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goldenfox [79]
3 years ago
9

A spacecraft at rest has moment of inertia of 100 kg-m^2 about an axis of interest. If a 1 newton thruster with a 1 meter moment

arm is fired orthogonal to the axis of interest for 1 minute, how far has the vehicle rotated when the firing is complete?
Physics
1 answer:
snow_lady [41]3 years ago
3 0

Answer:

18 radians

Explanation:

The computation is shown below:

As we know that

Torque = Force × Moment arm

= 1N × 1M

= 1N-M

Torque = I\alpha

\alpha = \frac{torque}{I}\\\\= \frac{1}{100}\\\\= 0.01 rad/s^2

Now

\theta = w_ot + \frac{1}{2} \alpha t^2\\\\w_o = 0\\\\\theta = 0 \times 60 + \frac{1}{2} \times 0.01 \times 60^2\\\\= 18\ radians

Here t = 1 minutes = 60 seconds

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Which ocean borders the west coast of this United States
Gennadij [26K]
The Pacific.
Hope it helps!
3 0
3 years ago
A 100 kg mass is pulled along a frictionless surface by a horizontal force F such that its acceleration is 10.0 m/s2. A 20 kg ma
DIA [1.3K]

Answer

given,

mass = 100 kg

acceleration = 10 m/s²

A mass 20 kg slides over 100 kg block

acceleration = 3 m/s²

horizontal friction exerted by the 100 kg block on 20 kg

using newton's second law

F - f = 0

F = f

f = ma

f = 20 × 3

f = 60 N

now net force acting on the 100 kg block

F_net = m a

F_net = 100 x 10

F_net = 1000 N

after 20 kg block falls the acceleration of the bock

F = 1000 +60

F = 1060 N

acceleartion on the block

a = \dfrac{F}{m}

a = \dfrac{1060}{100}

a = 10.60 m/s²

3 0
3 years ago
17,874,000 what is the value of 1
kotegsom [21]
Salutations!

17,874,000 what is the value of 1?

The value of 1 is 10 million. The place value would be 10,000,000.

Hope I helped.
3 0
2 years ago
Read 2 more answers
A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr
VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

rR:\\\\E=k\frac{Q}{r^2}

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}

with this values of a you can use the following formula:

a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0
2 years ago
On a day when the temperature reaches 50°F, the temperature in degrees Celsius is: 20°C
anzhelika [568]

Answer:

10°C

Explanation:

To convert °F to °C, we use the formula:

°C =  (°F - 32) * ( 5/9)

So, to convert 50°F to the equivalent  in °C, we can proceed as follows:  

°C = ( 50 - 32 ) * (5/9)  

°C = ( 18 ) *  (5/9), which is, approximately,

°C = 9.999999999... ≈ 10 (5/9 ≈0.555555...)

So, 50°F is equivalent to 10°C.

3 0
2 years ago
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