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2 years ago

You are falling off the edge.. What should you do to avoid falling..

Physics

2 answers:

FrozenT [24]2 years ago

4 0

You could try not to panic and try ur best to get up. If not, call for help.

Ugo [173]2 years ago

3 0

If it is a matter of which way you are going you could lean forward. It would help to put all the weight opposite of where you are falling.

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A car starting from rest accelerates at a constant 2.0 m/s2 for 10 s. It then travels with constant speed it has achieved for an

Anon25 [30]

Answer:

The total distance will be 400 m.

Explanation:

For portion AB:

Acceleration = 2 $m/s^2$

t= 10 s

Car start from rest , u=0 m/s

We know that

$S=ut+\dfrac{1}{2}at^2$

$S=0+\dfrac{1}{2}\times 2\times 10^2$

S= 100 m.

For portion BC:

V= u + at

V=0 + 2 x 10

V= 20 m/s

In this portion car moves with constant velocity 20 m/s for 10 s.

So distance S= V x t

S=20 x 10 =200 m.

For portion CD:

The velocity at point C will be 20 m/s

In this portion the final speed of car will be zero because given that at final car come to rest.

So the acceleration will be in the negative direction to stop the car.

We know that

$v^2=u^2-2aS$

$0=20^2-2\times 2\times S$

S=100 m

The total distance AD=AB + BC+ CD

AD=100 +200 + 100 m

AD=400 m.

The total distance will be 400 m.

4 0

2 years ago

A circular loop of radius 11.9 cm is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane

djverab [1.8K]

Answer:

(a). The strength of the magnetic field is 0.1933 T.

(b). The magnetic flux through the loop is zero.

Explanation:

Given that,

Radius = 11.9 cm

Magnetic flux $\phi=8.60\times10^{-3}\ T m^2$

(a). We need to calculate the strength of the magnetic field

Using formula of magnetic flux

$\phi=BA\cos\theta$

$\phi=BA\cos0$

$\phi=BA$

$B=\dfrac{\phi}{A}$

$B=\dfrac{\phi}{\pi r^2}$

Put the value into the formula

$B=\dfrac{8.60\times10^{-3}}{\pi\times(11.9\times10^{-2})^2}$

$B=0.1933\ T$

(b). If the magnetic field is directed parallel to the plane of the loop,

We need to calculate the magnetic flux through the loop

Using formula of flux

$\phi=BA\cos\theta$

Here, $\theta=90^{\circ}$

$\phi=BA\cos90$

$\phi=0$

Hence, (a). The strength of the magnetic field is 0.1933 T.

(b). The magnetic flux through the loop is zero.

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Explanation:

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Answer:

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