Why does the amount of damage to a car depend on speed?

I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block

borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

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

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

$T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\$

Now, determine the amplitude of oscillation, A;

$E = \frac{1}{2} kA^2$

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

$E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm$

Therefore, the amplitude of the oscillation is 2.82 cm

8 0

3 years ago

Help please ! USA test prep

Illusion [34]

C) 2 ohms I think don't fully trust my answer I suck at math but I think it's c

3 0

3 years ago

Two long parallel wires that are 0.30 m apart carry currents of 5.0 A and 8.0 A in the opposite direction. Find the magnitude of

olya-2409 [2.1K]

To solve this problem we will apply the concepts related to electromagnetic force. This force is defined as the product between the free space permeability constant, the product of the two currents found, at the rate of the distance that separates them. In this way mathematically it can be given as,

$F = \frac{\mu_0 I_1I_2 }{2\pi d}$

Here,

$\mu_0$ = Permeability of free space

I = Each current

d = Distance

Replacing with our values,

$F = \frac{(4\pi *10^{-7})(5)(8)}{2\pi (0.3)}$

$F = 2.7*10^{-5} N$

Therefore the force is $2.7*10^{-5}N$

5 0

3 years ago

An object moves with velocity v(t)=t^2-8t+7

solong [7]

a) write a polynomial expression for the position of the particle at any time t greater or equal to zero.
Position is found by integrating velocity:

s(t) = (t^3)/3 - 4t^2 + 7t + c
where c is a constant corresponding to the position at t=0. 

b) at what time(s) is the particle changing direction
the particle changes direction whenever the velocity is zero; the velocity function equals

(t-1)(t-7) a difference of squares so the zeros are 1 and 7, it changes direction at 1 second and 7 seconds. 

c) find the total distance traveled by the particle from t=0 and t=4
s(0) = c
s(1) = 8/3 + c
s(4) = 64/3 - 64 + 28 + c.

from 0 to 1 the particle travels 8/3 units. From 1 to 4 it travels -(64/3 - 36 - 8/3) = (-(56/3 - 108/3))
=-(-52/3) = 52/3 units

so in total it travels 52/3 + 8/3 =20 units

6 0

3 years ago

A proton of mass is released from rest just above the lower plate and reaches the top plate with speed . An electron of mass is

xenn [34]

Answer:

v = √ 2e (V₂-V₁) / m

Explanation:

For this exercise we can use the conservation of the energy of the electron

At the highest point. Resting on the top plate

Em₀ = U = -e V₁

At the lowest point. Just before touching the bottom plate

Emf = K + U = ½ m v² - e V₂

Energy is conserved

Em₀ = Emf

-eV₁ = ½ m v² - e V₂

v = √ 2e (V₂-V₁) / m

Where e is the charge of the electron, V₂-V₁ is the potential difference applied to the capacitor and m is the mass of the electron

3 0

4 years ago