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4 years ago

How many vibrations per second are represented in a radio wave of 101.7 MHz. (1 MHz = 106Hz)?

Physics

1 answer:

Ne4ueva [31]4 years ago

3 0

Answer:

$n=101.7\times 10^6$

Explanation:

It is given that,

Frequency of the radio wave, $f=101.7\ MHz=101.7\times 10^6\ Hz$

We know that the number of vibrations per second is called frequency of an object. We need to find the number of vibrations per second. Clearly, the number of vibrations per second represented in a radio wave is $101.7\times 10^6$ . Hence, this is the required solution.

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NASA operates a 2.2-second drop tower at the Glenn Research Centre in Cleveland, Ohio. At this facility, experimental packages a

alukav5142 [94]

Answer:

23.7402 m

21.582 m/s

310.521816 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of motion

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 2.2^2\\\Rightarrow s=23.7402\ m$

The drop distance is 23.7402 m

$v=u+at\\\Rightarrow v=0+9.81\times 2.2\\\Rightarrow v=21.582\ m/s$

When they hit the air bags at the bottom of the tower the speed of the experiments is 21.582 m/s

The final speed of the fall will be the initial velocity of stopping

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-21.582^2}{2\times 0.75}\\\Rightarrow a=-310.521816\ m/s^2$

The average stopping acceleration is 310.521816 m/s²

3 0

4 years ago

A wall that is 12 feet wide and 10 feet high is to be painted. A blackboard that is 5 feet wide and 3 feet high is affixed to th

levacccp [35]

Answer:

The area of the portion of the wall that will be painted is 105 square feet

Explanation:

Total area of wall = 12 × 10 = 120 ft²

The area of the wall that will not be painted = area of the blackboard = 5 × 3 = 15 ft²

So, the area of the wall to be painted = Total area of wall - Area of the wall that will not be painted = 120 - 15 = 105 ft²

4 0

4 years ago

A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that

gavmur [86]

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken

g = Acceleration due to gravity = 9.81 m/s²

The equation

$v(t)=At+Bt^2$

Differentiating with respect to time

$\frac{dv}{dt}=\frac{d(At+Bt^2)}{dt}\\\Rightarrow 1.4=A+2Bt$

At t = 0

$1.4=A$

Hence, A = 1.4 m/s²

$B=\frac{v-At}{t^2}\\\Rightarrow B=\frac{2.18-1.4\times 1.8}{1.8^2}\\\Rightarrow B=-0.10493\ m/s^3$

B = -0.10493 m/s³

At t = 5 seconds

$a=1.4+2\times -0.010493\times 5=1.29507\ m/s^2$

a = 1.29507 m/s²

$T=m(a+g)\\\Rightarrow T=2530(1.29507+9.81)\\\Rightarrow T=28095.8271\ N$

T = 28095.8271 N

Weight of rocket

$W=2530\times 9.81=24819.9\ N$

$\frac{T}{W}=\frac{28095.8271}{24819.9}\\\Rightarrow \frac{T}{W}=1.13198\\\Rightarrow T=1.13198W$

T = 1.13198 W

3 0

4 years ago

John pushes forward on a car with a force of 125n while bob pushes backward on the car with a force of 225n. what is the net for

Pie

Answer:

100N

Explanation:

because 225-125= 100

5 0

3 years ago

Read 2 more answers

What is the mass of a cannonball if a force of 2,500 N gives the cannonball an acceleration of 200m/s2?

pentagon [3]

Newton's 2nd law of motion:

   Force = (mass) x (acceleration)

Divide each side
by 'acceleration': Mass = (force) / (acceleration)

   = (2,500 N) / (200 m/s²)

   = 12.5 kg

8 0

4 years ago

Read 2 more answers