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3 years ago

The waste products of a nuclear fission powerplant can best be described as

2 answers:

7 0

The waste products of a nuclear fission power plant can best be described as radioactive waste.
These are the by-products from the processes carried out that produce nuclear energy. This type of waste is highly dangerous. A lot of attention has to be paid to the collection and disposal of this waste as it must not reach any near by water bodies for example. It can be deadly for life.

Serggg [28]3 years ago

6 0

Answer:

D small in quantity and very dangerously radioactive.

Explanation: i had the same question

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According to recent research, ice skaters are able to glide smoothly across the ice because _____.

nekit [7.7K]

In the blank should go of friction.

8 0

3 years ago

How much heat is required to raise the temperature of 0.25 kg of water from 20°C to 30°C

pentagon [3]

Answer:

10500 J/kg/*C

Explanation:

Quantity of heat required=mass of substance x specific heat capacity x change in temperature

Quantity of heat required=0.25 x 4200 x [30-20]

Quantity of heat required=0.25 x 4200 x 10

Quantity of heat required=10500 J/kg/*C

3 0

2 years ago

A 500 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an avera

Fantom [35]

In order to accelerate the dragster at a speed $v_f = 100 m/s$ , its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
$W= K_f - K_i = K_f = \frac{1}{2}mv_f^2=2.5 \cdot 10^6 J$

however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
$W_f = F_f d = -(1200 N)(400 m)= -4.8 \cdot 10^5 J$
and the sign is negative because the frictional force acts against the direction of motion of the dragster.

This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is $-W_f$ :
$W_t = W + (-W_f)=2.5 \cdot 10^6 J+4.8 \cdot 10^5 J=2.98 \cdot 10^6 J$

So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
$P= \frac{W}{t}= \frac{2.98 \cdot 10^6 J}{7.30 s}=4.08 \cdot 10^6 W$

And since 1 horsepower is equal to 746 W, we can rewrite the power as
$P=4.08 \cdot 10^6 W \cdot \frac{1 hp}{746 W} =547 hp$

3 0

3 years ago

Read 2 more answers

How much work do you need to do if you use a force of 3 Newtons to move a table 25 meters?

Harlamova29_29 [7]

The work needed if a force of 3N is used to move a table 25 metres is 75Nm (joules).

<h3>How to calculate work done?</h3>

Work is a measure of energy expended in moving an object. This means that no work is done if the object does not move.

Force is a physical quantity that denotes the ability to push, pull, twist or accelerate a body and which has a direction and is measured in Newtons.

Work done on an object can be calculated by multiplying the force applied to move the object by the distance the object moves as follows:

Work done = Force (N) × distance (m)

According to this question, a force of 3 Newtons is used to move a table over a distance of 25 meters. The work done on the table can be calculated as follows;

W = 3N × 25metres

W = 75Nm

Therefore, 75Nm is the work done on the table.

Learn more about work done at: brainly.com/question/2117638

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6 0

8 months ago

How fast will it be traveling after it goes 87 m ?

swat32

To solve this we can use one of the main four kinematic equations, preferably we want something that we can use directly without solving for other values.

So in this problem we know that:

Inital velocity or Vi = 0 (started from rest)

The acceleration of the object = 7.5

The distance the object travels Δx = 87m

So, our best equation here with what we have would be:

Vf² = Vi² + 2aΔx

Now we now that Vi is 0 so:

Vf² = 2aΔx

We need final velocity so lets take the square root of Vf to isolate it:

Vf = √2aΔx

Now we plug in

Vf = √2(7.5)(87)

Vf = 36.124 m/s or 36.1 m/s

6 0

2 years ago