than the mass of any object. Thus,

$\text {Gravitational force of planet}=\frac{G M m}{R^{2}}$

Here, G is the gravitational constant, R is the radius of Mars and M, m is the mass of Mars and the object respectively..

Also, according to Newton’s second law of motion, the acceleration of any object will be equal to the ratio of force exerted on it to the mass of the object.

So in order to determine the acceleration due to gravity of Mars, divide the gravitational force of Mars by mass of object on the surface of Mars.

$Acceleration\ due\ to\ gravity\ of\ mars =\frac{\text {Gravitational force of Mars}}{m \text { of object }}$

$Acceleration\ due\ to\ gravity\ of\ mars =\frac{G M m}{R^{2} \times m}=\frac{G M}{R^{2}}$

$\text { Acceleration due to gravity of mars }=\frac{6.67259 \times 10^{-11} \times 6.1 \times 10^{23}}{3.4 \times 3.4 \times 10^{12}}=\frac{40.703 \times 10^{12}}{11.56 \times 10^{12}}$

$\text { Acceleration }=3.5\ \mathrm{m} / \mathrm{s}^{2}$

3 0

3 years ago

How many electrons does copper borrow

Katen [24]

3....................................................

4 0

4 years ago

6. A garden hose attached to a nozzle is used to fill a 15-gal bucket. The inner diameter of the hose is 1.5 cm, and it reduces

Vesna [10]

Answer: 1.135 L/s; 1.35 kg/s, 22.57 m/s

Explanation:

Given

Volume of bucket $V=15\ gal\approx 56.78\ L$

time to fill it $t=50\ s$

Volume flow rate

$\dot{V}=\dfrac{56.78}{50}=1.135\ L/s\approx 1.135\times 10^{-3}\ m^3/s$

The inner diameter of the hose $D=1.5\ cm$

diameter of the nozzle exit $d=0.8\ cm$

we can volume flow rate as

$\Rightarrow \dot{V}=Av\quad \quad \text{v=average velocity through nozzle exit}\\\\\Rightarrow 1.135\times 10^{-3}=\frac{\pi }{4}d^2\times v\\\\\Rightarrow 1.135\times 10^{-3}=\frac{\pi }{4}(0.8\times 10^{-2})^2\times v\\\\\Rightarrow v=\dfrac{4\times 1.135\times 10^{-3}}{\pi \times 64\times 10^{-6}}=22.57\ m/s$

Mass flow rate

$\Rightarrow \dot{m}=\rho \times \dot{V}\\\Rightarrow \dot{m}=1\ kg/L\times 1.135\ L/s=1.35\ kg/s$

5 0

3 years ago

What is the frequency of the fundamental mode of vibration of a steel piano wire stretched to a tension of 440 N? The wire is 0.

Rasek [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

The tension is $T = 440 \ N$

The length of the wire is $l = 0.600 \ m$

The mass is $m = 5.60 \ g = 0.0056 \ kg$

Generally the frequency is mathematically represented as

$f = \frac{1}{2l} * \sqrt{\frac{T}{\frac{m}{l} } }$

=> $f = \frac{1}{2 * 0.600 } * \sqrt{\frac{440}{\frac{0.0056}{0.600} } }$

=> $f = 181 \ Hz$

4 0

3 years ago

Please show work : A particle with mass 2.00 μg and a charge of – 200 nC has a velocity of 3000 m/s in the x-direction. There is

irga5000 [103]

Answer:

x =4.5 10⁴ m

Explanation:

To find the distance that the particle moves we must use the equations of motion in one dimension and to find the acceleration of the particle we will use Newton's second law

m = 2.00 mg (1 g / 1000 ug) (1 Kg / 1000g) = 2.00 10-6 Kg

q = -200 nc (1C / 10 9 nC) = -200 10-9 C

Let's calculate the acceleration

F = ma

F = q E

a = qE / m

a = -200 10⁻⁹ 1000 / 2.00 10⁻⁶

a = 1 10² m / s²

Let's use kinematics to find the distance traveled before stopping, where it has zero speed (Vf = 0)

Vf² = Vo² -2 a x

0 = Vo² - 2 a x

x = Vo² / 2a

x = 3000²/ 2100

x =4.5 10⁴ m

This is the distance the particule stop, after this distance in the field accelerates in the opposite direction of the initial

Second part

In this case Newton's second law is applied on the y axis

F -W = 0

F = w = mg

E q = mg

E = mg / q

E = 2.00 10⁻⁶ 9.8 / 200 10⁻⁹

E = 9.8 10⁵ C

The direction of the field is such that the force on the particle is up, as the particle has a negative charge, the field must be directed downwards F = qE = (-q) E

7 0

4 years ago