All categories

3 years ago

Why is it important to develop alternative sources of energy?

1 answer:

7 0

We need to use differant energy soureces that are renewable. THis is because at the moment most of the enrgy that we use is unrenewable, coal, oil, petrol etc. It takes millions of year to create and is burnt in a matter of minuetes. We need energy that is renewable so we can supply it quickly when we need it because all nonrenewable enrgy's are running out. Only in 1 million years if a fish dies now will it be compatable to turn into fuel. Wind power, solar power, movement energy are all renewable. THis is why it is very important we use those things. Hope this helped :-) Bonnie Year 7

You might be interested in

The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiato

djverab [1.8K]

Answer:571.09 kJ

Explanation:

Given

Temperature of cooling water from engine exit $=240^{\circ} F\approx 115.55^{\circ}C$

After Passing through the radiator its temperature decreases to $175^{\circ}F\approx 79.44^{\circ}F$

specific heat of water $=4.184 J/g^{\circ}C$

Volume of water $= 1 gallon\approx 3.78 L$

density of water $\rho =1 gm/mL$

Thus mass of water $=\rho \times V=3.78\times 1=3.78 kg$

Heat transferred to the surrounding is equal to heat absorbed by cooling water

$Q=m\cdot c\cdot \Delta T$

$Q=3.78\times 4.184\times 1000\times (115.55-79.44)$

$Q=3.78\times 4.184\times 1000\times (36.11)$

$Q=571.09 kJ$

4 0

3 years ago

Find the circumference of a circle whose area is 452.16 square meters. (Use 3.1416 as the value of .)

kakasveta [241]

1) The area of a circle is given by

$A=\pi r^2$

where r is the radius of the circle. In this problem, we are told that the area is

$A=452.16 m^2$

so we can rearrange the previous equation to calculate the radius:

$r=\sqrt{\frac{A}{\pi}}=\sqrt{\frac{452.16 m^2}{3.1416}}=12.0 m$

And now we can find the circumference, by using the formula:

$c=2 \pi r=2 \pi (12.0 m)=75.3984 m$

Therefore, the correct answer is D. 75.3984 meters.

2) The volume of a cube is given by:

$V=L^3$

where L is the length of the edge. In this problem, L=6 cm, therefore the volume of the cube is

$V=L^3=(6 cm)^3=216 cm^3$

and so, the correct answer is A. 216 cu. cm.

5 0

4 years ago

Hello guys! Can you please help me with physics. I translated it in English. Can yall help me please how much you can!!

Alex777 [14]

Answer:

1. 60m

4. its speed is 20m/s . gravity is 10m/s . 20 ÷ 10 = 2 seconds after these 2 sec are over the velocity will be zero thus it would have fallen.

Explanation:

Am sorry these is what i could answer but i will try to research and get answers for the rest of the questions, hope this helps

6 0

3 years ago

A) An electron is moving with a speed of 3.5 x105 m/s when it encounters a magnetic field of 0.60T. The direction of the magneti

kirza4 [7]

Given Information:

v = 3.5x10⁵ m/s

θ = 60°

Magnetic field = 0.60 T

no. of electrons = 6.242x10¹⁸

radius = 0.750 m

Required Information:

Magnetic force = F = ?

Magnetic field = B = ?

Answer:

Part A) F = 2.909x10⁻¹⁴ N

Part B) B = 5.34x10⁻⁷ T

Explanation:

Part A:

The magnetic force can be found using

F = qvBsin(θ)

Where q = 1.60x10⁻¹⁹ C and v is the speed of electron, B is the magnetic field and θ is the angle between v and B.

F = 1.60x10⁻¹⁹*3.5x10⁵*0.60sin(60°)

F = 2.909x10⁻¹⁴ N

Part B:

From the Ampere's law, the magnetic field can be found using

B = μ₀I/2πr

Where μ₀ = 4πx10⁻⁷ is the permeability of free space, I is the current, and B is the magnitude of the magnetic field produced.

1 ampere of current has 6.242x10¹⁸ electrons per second so

I = 1.25x10¹⁹/6.242x10¹⁸

I = 2.0032 A

B = 4πx10⁻⁷*2.0032/2π*0.750

B = 5.34x10⁻⁷ T

6 0

4 years ago

Two capacitors of capacitances 1.0 microfarad and 2.0 microfarads are each charged by being connected across a 5.0-volt battery.

Svetach [21]

Answer:

1.67 V

Explanation:

Q = Capacitance

Q' = Capacitance after connecting in series

V = Voltage

$V_t$ = Voltage of battery = 5 V

$Q_1=C_1\times V_t\\\Rightarrow Q_1= 1\times 5=5\ \mu F$

$Q_2=C_2\times V_t\\\Rightarrow Q_2= 2\times 5=10\ \mu F$

As the plates opposite charge are connected together they are in series

$Q_1'=Q_2'=Q_t$

$\frac{1}{C_t}=\frac{1}{C_1}+\frac{1}{C_2}\\\Rightarrow \frac{1}{C_t}=\frac{1}{1}+\frac{1}{2}\\\Rightarrow \frac{1}{C_t}=\frac{3}{2}\\\Rightarrow C_t=\frac{2}{3}$

$Q_2'=C_tV_t\\\Rightarrow Q_2'=\frac{2}{3}\times 5\\\Rightarrow Q_2'=\frac{10}{3}\ \mu F$

$V_2=\frac{Q_2'}{C_2}\\\Rightarrow V_2=\frac{\frac{10}{3}}{2}\\\Rightarrow V_2=1.66\ V$

The voltage across the 2 microfarad capacitor is 1.67 V

4 0

4 years ago