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2 years ago

If you drive through water, your brakes may become slippery and ineffective. To dry the brakes off, __________.

Physics

2 answers:

GenaCL600 [577]2 years ago

6 0

Answer:

I am not a driver, but I think it's C.

Explanation:

Tom [10]2 years ago

6 0

Answer:

Option-(C): If a person is driving through water, then he or she should avoid any accident by keep on accelerating by means of not taking off foot from the accelerator. While, at the same time keep on applying brakes slightly to the car in order to avoid slipping off the track or line.

Explanation:

Most of the road accidents happens just because the driver is unable to control his senses well enough which results in driving the car on the wrong direction or track. To avoid such incidents one must be able to master his or her skills to control the car in different situations. When a person is driving through water, one must then avoid to apply brakes all at the same time but the driver should rather keep on accelerating as the brakes are applied to the car. And when the car is all in control then any further actions should be considered depending upon the situations.

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The magnetic field around a current-carrying wire is ________proportional to the current and _________proportional to the distan

PSYCHO15rus [73]

Answer:Thus, The magnetic field around a current-carrying wire is directly proportional to the current and inversely proportional to the distance from the wire. If the current triples while the distance doubles, the strength of the magnetic field increases by one and half (1.5) times.

Explanation:

Magnetic field around a long current carrying wire is given by

$B=\frac{\mu _o I}{2\pi r}$

where B= magnetic field

$\mu _o=$ permeability of free space

I= current in the long wire and

r= distance from the current carrying wire

Thus, The magnetic field around a current-carrying wire is directly proportional to the current and inversely proportional to the distance from the wire.

Now if I'=3I and r'=2r then magnetic field B' is given by

$B'=\frac{\mu _oI'}{2\pi r'}=\frac{\mu _o3I}{2\pi 2r}=1.5B$

Thus If the current triples while the distance doubles, the strength of the magnetic field increases by one and half (1.5) times.

7 0

2 years ago

Read 2 more answers

What is the absolute value of the horizontal force that each athlete exerts against the ground?

alexandr402 [8]

Refer to the diagram shown below.

When an athlete is in motion, he/she exerts a vertical force (the person's weight, W) on the ground. The ground exerts an equal and opposite force, N, the normal reaction on the athlete, so that W = N.

At the same time, the ground exerts a horizontal force, F, o n the athlete so that he/she does not slip.
The magnitude of the horizontal force is
F = μN = μW
where μ = the dynamic coefficient of friction.

Answer:
The horizontal force is μW,
where
W = the weight of the athlete and,
μ = the dynamic coefficient of friction.

6 0

2 years ago

You will now use Graphical Analysis to calculate the slope of the line. Choose Analyze from the menu. Select Automatic Curve Fit

Vlad1618 [11]

The value for the slope is M=1.13

3 0

1 year ago

. Emergency rations are to be dropped from a plane to some stranded hikers. The search and rescue plane is flying at an altitude

almond37 [142]

Answer:

35 m/s down

Explanation:

The horizontal speed of the package is 70 m/s. So the time needed to reach the hikers is:

1000 m / (70 m/s) = 14.28 s

Taking down to be positive, the initial velocity needed is:

Δy = v₀ t + ½ at²

1500 m = v₀ (14.28 s) + ½ (9.8 m/s²) (14.28 s)²

v₀ = 35 m/s

The package must be launched down with an initial velocity of 35 m/s.

3 0

2 years ago

Read 2 more answers

A 385-g tile hangs from one end of a string that goes over a pulley with a moment of inertia of 0.0125 kg ⋅ m2 and a radius of 1

dimaraw [331]

Answer:

the tension in the string is 5.59 N

Explanation:

Here ,

m_1 = 0.385 Kg

m_2 = 0.710 Kg

Using second law of motion ,

a = F_net / effective mass

a = (0.710- 0.385)×9.8/(0.710 + 0.385 + 0.0125/0.15^2)

a = 1.93 m/s^2

Now , let tension be T ,

then,

mg-T=ma

0.710×g - T = 0.710×1.93

T = 5.59 N

the tension in the string is 5.59 N

7 0

1 year ago