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viktelen [127]
3 years ago
13

Which type of friction force acts when there is no motion?

Physics
1 answer:
rosijanka [135]3 years ago
3 0
Its called static friction.
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A complete circuit contains two parallel connected devices and a generator for providing the electromotive force. The resistance
irga5000 [103]
We are asked to solve and determine the magnitude of the current flowing through the first device. In order for us to have a better understanding of the problem, we can refer to the attached picture which contains electric circuit diagram. Since it the problem we are already given with an electromotive source or the voltage supply and since the two resistance is in parallel, it would clearly mean that the voltage drop in each resistance is just the same. The resistance 1 uses the 40 volts at the same time the resistance 2 uses 40 volts also. Solving further for the current, we can apply Ohm's law which V = IR where "V" represents the voltage, the "I" represents the current and "R" represents the resistance.

Such as the solution in obtaining current is shown below:
I = V / R, substitute values we have it
I = 40 volts / 1208 ohms
I = 0.0331 Amperes

Therefore, the current flowing in the first device is 0.033 Amperes or 33 milliAmperes.

6 0
3 years ago
A 100 kg person rides in an elevator moving at a constant speed of
Airida [17]

Answer:

Explanation:

All this information only applies to the person. There is an extra tension force if we are talking about the elevator, but we are not. Dont forget to apply the units

Acceleration means change in speed or velocity. The elevator is moving at a constant speed of 3 meters. You wont even know you are moving because there is no change in acceleration. It equals 0

The forces ONLY acting on the person would be the force of gravity pulling them down, and the normal force that the elevator is reciprocating from the person standing on it.

Force = mass x acceleration. You have 100 kg and you are accelerating at 0 m/s. The force is 0. Which makes sense because the force of gravity and the net force completely cancel each other out.

8 0
3 years ago
This exercise uses the radioactive decay model. After 3 days a sample of radon-222 has decayed to 58% of its original amount. (a
Vadim26 [7]

Answer: a) 3.85 days

b) 10.54 days

Explanation:-

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = time taken for decomposition  = 3 days

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{58}{100}\times 100=58g

First we have to calculate the rate constant, we use the formula :

Now put all the given values in above equation, we get

k=\frac{2.303}{3}\log\frac{100}{58}

k=0.18days^{-1}

a) Half-life of radon-222:

t_{\frac{1}{2}}=\frac{0.693}{k}

t_{\frac{1}{2}}=\frac{0.693}{0.18}=3.85days

Thus half-life of radon-222 is 3.85 days.

b) Time taken for the sample to decay to 15% of its original amount:

where,

k = rate constant  = 0.18days^{-1}

t = time taken for decomposition  = ?

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{15}{100}\times 100=15g

t=\frac{2.303}{0.18}\log\frac{100}{15}

t=10.54days

Thus it will take 10.54 days for the sample to decay to 15% of its original amount.

3 0
3 years ago
Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 23.5 m above water wit
Elodia [21]

As stated in the statement, we will apply energy conservation to solve this problem.

From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as

\Delta KE = \Delta PE

\frac{1}{2}mv_f^2-\frac{1}{2} mv_0^2 = mgh_2-mgh_1

Where,

m = mass

v_{f,i} = initial and final velocity

g = Gravity

h = height

As the mass is tHe same and the final height is zero we have that the expression is now:

\frac{1}{2}v_f^2-\frac{1}{2} v_0^2 = gh_2

\frac{1}{2} (v_f^2-v_0^2) = gh_2

(v_f^2-v_0^2) = 2gh_2

v_f = \sqrt{2gh_2+v_0^2}

v_f = \sqrt{2(9.8)(23.5)+13.6^2}

v_f = 25.4m/s

7 0
3 years ago
If you fall from a building onto a net which extends the time of impact by 10 times, what happens to the force you experience?
zubka84 [21]

Answer:

A larger impulse. A 1-kg ball has twice as much speed as a 10-kg ball.

Explanation:

6 0
3 years ago
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