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Ainat [17]
4 years ago
9

How long have us been attempting space travel?

Physics
1 answer:
chubhunter [2.5K]4 years ago
7 0
The first attempt was made in 1957
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A disk 7.90 cm in radius rotates at a constant rate of 1 190 rev/min about its central axis. (a) Determine its angular speed. 12
Tanya [424]

Answer:

124.62\ \text{rad/s}

3.71\ \text{m/s}

1.23\ \text{km/s}^2

20.28\ \text{m}

Explanation:

r = Radius of disk = 7.9 cm

N = Number of revolution per minute = 1190 rev/minute

Angular speed is given by

\omega=N\dfrac{2\pi}{60}\\\Rightarrow \omega=1190\times \dfrac{2\pi}{60}\\\Rightarrow \omega=124.62\ \text{rad/s}

The angular speed is 124.62\ \text{rad/s}

r = 2.98 cm

Tangential speed is given by

v=r\omega\\\Rightarrow v=2.98\times 10^{-2}\times 124.62\\\Rightarrow v=3.71\ \text{m/s}

Tangential speed at the required point is 3.71\ \text{m/s}

Radial acceleration is given by

a=\omega^2r\\\Rightarrow a=124.62^2\times 7.9\times 10^{-2}\\\Rightarrow a=1226.88\approx 1.23\ \text{km/s}^2

The radial acceleration is 1.23\ \text{km/s}^2.

t = Time = 2.06 s

Distance traveled is given by

d=vt\\\Rightarrow d=\omega rt\\\Rightarrow d=124.62\times 7.9\times 10^{-2}\times 2.06\\\Rightarrow d=20.28\ \text{m}

The total distance a point on the rim moves in the required time is 20.28\ \text{m}.

8 0
3 years ago
A gymnast of mass 63.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
Sergio [31]

Answer:

Explanation:

A ) When gymnast is motionless , he is in equilibrium

T = mg

= 63 x 9.81

= 618.03 N

B )

When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.

T = mg

= 618.03 N

C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2

Net force on it = T - mg   , acting in upward direction

T - mg = m a

T =  mg + m a

= m ( g + a )

= 63 ( 9.81 + .6)

= 655.83 N

D )  If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2

Net force acting in downward direction

mg - T = ma

T = m ( g - a )

= 63 x ( 9.81 - .6 )

= 580.23 N

6 0
3 years ago
What is the relationship between the normal force and weight
GalinKa [24]

Answer: they have the same magnitude.

Explanation:

normal force = mg

weight = mg

4 0
3 years ago
A 946.4 kg ( 2083 lb ) car is moving at 14.8 m / s ( 33.0 mph ) . Calculate the magnitude of its momentum.
IgorLugansk [536]

Answer:

14009. 72 kgms^-1

Explanation:

Momentum is the product of an objects mass and velocity

7 0
4 years ago
A baseball m=.34kg is spun vertically on a massless string of length l=.52m. the string can only support a tension of tmax=9.9n
larisa86 [58]
<span>4.5 m/s This is an exercise in centripetal force. The formula is F = mv^2/r where m = mass v = velocity r = radius Now to add a little extra twist to the fun, we're swinging in a vertical plane so gravity comes into effect. At the bottom of the swing, the force experienced is the F above plus the acceleration due to gravity, and at the top of the swing, the force experienced is the F above minus the acceleration due to gravity. I will assume you're capable of changing the velocity of the ball quickly so you don't break the string at the bottom of the loop. Let's determine the force we get from gravity. 0.34 kg * 9.8 m/s^2 = 3.332 kg m/s^2 = 3.332 N Since we're getting some help from gravity, the force that will break the string is 9.9 N + 3.332 N = 13.232 N Plug known values into formula. F = mv^2/r 13.232 kg m/s^2 = 0.34 kg V^2 / 0.52 m 6.88064 kg m^2/s^2 = 0.34 kg V^2 20.23717647 m^2/s^2 = V^2 4.498574938 m/s = V Rounding to 2 significant figures gives 4.5 m/s The actual obtainable velocity is likely to be much lower. You may handle 13.232 N at the top of the swing where gravity is helping to keep you from breaking the string, but at the bottom of the swing, you can only handle 6.568 N where gravity is working against you, making the string easier to break.</span>
7 0
4 years ago
Read 2 more answers
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