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kherson [118]
4 years ago
14

Can someone tell me what my displacement is?

Physics
1 answer:
natulia [17]4 years ago
3 0
60017 is or answer ok you got it
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a college student produces about 100 kcal of heat per hour on the average what is the rate of energy production and joules
Bond [772]

Given:

Amount of heat produced = 100 kcal per hour

Let's find the rate of energy production in joules.

We know that:

1 calorie = 4.184 Joules

1 kcal = 4.184 Joules

To find the rate of energy production in Joules, we have:

\begin{gathered} Rate=100\ast4.184 \\  \\ \text{Rate}=418.4\text{ KJ/hour} \end{gathered}

Therefore, the rate of energy production in joules is 418.4 kJ/h which is equivalent to 418400 Joules

ANSWER:

418.4 kJ/h

6 0
1 year ago
To live a good life and be the sort of people we ought to be, we need to develop a virtuous character that
nordsb [41]

Answer:

I belive its THAT HELPS US BE A BETTER PERSON

3 0
3 years ago
In a pool game, the cue ball, which has an initial speed of 3.0 m/s, make an elastic collision with the eight ball, which is ini
zhenek [66]

Explanation:

Given

initial speed(u)=3 m/s

mass of each ball is m

Let the cue ball is moving in x direction initially

In elastic collision Energy and momentum is conserved

Let u be the initial velocity and v_1 , v_2 be the final velocity of 8 ball and cue ball respectively

\frac{mu^2}{2}+0=\frac{mv^2_1}{2}+\frac{mv^2_2}{2}

The angle after which cue ball is deflected is given by

\theta _1=90-40=50^{\circ}

Conserving momentum in x direction

mu=mv_1cos40+mv_2cos50

3=v_1cos40+v_2cos50

Along Y axis

0+0=v_1sin40-v_2sin50

v_1sin40=v_2sin50

substitute the value of v_1

we get v_2=1.912 m/s

v_1=2.27 m/s

5 0
3 years ago
The Newton is the SI unit for ____.<br> temperature<br> mass<br> weight<br> density
mel-nik [20]

Answer:

weight

Explanation:

weight is measurd in newtons (N)

7 0
3 years ago
Now suppose you carry out a second Thomson experiment with a different beam that contains two types of particles. In particular,
Tems11 [23]

Answer:

Two off-centered spots in the first phase of the experiment; one centered spot in the second phase of the experiment.

Explanation:

If two particles are selected in which both have the same electron mass and the same velocity, but one of the particles has a charge and the other particle has a charge of 2e. During the first stage of the experiment, the two particles have an electric force equal to F = Eq in the entire vertical direction. The acceleration of particle is equal to a = (Eq)/m.

In the second part of the experiment, the magnetic field cancels the electric field. In this way, the electric force and the magnetic force cancel each other out. Therefore, the net force acting on each particle is equal to zero.

Because these two forces cancel each other out, the particles fail to create two off-center points on the screen in the second part of the experiment. Also, if the loads are different, the deviation is also different. In this way, an off-center point cannot be achieved in the first part of the experiment. There will be two off-center points.

8 0
3 years ago
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