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3 years ago

The power of the kettle was 2.6 kW

Physics

1 answer:

faltersainse [42]3 years ago

7 0

Answer:

Cp = 4756 [J/kg*°C]

Explanation:

In order to calculate the specific heat of water, we must use the equation of energy for heat or heat transfer equation.

Q = m*Cp*(T_f - T_i)/t

where:

Q = heat transfer = 2.6 [kW] = 2600[W]

m = mass of the water = 0.8 [kg]

Cp = specific heat of water [J/kg*°C]

T_f = final temperature of the water = 100 [°C]

T_i = initial temperature of the water = 18 [°C]

t = time = 120 [s]

Now clearing the Cp, we have:

Cp = Q*t/(m*(T_f - T_i))

Now replacing

Cp = (2600*120)/(0.8*(100-18))

Cp = 4756 [J/kg*°C]

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What are 3 pros and cons of nuclear transmutation?

Tanya [424]

Pros -

It doesn't cost much
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Transportation of the material is easy

Cons-

It's not safe
Nuclear waste take 200 years to degrade
Greater risk of explosion

-IF THIS HELPED WHICH I HOPE COMMENT BELOWw ↓PLEASEX . THANKs!

6 0

3 years ago

Read 2 more answers

The era of planet formation ended when the remaining hydrogen and helium gas of the solar nebula was swept into interstellar spa

artcher [175]

Answer:

The era of planet formation ended when the remaining hydrogen and helium gas of the solar nebula was swept into interstellar space by the solar winds.

Explanation:

The Solar System is formed from a molecular cloud (compound by gas and dust). If there is a near perturbation to the cloud, maybe due to a supernova explosion, the molecular cloud will collapse under its own gravity. Then, in some point it starts to rotate and will accrete all the material in a disk around the protostar¹.

Inside the disk, dust particles start to collide and accrete until they form planetesimals². As a consequence of the gravitational force of the star, rocky and metallic particles will be more attracted to the inner part of the Solar System (close to the Sun) since they have more mass than gas.

Then, when the star has the necessary pressure and temperature to initiate nuclear reactions in its core, it will be able to emit huge amounts of energy, better known as solar winds. These winds will expel gas (hydrogen and helium) from the Solar System more easily than the rocky and metallic particles.

Notice that when such event occurs, rocky and gaseous planets were already formed.

Key terms:

¹Protostar: A young star.

²Planetesimals: Object formed by many fragments due to the gravitational attraction between them.

5 0

3 years ago

A football field lies so one endzone is to the east and one is to the west. A running back gets the ball and starts running 20o

Jobisdone [24]

Answer:

( 80.87 i - 12.96 j ) m

Explanation:

Running back movements :

20⁰ north of east for 10 meters,

Runs straight east for 60 meters

runs 35⁰ east of south for 20 meters

A) show vector pointing from the starting point to the end ( where he scored )

The final vector displacement : ( 80.87 i - 12.96 j ) m

which is : 81.90 m, 9.10⁰ south of east

attached below is the required diagram

8 0

3 years ago

A baseball is hit nearly straight up into the air with a speed of 22 m/s. (a) how high does it go? (b) how long is it in the air

Mariulka [41]

Refer to the diagram shown below.

When the ball attains maximum height, it will have zero vertical velocity.
The maximum height, h. obeys the equation
0 = (22 m/s)² - 2*(9.8 m/s²)*(h m)
h = 22²/(2*9.8) = 24.694 m

Answer: The maximum height attained is 24.7 m (nearest tenth)

Part b.
The vertical height traveled by the ball obeys the equation
h = (22 m/s)t - (1/2)*(9.8 m/s²)*(t s)²
where
h = vertical height, m
g = 9.8 m/s², acceleration due to gravity
t = time, s

To find how long the ball stays in the air, set h = 0 to obtain
4.9t² - 22t = 0
t(4.9t - 22) = 0
t = 0, or t = 22/4.9 = 4.49 s
t = 0 corresponds to the launch, and t = 4.49 s corresponds to when the ball retuns to the ground.

Answer: The ball stays in the air for 4.5 s (nearest tenth)

3 0

4 years ago

A 1.8 kg uniform rod with a length of 90 cm is attached at one end to a frictionless pivot. It is free to rotate about the pivot

Leokris [45]

Answer:

a. 32.67 rad/s² b. 29.4 m/s²

Explanation:

a. The initial angular acceleration of the rod

Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.

So, Iα = WL

mL²α/3 = mgL

dividing through by mL, we have

Lα/3 = g

multiplying both sides by 3, we have

Lα = 3g

dividing both sides by L, we have

α = 3g/L

Substituting the values of the variables, we have

α = 3g/L

= 3 × 9.8 m/s²/0.9 m

= 29.4/0.9 rad/s²

= 32.67 rad/s²

b. The initial linear acceleration of the right end of the rod?

The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²

5 0

3 years ago