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Vanyuwa [196]
1 year ago
6

You are driving on a road where rain has left large pools of water, and you have driven through water that was several inches de

ep. Now your brakes are not working. You may have
Physics
2 answers:
makvit [3.9K]1 year ago
3 0
In would say that you may have water in your brakes which may have gotten in the brake lines or in the brake discs so that could cause the brakes to malfunction due to driving through the pools of water so the brakes should be examined as soon as possible.
Jobisdone [24]1 year ago
3 0

<u>You're driving on a road where rain has left large pools of water, and you have driven through water that was several inches deep. Now your brakes are not working</u>. You may have the brakes to be examined ASAP. If the brakes become very wet, they just won`t work as well as they would when they are dry (the brakes will lose effectiveness, increasing your stopping distance). To un-soak the brakes, drive/pump them intermittently while accelerating the heat/friction generated will remove the water. It is dangerous to drive with wet brakes.

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An uncharged 30.0-µF capacitor is connected in series with a 25.0-Ω resistor, a DC battery, and an open switch. The battery has
kolezko [41]

Answer:

(a) Maximum current through resistor is 1.43 A

(b) Maximum charge capacitor receives is 1.50\times 10^{-3}\text{ C}.

Explanation:

(a)

In an RC (resistor-capacitor) DC circuit, when charging, the current at any time, <em>t</em>, is given by

I(t) = I_0e^{-t/\tau}

Here, I_0 is the maximum current and <em>τ</em> represents time constant which is given by RC (the product of the resistance and capacitance).

The maximum current is given by

I = \dfrac{V}{R_\text{eff}}

<em>V</em> is the emf of the battery and R_\text{eff} is the effective resistance.

In this question, R_\text{eff} = 10.0 Ω + 25.0 Ω = 35.0 Ω

I = \dfrac{50.0\text{ V}}{35.0\ \Omega} = 1.43\text{ A}

(b) The maximum charge is given

<em>Q</em> = <em>CV</em>

where <em>C</em> is the capacitance of the capacitor

Q = (30.0\times10^{-6}\text{ F})(50.0\text{ V}) = 1.50\times 10^{-3}\text{ C}

3 0
2 years ago
Read 2 more answers
The ammeter displays a reading of 0.10 A. Calculate the potential difference across the 45 Ω resistor.
NeX [460]

Answer:

V = 45× 0.10= 4.5 volt

...........

5 0
1 year ago
Why are fossil fuels considered nonrenewable resources if they are still forming beneath the surface today?
AleksandrR [38]

B is the answer because it takes millions of years to form these fossil fuels and everyday we use way more than we can find we may have a surplus for now but we may run out sooner than some think

7 0
2 years ago
Imagine you derive the following expression by analyzing the physics of a particular system: v2=v20+2ax. The problem requires so
lisabon 2012 [21]

As per kinematics equation we are given that

v^2 = v_o^2 + 2ax

now we are given that

a = 2.55 m/s^2

v_0 = 21.8 m/s

v = 0

now we need to find x

from above equation we have

0^2 = 21.8^2 + 2(2.55)x

0 = 475.24 + 5.1 x

x = 93.2 m

so it will cover a distance of 93.2 m

7 0
2 years ago
Read 2 more answers
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin
Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x^{2}, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

     W = \int_{0}^{0.54} F dx

         = \int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx

         = [14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}

         = 7560 + 1458 - 1364.69

         = 7653.31 J

or,      = 7.65 kJ       (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0
2 years ago
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