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gulaghasi [49]
3 years ago
6

As denizens of the surface of a spinning planet, we are always in uniform circular motion. Imagine you are in Nairobi (on the Ea

rth's equator) at noon on a Monday. Answer the following questions only considering the rotation of the earth and NOT the Earth's circular motion around the sun. The radius of the earth is 6371 km. A day is 24 hours.
Assuming that you are at at theta = 0 rad at noon on Monday and moving in the positive direction, what is your position theta in rad on Tuesday at 11:00 am?
Physics
1 answer:
Klio2033 [76]3 years ago
7 0

Answer:

6.02 radian

Explanation:

At 12 noon position is zero radian on Monday

at 12 midnight the position is π/2

at 11 am position on Tuesday is 2π/24 x 23 ( after 23 hours ) = 6..02 radian.

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Explanation:

primary voltage Vp=240V

secondary voltage Vs=60V

resistance 5Ω

primary current in the circuit is

I_{P} =I_{S} \frac{V_{S}}{V_{P}}\\I_{P} =\frac{V_{S}}{R}  \frac{V_{S}}{V_{P}}\\I_{P}=\frac{60\times 60}{5\times 240\\}\\I_{P}=3 A

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One horse is pulling a 755 kg sled straight ahead applying a force of 1988 N. If the acceleration of the sled is 1.36 m/s2, what
Inessa [10]

Answer:

The coefficient of kinetic friction is 0.13

Explanation:

Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

F_y=ma_y (2)

F_x=ma_x (3)

On y we have that the acceleration is zero because the sled is not moving upward or downward, remember that the net force on y is the weight (W) pointing downward and the normal force pointing upward:

F_y=W+n=0

Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

F_y=(755)(-9.81)+n=0

n=7406.55 N (4)

Now on the x direction we have the sum of the forces is the pulling force (T) and friction force (f)

F_x=F+f=ma_x

Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

1988+f=m(1.36)

f=(755)(1.36)-1988=-961.2 N

The negative sign means it's in the opposite direction the horse is pulling

The frictional force is related with the coefficient of kinetic friction in the next way:

|f|=\mu_k n

with μk the coefficient of kinetic friction, and n the normal force that we already found on (4), so we simply solve the last equation for μk:

\mu_k=\frac{|f|}{n}=\frac{961.2}{7406.55}=0.13

4 0
3 years ago
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