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tigry1 [53]
3 years ago
9

When the car carrying the unbelted driver crashes, what brings the driver to a stop

Physics
1 answer:
pychu [463]3 years ago
4 0
When the driver is not wearing a seat belt and the car hits a brick wall,
the driver keeps going and doesn't stop, (because of the inertia of the
driver's body). 

-- If the driver is lucky, the car is going slow enough, AND it has a
driver-side airbag, AND the airbag works, AND it inflates fast enough,
AND it cushions the driver well enough to avoid the driver's death.

-- If an airbag doesn't stop the driver, then one or more of these brings
the driver to a stop:

... the car's engine block crashing in through the dashboard 
... the steering wheel
... the windshield
... the same brick wall that stopped the car.

Any one of these is probably fatal.
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What will be the time period of simple pendulum at the center of earth​
Furkat [3]

Answer:

As g=0 at the centre of earth the time period becomes infinite as T=2pi/underoot g. At center of earth, r = 0 & hence g = 0. Time period of a simple pendulum is T = 2π sqrt (1/g).

Explanation:

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3 years ago
Dharna is said to be concentration it is true or false​
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5 0
2 years ago
Base your answers to questions 4 and 5 on the informa- tion below
emmasim [6.3K]

If the sphere is positively charged, the positive rod is repelled by the sphere while the negative rod is attracted by the sphere.

<h3>What is an electrical charge?</h3>

An electrical charge can be positive or negative. From the laws of electrostatics, unlike charges attract while like charges repel. As such the effect observed when the rods are individually brought near the sphere will decide the charge on the sphere.

If the sphere is neutral, there is no effect observed when each rod is brought near the sphere. If the sphere is positively charged, the positive rod is repelled by the sphere while the negative rod is attracted by the sphere.

Learn more about electrostatics: brainly.com/question/9774180

3 0
2 years ago
A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 w
steposvetlana [31]

Explanation:

a) The height of the ball h with respect to the reference line is

h = L - L\cos{31°} = L(1 - \cos{31°})

so its initial gravitational potential energy U_0 is

U = mgh = mgL(1 - \cos{31°})

\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})

\:\:\:\:\:=0.23\:\text{J}

b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U

We know that the initial kinetic energy K_0, as well as its final gravitational potential energy U are zero so we can write the conservation law as

mgL(1 - \cos{31°}) = \frac{1}{2}mv^2

Note that the mass gets cancelled out and then we solve for the velocity v as

v = \sqrt{2gL(1 - \cos{31°})}

\:\:\:\:\:= \sqrt{2(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31°})}

\:\:\:\:\:= 1.3\:\text{m/s}

5 0
2 years ago
Read 2 more answers
An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.275 T. If the kinetic energy of the electr
xxMikexx [17]

Answer:

Radius, r=2.14\times 10^{-5}\ m

Explanation:

It is given that,

Magnetic field, B = 0.275 T

Kinetic energy of the electron, E=4.9\times 10^{-19}\ J

Kinetic energy is given by :

E=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{2E}{m}}

v=\sqrt{\dfrac{2\times 4.9\times 10^{-19}}{9.1\times 10^{-31}}}            

v = 1037749.04 m/s

The centripetal force is balanced by the magnetic force as :

qvB\ sin90=\dfrac{mv^2}{r}

r=\dfrac{mv}{qB}

r=\dfrac{9.1\times 10^{-31}\times 1037749.04}{1.6\times 10^{-19}\times 0.275 }

r=2.14\times 10^{-5}\ m

So, the radius of the circular path is 2.14\times 10^{-5}\ m. Hence, this is the required solution.

3 0
2 years ago
Read 2 more answers
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