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2 years ago

15

What unit is used to measure the amount of energy that people get from food?

1 answer:

3241004551 [841]2 years ago

4 0

Calories... I think Hopefully It's the right answer.

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Which of the following is the most appropriate description of an electric current?

Olenka [21]

Electric current is the flow of charge due to the potential difference between two terminals per unit time. It is denoted by I and its unit is amp. It can be mathematically expressed as I=Q/t.

If this isn’t what your looking for I need the options they gave you

6 0

2 years ago

Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time accordin

Vera_Pavlovna [14]

Answer:

θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²

Explanation:

This is an angular kinematic exercise the equation for the angular position

the particle A

θ = θ₀ + ω₀ t + ½ α t²

They say for the particle B

w₀B = ½ w₀

αB = 2 α

In addition, the particle begins at a time t_1 after particle A, in order to use the same timer, we must subtract this time from the initial

t´ = t - t_1

l

et's write the equation of particle B

θ = θ₀ + w₀B t´ + ½ αB t´2

replace

θ = θ₀ + ½ w₀ (t -t_1) + ½ 2α (t -t_1)²

θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²

4 0

2 years ago

This time particle A starts from rest and accelerates to the right at 65.5 cm/s

FrozenT [24]

Answer:

t = 4 s

Explanation:

As we know that the particle A starts from Rest with constant acceleration

So the distance moved by the particle in given time "t"

$d = v_i t + \frac{1}{2}at^2$

$d = 0 + \frac{1}{2}(65.5)t^2$

$d_1 = 32.75 t^2 cm$

Now we know that B moves with constant speed so in the same time B will move to another distance

$d_2 = 44 \times t$

now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

$d_1 = 349 + d_2$

$32.75 t^2 = 349 + 44 t$

on solving above kinematics equation we have

$t = 4 s$

4 0

3 years ago

Highest density if electrostatic charge in a metal is found in

arsen [322]

There’s no picture so how r we supposed to answer it

8 0

2 years ago

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

4 0

3 years ago