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diamong [38]
3 years ago
8

What is the amount of charge when 13.5a is flowing for 2 1/2 hours

Physics
2 answers:
abruzzese [7]3 years ago
4 0

Answer:

Quick Answer:

a.  8.9Ω

b.  1.2×104 C

Explanation:

Illusion [34]3 years ago
4 0

Answer:

Answer provided in explanation

Explanation:

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Consider an embedded system which uses a battery with a 17.15-Amp-Hour capacity. What is the maximum average current draw (in mi
Marianna [84]

Answer:

<em>85.12 μAmp</em>

<em></em>

Explanation:

The battery power output = 17.15 Amp-hr

If the battery is to last 23 years, we have to calculate how many hours there are in 23 years

in one year there are 24 hours x 365 day = 8760 hrs

in 23 years there are 23 x 8760 = 201480 hours

maximum current to be drawn from the battery = (17.15 Amp-hr) ÷ (201480 hours) = 85.12 x 10^-6 Amp = <em>85.12 μA</em>

7 0
2 years ago
Define momentum in terms of football.
inna [77]
The law of conservation of momentum basically means that energy is always conserved and never lost when a collision happens.

Using the formula p=mv ...
Player A would have a momentum of 220 N•S
Player B would have a momentum of 0 because he is not moving

After the collision, the total momentum is still 220 N•S because energy is never lost, but now player A is at 0 and player B took his momentum. Think about it this way, if you bumped into something that wasn’t moving, it would fall and you most likely wouldn’t keep moving.

Elastic collisions are where the objects bounce each other and in inelastic collisions they stick together. I don’t watch much football but if you do this should make sense.
If the players fall down together (they tackle each other and fall? I think) it should be inelastic.

Sorry if this was long and confusing but I really hope this helps! ☺️
7 0
3 years ago
found the potential spring energy. not sure how to find the height of the block by knowing only the mass. found the weight of it
evablogger [386]

(h + .16) m g = 1/2 k x^2   total PE of block relative to where it stops

(h + .16) .82 * 9.8 = .5 * 120 * .16^2    PE released = PE of  spring

8.04 h + 1.29 = 1.536

h = (1.536 - 1.29) / 8.04 = .031 m = 3.1 cm

7 0
3 years ago
A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the grou
True [87]

Answer:

At the highest point the velocity is zero, the acceleration is directed downward.

Explanation:

This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.

I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.

At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.

6 0
3 years ago
Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water
Ksenya-84 [330]

To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

v_2^2 = v_1^2 + 2gh

Where,

V_i = Velocity in each state

g= Gravity

h = Height

Our values are given as,

A_1 = 2.4*10^{-4} m^2

v_1 = 0.8 m/s

h = 0.11m

Replacing at the kinetic equation to find V_2 we have,

v_2 = \sqrt{v_1^2 + 2gh}

v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}

v_2= 1.67 m/s

Applying the concepts of continuity,

A_1v_1 = A_2v_2

We need to find A_2 then,

A_2= \frac{A_1v_1 }{v_2}

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

A_2= \frac{A_1v_1 }{v_2}

A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}

A_2= 1.14*10^{-4} m2

Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is 1.14*10^{-4} m2

8 0
3 years ago
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