Which of the following is associated with fission but not fusion?

Inessa [10]

The watt is a rate, similar to something like speed (miles per hour) and other time-interval related measurements.

Specifically, watt means Joules per Second. We are given that the electrical engine has 400 watts, meaning it can make 400 joules per second. If we need 300 kJ, or 3000 Joules, then we can write an equation to solve the time it would take to reach this amount of joules:

w * t = E

w: Watts

t: Time

E: Energy required

(Watts times time is equal to the energy required)

Input our values:

400 * t = 3000

(We need to write 3000 joules instead of 300 kilojoules, since Watts is in joules per second. It's important to make sure your units are consistent in your equations)

Divide both sides by 400 to isolate t:

$\frac{400t}{400}$ = $\frac{3000}{400}$

t = 7.5 (s)

It will take 7.5 seconds for the 400 W engine to produce 300 kJ of work.

If you have any questions on how I got to the answer, just ask!

- breezyツ

6 0

3 years ago

A beam of protons enter the electric field of magnitude E = 0.5 N/C between a pair of parallel plates. There is a magnetic field

HACTEHA [7]

Answer:

0.217 m/s

Explanation:

The protons in the beam passes undeflected when the electric force is equal to the magnetic force:

qE = qvB

where

q is the proton's charge

E is the magnitude of the electric field

v is the speed of the protons

B is the magnitude of the magnetic field

Re-arranging the equation,

$v=\frac{E}{B}$

And by substituting

E = 0.5 N/C

B = 2.3 T

We find

$v=\frac{0.5}{2.3}=0.217 m/s$

3 0

3 years ago

Where does the energy that is used to ride a bicycle up a hill come from and how is it classified?

MariettaO [177]

Well when you eat you get energy and then you use it to ride up the hill sorry I don't know how it is classified

7 0

3 years ago

Read 2 more answers

In this first example of constant accelerated motion, we will simply consider a car that is initially traveling along a straight

algol13

Answer:

$v_{f} =25m/s$

Explanation:

Kinematics equation for constant acceleration:

$v_{f} =v_{o} + at=15+2*5=25m/s$

4 0

3 years ago

A running student has half the kinetic energy that his brother has. The student speeds up by 1 m/s, at which point he has the sa

hoa [83]

Answer:

V = (√2) + 1) m/s

Explanation:

Let the mass and speed of the running student be M and V respectively.

We are told that when he speeds up by 1 m/s, he has the same kinetic energy as his brother.

Thus, his speed at which he mow has the same kinetic energy as his brother is (V + 1) m/s

Now, we are told that the mass of the student is twice as large as that of his brother. Thus, his brother's mass is; M/2

Since kinetic energy is given by the formula K.E = ½mv²

Therefore, since we want to find the original speed of both students and that the initial condition says that the running student had half the kinetic energy of the brother, we now initial condition as;

½MV²= ½(½(M/2)V²) - - - - (eq 1)

Since he has sped up by 1 m/s, and has a kinetic energy now equal to that of his brother, we have;

(½M(V + 1)²) = (½(M/2)V²) - - - - (Eq2)

Dividing eq 1 by eq 2 gives;

V²/(V + 1)²= 1/2

Taking square root of both sides gives;

V/(V + 1) = 1/√2

Cross multiply to give;

(√2)V = V + 1

(√2)V - V = 1

V((√2) - 1) = 1

V = 1/((√2) - 1)

Simplifying this using surfs gives;

V = [1/((√2) - 1)] × ((√2) + 1))/((√2) + 1))

V = ((√2) + 1))/1

V = (√2) + 1) m/s

8 0

3 years ago