What are the two opposing forces that keep stars between collapsing and inflating?

An fm radio station broadcasts at 98. 6 mhz. What is the wavelength of the radiowaves?.

victus00 [196]

The wavelength of the radio waves is 3.04 cm.

<h3>Calculation:</h3>

λf = c

λ = c/f

where,

λ = wavelength

c = speed of light

f = frequency

Given,

f = 98.6 MHz = 98.6 × 10⁶

c = 3 × 10⁸

To find,

λ =?

Put the values in the formula,

λ = c/f

λ = 3 × 10⁸/98.6 × 10⁶

= 0.0304 × 10² m

= 3.04 cm

Therefore, the wavelength of the radio waves is 3.04 cm.

Learn more about the calculation of wavelength here:

brainly.com/question/8422432

#SPJ4

3 0

1 year ago

If a builder of mass 75kg climbs a vertical ladder of 25m how much energy has she gained ?

erica [24]

So,

GPE (graviational potential energy) = mass x g x height

GPE is depends on where zero height is defined. In this situation, we define h = 0 as the initial height.

$GPE = 75 \ kg*9.8 \ \frac{m}{s^2}*25 \ m$

$GPE = 18,375 \frac{kg*m^2}{s^2}$

$GPE = 18,375 \ joules(J) \ or \ 18.375 \ kilojoules(kJ)$

The builder has gained 18.375 kJ of PE.

4 0

2 years ago

A block of mass 57.1 kg rests on a slope having an angle of elevation of 28.3°. If pushing downhill on the block with a force ju

stira [4]

Answer:

The coefficient is 0.90

Explanation:

Drawing a diagram makes thing easier, we will assume that the acceleration tends to zero because it start barely moving.

$-F_s+mg*sin(\theta)+F=0\\F_s=57.1kg*9.8m/s^2*sin(28.3)+177N\\F_s=442N\\F_s=\µ*N\\N=m*g*cos(\theta)\\N=57.1*9.8*cos(28.3)=493N\\\\\µ=\frac{442N}{493N}=0.90$

3 0

3 years ago

A meteoroid is moving towards a planet. It has mass m = 0.78×109 kg and speed v1 = 4.1×107 m/s at distance R1 = 2.8×107 m from t

wariber [46]

Answer:

$PE=81.755\, J$

Explanation:

Given that:

mass of meteoroid, $m=7.8\times 10^8 \,kg$

radial distance from the center of the planet, $R= 2.8\times 10^7 m$

mass of the planet, $M=4.4\times 10^{25}\, kg$

<u>For gravitational potential energy we have:</u>

$PE=G\frac{M.m}{R}$

substituting the respective values:

$PE=6.67\times 10^{-11}\times \frac{4.4\times 10^{25}\times 7.8\times 10^8}{2.8\times 10^7}$

$PE=81.755\, J$

5 0

3 years ago

The traffic on the freeway is moving at a constant speed of 24 m/sm/s. What distance does the traffic travel while the car is mo

ExtremeBDS [4]

Incomplete question as there is so much information is missing.The complete question is here

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Answer:

Distance traveled=240 m

Explanation:

Given data

Initial velocity of car v₀=0 m/s

Final velocity of car vf=24 m/s

Distance traveled by car S=120 m

To find

Distance does the traffic travel

Solution

To find the distance first we need to find time, for time first we need acceleration

So

$(V_{f})^{2}=(V_{o})^{2}+2aS\\ So\\a=\frac{(V_{f})^{2}-(V_{o})^{2} }{2S}\\ a=\frac{(24m/s)^{2}-(0m/s)^{2} }{2(120)}\\a=2.4 m/s^{2}$

As we find acceleration.Now we need to find time

So

$V_{f}=V_{i}+at\\t=\frac{V_{f}-V_{i}}{a}\\t=\frac{(24m/s)-(0m/s)}{(2.4m/s^{2} )}\\t=10s$

Now for distance

So

$Distance=velocity*time\\Distance=(24m/s)*(10s)\\Distance=240m$

7 0

3 years ago