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borishaifa [10]

1 year ago

15

What speed do all electromagnet waves travel

1 answer:

Grace [21]1 year ago

3 0

3x10^8 m/s, or the ‘speed of light’ - this applies to the entire spectrum

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A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equ

DochEvi [55]

Answer:

a) $k=19.6N/m$

b) $V_m=0.81m/s$

c) $a_m=6.561m/s^2$

d) $K.E=0.096J$

e) $T=0.78sec$ & $F=1.29sec$

f) $mx'' + kx' =0$

Explanation:

From the question we are told that:

Stretch Length $L=0.150m$

Mass $m=0.30kg$

Total stretch length $L_t=0.150+0.100=>0.25$

a)

Generally the equation for Force F on the spring is mathematically given by

$F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}$

$k=19.6N/m$

b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

$V_m=A\omega$

Where

A=Amplitude

$A=0.100m$

And

$\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s$

Therefore

$V_m=A\omega\\\\V_m=8.1*0.1$

$V_m=0.81m/s$

c)

Generally the equation for Max Acceleration of Mass on the spring is mathematically given by

$a_m=\omega^2A$

$a_m=8.1^2*0.1$

$a_m=6.561m/s^2$

d)

Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}*0.3*0.8^2$

$K.E=0.096J$

e)

Generally the equation for the period T is mathematically given by

$\omega=\frac{2\pi}{T}$

$T=\frac{2*3.142}{8.1}$

$T=0.78sec$

Generally the equation for the Frequency is mathematically given by

$F=\frac{1}{T}$

$F=1.29sec$

f)

Generally the Equation of time-dependent vertical position of the mass is mathematically given by

$mx'' + kx' =0$

Where

'= signify order of differentiation

7 0

3 years ago

HELLLP PLEASE || the graph below shows a conversion of energy for a skydive jumping out of a plane and landing safely on the gro

fenix001 [56]

Answer: I maybe wrong but i'm pretty sure its C) Kinetic energy

5 0

3 years ago

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh

n200080 [17]

Answer:

Part a)

$T_L = 155.4 N$

Part b)

$T_R = 379 N$

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

$T_L cos65 = T_R cos80$

$T_L = 0.41 T_R$

now we will have force balance in Y direction

$mg = T_L sin65 + T_Rsin80$

$514 = 0.906T_L + 0.985T_R$

Part a)

so from above equations we have

$514 = 0.906T_L + 0.985(\frac{T_L}{0.41})$

$514 = 3.3 T_L$

$T_L = 155.4 N$

Part b)

Now for tension in right string we will have

$T_R = \frac{T_L}{0.41}$

$T_R = 379 N$

3 0

3 years ago

Người ta đúc một khối nhóm (A) và một khối sắt ( (Fe) hình khối với cũng khói lượng 5 kg Khối lượng riêng (2) của Al và Fe là 2

IrinaVladis [17]

Translate to English please

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2 years ago

How do unicellular organisms grow?

Snezhnost [94]

Answer:

C is your answer good luck

Explanation:

3 0

2 years ago

Read 2 more answers