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3 years ago

15

Why do crumple zones reduce the force in a collision.

1 answer:

Stels [109]3 years ago

4 0

Crumple zones are areas of a vehicle that are designed to crush in a controlled way in the collision

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Diffusion occurs faster in gases than in liquids because _____.

jolli1 [7]

The correct answer to go in the blank would be A) The particles are moving faster.

5 0

3 years ago

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An object of mass m is hung from a spring and set into oscillation. The period of the oscillation is measured and recorded as T.

sergejj [24]

Answer: $\sqrt{2}T$

Explanation:

Given

object of mass m is suspended from spring and set in oscillation with time Period T

We know Time period of a mass in oscillation is given by

$T=2\pi \sqrt{\frac{m}{k}}$

where k=spring constant

When mass m is replaced by a mass of 2 m time period is given by

$T'=2\pi \sqrt{\frac{2m}{k}}$

$T'=\sqrt{2}\times 2\pi \sqrt{\frac{m}{k}}$

$T'=\sqrt{2}T$

i.e. New time period becomes $\sqrt{2}$ times of previous one

7 0

2 years ago

A certain mass of a gas occupies 2 litres at 27 °C and 100

Zielflug [23.3K]

Answer:

75k

Explanation:

You can see solution in the picture

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2 years ago

A circular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux

Mila [183]

Answer:

C) 2.44 × 106 N/C

Explanation:

The electric flux through a circular loop of wire is given by

$\Phi = EA cos \theta$

where

E is the electric field

A is the cross-sectional area

$\theta$ is the angle between the direction of the electric field and the normal to A

The flux is maximum when $\theta=0^{\circ}$ , so we are in this situation and therefore $cos \theta =1$ , so we can write

$\Phi = EA$

Here we have:

$\Phi = 7.50\cdot 10^5 N/m^2 C$ is the flux

d = 0.626 m is the diameter of the coil, so the radius is

r = 0.313 m

and so the area is

$A=\pi r^2 = \pi (0.313 m)^2=0.308 m^2$

And so, we can find the magnitude of the electric field:

$E=\frac{\Phi}{A}=\frac{7.50\cdot 10^5 Nm^2/C}{0.308 m^2}=2.44\cdot 10^6 N/C$

3 0

3 years ago

A scientist at a space research center is doing research on near-Earth objects and determining the kinetic energy of a theoretic

Marysya12 [62]

<span>its kinetic energy is 7350kJ
</span>

Kinetic energy is given as = $\frac{1}{2} mv^{2}$

Now, m = 12 gms = 0.012 kg
And, velocity = 35 kilometers/second = 35000 m/sec

Kinetic energy is given as = $\frac{1}{2} 0.012 kg * 35000*35000 m/[tex] s^{2}$
= 6 $10^{-2}$ ×1225 × $10^{6}$ m/ $s^{2}$
= 7350 kJ

6 0

3 years ago

Read 2 more answers