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3 years ago

Does earths Roation causes gravity?

Physics

1 answer:

gogolik [260]3 years ago

8 0

Answer:

he gravity of Earth, denoted by g, is the net acceleration that is imparted to objects due to the combined effect of gravitation (from mass distribution within Earth) and the centrifugal force (from the Earth's rotation )

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Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of

Alchen [17]

Answer:

$\theta=10.20^{\circ}$

$\Delta l=0.10 ft$

Explanation:

First of all, we analyze the system of blocks before starting to move.

$\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0$

$\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0$

$11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0$

$11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0$

$16sin(\theta)-2.91cos(\theta)=0$

$tan(\theta)=0.18$

$\theta=arctan(0.18)$

$\theta=10.20^{\circ}$

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.

$P_{A}sin(\theta)-F_{fA}-F_{spring}=0$

Where:

$F_{spring} = k\Delta l=2.1\Delta l$

$P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0$

$\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}$

$\Delta l=0.10 ft$

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

4 0

3 years ago

A 0.0223 m diameter coin rolls up a 12.0◦ inclined plane. The coin starts with an initial angular speed of 49.3 rad/s and rolls

zlopas [31]

Answer:

$h = 0.0231 m$

Explanation:

The movement of the coin is modelled after the Principle of Energy Conservation. The kinetic energy of the coin is the sum of the components associated with translation and rotation and there are no non-conservative forces. In addition, the coin starts moving at height of zero

$K_{rot} + K_{tr} = U_{g}$

$\frac{1}{2} \cdot m_{coin} \cdot \omega^{2} \cdot R^{2} + \frac{1}{4} \cdot m_{coin} \cdot R^{2} \cdot \omega^{2} = m_{coin} \cdot g \cdot h\\$

$\frac{3}{4} \cdot R^{2} \cdot \omega^{2} = g \cdot h$

The maximum vertical height is isolated in the previous equation:

$h = \frac{3 \cdot R^{2}\cdot \omega^{2}}{4\cdot g}$

$h = \frac{3 \cdot (0.01115 m)^{2} \cdot (49.3 \frac{rad}{s} )^{2}}{4 \cdot (9.807 \frac{m}{s^{2}} )} \\h = 0.0231 m$

5 0

4 years ago

What would happen if i build a bomb and destroy Pluto

VLD [36.1K]

Answer:

a what

well it wouldn't matter because it's not counted a planet anyway

Explanation:

8 0

3 years ago

a voltmeter and an ammeter are used to respectively monitor the voltage across the current through the resistor

dusya [7]

When a current is established in a closed conducting loop , we use ammeter. Due to this resultant potential developed detected by voltmeter.

When current flows through loop , it gets conducted.
Due to this Electric field is generated . This happens due to induced magnetic field.
The free electrons were at rest until current is there , still ammeter detects some deflection.
This deflection is due to zero current.
Due to magnetic field , current in coil changes.
Mathematically, $\alpha =-\frac{d\beta }{dt}$
The presence of conducting loop is not necessary for having electric field.