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2 years ago

10

Head gear is required for many sports and can be the most important piece of safety equipment you can have. Question 2 options:

True False

2 answers:

Crank2 years ago

6 0

Answer:true

Explanation:

bija089 [108]2 years ago

4 0

It’s true because head injuries can be the most harmful, and possible fatal, therefore it is the most important area to protect

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An undamped spring-mass system contains a mass that weighs and a spring with spring constant . It is suddenly set in motion at b

balandron [24]

Answer:

Explanation:

When all other forces acting on the mass in a damped mass-spring system are grouped together into one term denoted by F(t), the differential equation describing

motion is

Mx''+ βx' + kx = F(t).

Note for an undamped system

β=0,

Then, the differential equation becomes

Mx'' + kx = F(t).

The force is in the form

F=Fo•Sinωo•t

Let solved for the homogeneous or complementary solution, I.e f(t) = 0

Using D operator

MD² + k = 0

MD²=-k

D²=-k/M

Then, D= ±√(-k/m)

D=±√(k/m) •i

So we have a complex root

Therefore, the solution is

x= C1•Cos[√(k/m)t] + C2•Sin[√(k/m)]

This is simple harmonic motion that once again we prefer to write in the form

x(t) = A•Sin[ √(k/M)t + φ]

Where A=√(C1²+C2²)

and angle φ is defined by the equations

sin φ = C1/A and cos φ = C2/A.

Quantity √(k/M), often denoted by ω, is called the angular frequency.

This is called the natural frequency (ωn) of the system

ωn=√(k/M)

ωn²= k/M

Now, for particular solution

Xp=DSinωo•t

Xp' = Dωo•Cosωo•t

Xp"=-Dωo²•Sinωo•t

Now substituting this into

Mx'' + kx = F(t).

M(-Dωo²•Sinωo•t) + k(DSinωo•t)=FoSinωo•t

Now, let solve for D

D(-Mωo²•Sinωo•t +kSinωo•t) = FoSinωo•t

D=Fo•Sinωo•t/(-Mωo²•Sinωo•t +kSinωo•t)

D=Fo•Sinωo•t / Sinωo•t(-Mωo²+k)

D=Fo / (-Mωo²+k)

D=Fo / (k-Mωo²)

Divide through by k

D=Fo/k ÷ (1 -Mωo²/k)

Note from above

ωn²= k/M

Therefore,

D=Fo/k ÷ (1-ωo²/ωn²)

D=Fo/k ÷ [1-(ωo/ωn)²]

Then,

Xp=DSinωo•t

Xp=(Fo/k ÷ [1-(ωo/ωn)²]) Sinωo•t

Then the general solution is the sum of the homogeneous solution and particular solution

Xg(t)=(Fo/k ÷ [1-(ωo/ωn)²]) Sinωo•t + A•Sin[ √(k/M)t + φ]

Check attachment for the graph of homogeneous, particular and general solution.

Also, check for better way of writing the equations.

8 0

3 years ago

A positive point charge (q = +8.65 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 2.97 m. A p

ella [17]

Answer:

$r_B = 1.88 m$

Explanation:

As we know that work done by electric force is given as

$W_e = -q\Delta V$

so here we know that charge is moving from

$r_A = 2.97 m$

to another position

so we will have

$W_{AB} = \frac{kq_1q_2}{r_A} - \frac{kq_1q_2}{r_B}$

$-6.95\times 10^{-9} = (9\times 10^9)(8.65\times 10^{-8})(4.56\times 10^{-11})(\frac{1}{2.97} - \frac{1}{r_B})$

$-0.196 = (\frac{1}{2.97} - \frac{1}{r_B})$

$r_B = 1.88 m$

7 0

3 years ago

Set the initial bead height to 3.00 m. Click Play. Notice that the ball makes an entire loop. What is the minimum height require

strojnjashka [21]

Answer:

h> 2R

Explanation:

For this exercise let's use the conservation of energy relations

starting point. Before releasing the ball

Em₀ = U = m g h

Final point. In the highest part of the loop

Em_f = K + U = ½ m v² + ½ I w² + m g (2R)

where R is the radius of the curl, we are considering the ball as a point body.

I = m R²

v = w R

we substitute

Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R

em_f = m v² + 2 m g R

Energy is conserved

Emo = Em_f

mgh = m v² + 2m g R

h = v² / g + 2R

The lowest velocity that the ball can have at the top of the loop is v> 0

h> 2R

3 0

2 years ago

Find the ratio of the new/old periods of a pendulum if the pendulum were transported from earth to the moon, where the accelerat

vichka [17]

The period of a pendulum is given by
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the pendulum length and g is the gravitational acceleration.

We can write down the ratio between the period of the pendulum on the Moon and on Earth by using this formula, and we find:
$\frac{T_m}{T_e} = \frac{2 \pi \sqrt{ \frac{L}{g_m} } }{2 \pi \sqrt{ \frac{L}{g_e} } }= \sqrt{ \frac{g_e}{g_m} }$
where the labels m and e refer to "Moon" and "Earth".

Since the gravitational acceleration on Earth is $g_e = 9.81 m/s^2$ while on the Moon is $g_m=1.63 m/s^2$ , the ratio between the period on the Moon and on Earth is
$\frac{T_m}{T_e}= \sqrt{ \frac{g_e}{g_m} }= \sqrt{ \frac{9.81 m/s^2}{1.63 m/s^2} }=2.45$

3 0

2 years ago

Explain how you can use the factors of production to produce a fruit juice in a production company

frozen [14]

Answer:

The factors of production include Land, Labour, Capital and Enterpreneurship

Explanation:

The fruit could be apple, orange , pineapple etc which are usually grown on land . They are tended to by people to ensure there is maximum yield. These people provide the required labour needed.

The cost of planting and payment of workers usually comes from the capital which is often used in running the business by the owner which makes certain decisions to ensure the fruit company is in place. All these factors work hand in hand to ensure production of fruit in a production company is possible.

7 0

2 years ago