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2 years ago

Who has the primary responsibility to prepare for emergencies?

Physics

1 answer:

Zolol [24]2 years ago

5 0

Answer:

local government agencies

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What force is required to accelerate a body with a mass of 15kilograms at a rate of

Paladinen [302]

The force required is

(15 kg) x (the acceleration, in m/s²) newtons.

4 0

3 years ago

An object is released from rest and falls a distance h during the first second of time. How far will it fall during the next sec

Viefleur [7K]

Answer:

E. 3h

Explanation:

We know that

u = 0 m/s.

velocity after t = 1s

v = u+gt = 0+9.81 x 1s= 9.81 m/s

distance covered in 1st sec

= =>> ut+0.5 x g x t²

=>>0 + 0.5x 9.81 x 1 = 4.90m

Let 4.90 be h

distance travelled in 2nd second will now be used

So velocity after t = 1s

=>>1 x t+ 0.5 x g x t²

=>9.81x 1 + 0.5 x 9.81 x 1 = 3 x 4.90

So since h= 4.90

Then the ans is 3x h = 3h

3 0

2 years ago

igor_vitrenko [27]

Answer:

$a=40\ m/s^2$

Explanation:

Given that,

Initial speed of a shuttlecock, u = 30 m/s

Final speed of the shuttlecock, v = 10 m/s

Time, t = 0.5 s

We need to find its average acceleration. The acceleration of an object is equal to the change in speed divided by time taken. It is given by :

$a=\dfrac{v-u}{t}\\\\a=\dfrac{10-30}{0.5}\\\\a=-40\ m/s^2$

So, the average acceleration of badminton shuttlecock is $40\ m/s^2$ .

3 0

3 years ago

A diffraction grating produces a first-order bright fringe that is 0.18 m away from the central bright fringe on a flat screen.

Georgia [21]

Answer:

The wavelength of the light is 562.5 nm

Solution:

As per the question:

Order, n = 1

Slit separation, d = $2.5\times 10^{- 6}\ m$

Distance from the bright fringe, y = 0.18 m

Distance between the screen and the grating, D = 0.8 m

Now,

We know from the eqn for diffraction:

$n\lambda = dsin\theta$

n = 1

$\lambda = dsin\theta$ (1)

Also,

For very small angle, $\theta$ :

$sin\theta$ ≈ $tan\theta = \frac{y}{D} = \frac{0.18}{0.8} = 0.225$

Using the above value in eqn (1):

$\lambda = 2.5\times 10^{- 6}\times 0.225 = 5.625\times 10^{- 7}\ m = 562.5\ nm$

3 0

3 years ago

Gather data: Select Mercury from the Solar system menu at left. Turn on Additional data. In the table below, record Mercury’s Ma

ki77a [65]

Answer:

volume of substance of weight of mercury is 13593 kilograms

3 0

3 years ago