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3 years ago

If a 2 kg spring is compressed by exerting a force of 10 newtons a distance of 50 cm how much work is done?

1 answer:

5 0

Convert cm into meters,
50/100=0.5m

work done= 10x0.5
=5J

we dont consider the weight of the spring as it acts downwards.

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A satellite that goes around the earth once every 24 hours (86,400 s) is called a geosynchronous satellite. If a geosynchronous

Olegator [25]

Answer:

42244138.951 m

Explanation:

G = Gravitational constant = 6.667 × 10⁻¹¹ m³/kgs²

r = Radius of orbit from center of earth

M = Mass of Earth = 5.98 × 10²⁴ kg

m = Mass of Satellite

The satellite revolves around the Earth at a constant speed

Speed = Distance / Time

The distance is the perimeter of the orbit

$v=\frac{2\pi \times r}{24\times 3600}$

The Centripetal force of the satellite is balanced by the universal gravitational force

$m\frac{v^2}{r}=\frac{GMm}{r^2}\\\Rightarrow \frac{\left(\frac{2\pi \times r}{24\times 3600}\right)^2}{r}=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{r^2}\\\Rightarrow \left(\frac{2\pi \times r}{24\times 3600}\right)^2=6.667\times 10^{-11}\times 5.98\times 10^{24}\\\Rightarrow r^3=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\\\Rightarrow r=\left(\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\right)^{\frac{1}{3}}\\\Rightarrow r=42244138.951\ m$

The radius as measured from the center of the Earth) of the orbit of a geosynchronous satellite that circles the earth is 42244138.951 m

6 0

2 years ago

Little Timmy wants to measure how tall his house is. He doesn’t have a tape measure but does have a stopwatch. He recruits Big B

Komok [63]

The distance covered by an object accelerating from rest is

D = (1/2) · (acceleration) · (time)² .

In this particular case, 'acceleration' is 9.8 m/s² ... due to gravity.

D = (1/2) · (9.8 m/s²) · (1.67 s)²

D = (4.9 m/s²) · (2.789 s²)

D = 13.67 meters

6 0

3 years ago

Example of energy transfer from potential to kinetic and back

Alex_Xolod [135]

Swimmer and Divers. The Potential energy is transferred into Kinetic energy, and allows the diver to submerge into the water. The Kinetic energy then allows the diver to submerge and dive into the water. Potential energy however, is needed to allow the diver to get back out of the water after diving to get up and go and dive again, and then the Kinetic energy is transferred back to Potential energy to repeat the process.

Hope :) -Emilie Xo this is right and it helps! Xo

5 0

2 years ago

Yea, gonna need some help. Thanks

natulia [17]

Answer:

t = 3.48 s

Explanation:

The time for the maximum height can be calculated by taking the derivative of height function with respect to time and making it equal to zero:

$h(t) = -16t^2+v_ot+h_o\\\\\frac{dh(t)}{dt}=0=-32t+v_o\\\\v_o = 32t$

where,

v₀ = initial speed = 110 ft/s

Therefore,

$110 = 32t\\\\t = \frac{110}{32}\\\\$

8 0

2 years ago

A shot-putter released the shot at an angle of 41.5 degrees and a height of 1.9 m with an initial velocity of 13.3 m/s. How far

liq [111]

Answer:

x = 17.88[m]

Explanation:

We can find the components of the initial velocity:

$(v_{x})_{o} = 13.3*cos(41.5)=9.96[m/s]\\(v_{y})_{o} = 13.3*sin(41.5)=8.81[m/s]$

We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.

g = - 9.81[m/s^2]

Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.

$y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]$

With this time we can calculate the horizontal distance:

$x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]$

3 0

3 years ago