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Brrunno [24]
3 years ago
13

The tidal bulge on the side of the earth opposite the moon is due to _______ .

Physics
2 answers:
butalik [34]3 years ago
8 0
Due to the moon's gravitational force and inertias counterbalance.
Zepler [3.9K]3 years ago
5 0

Answer:

Due to the moon's gravitational force and inertias counterbalance.

:

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A flashing red traffic light at an intersection means
k0ka [10]
Just do what u would do if u were at a stop sign 
5 0
3 years ago
The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x
Diano4ka-milaya [45]

Answer:

Ff=m\times \dfrac{V_o^2}{2X_1}

Explanation:

Given that

At X=0 V=Vo

At X=X1  V=0

As we know that friction force is always try to oppose the motion of an object. It means that it provide acceleration in the negative direction.

We know that

V^2=U^2-2aS

0=V_o^2-2a X_1

a=\dfrac{V_o^2}{2X_1}

So the friction force on the box

Ff= m x a

Ff=m\times \dfrac{V_o^2}{2X_1}

Where m is the mass of the box.

4 0
3 years ago
Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part
Fantom [35]

a) Angular acceleration: 17.0 rad/s^2

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

W=mg

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

W_p =mg sin \theta

where \theta is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta

where L is the length of the pencil.

The relationship between torque and angular acceleration \alpha is

\tau = I \alpha (1)

where

I=\frac{1}{3}mL^2

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for \alpha, we find:

\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when \theta=10^{\circ} is

\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude mg

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

\tau = Fd

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

\tau =mg\frac{L}{2} cos \theta

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0
3 years ago
What is the expected wavelength (in cm) of 10.5 ghz microwaves in free space?
Sindrei [870]

Wavelength = (speed) / (frequency)

Wavelength = (300 thousand km per second) / (10.5 billion per second)

Wavelength = (300 / 10.5) (thousand km per second) / (billion per second)

Wavelength = (28.57) (million meters / second) / (thousand million / second)

Wavelength = (28.57) (meters / second) / (thousand / second)

Wavelength = (28.57) (meters / thousand)

<em>Wavelength = (28.57) (millimeters) </em>

5 0
3 years ago
The main water line enters a house on the first floor. The line has a gauge pressure of 2.10 x 105 Pa. (a) A faucet on the secon
Eva8 [605]

To answer this question is necessary to apply the concepts related to Bernoulli's equation. The Bernoulli-related concept describes the behavior of a liquid moving along a streamline. Pressure can be defined as the proportional ratio between height, density and gravity:

P = h\rho g

Where,

h = Height

\rho = Density

g = Gravity

Our values are

\rho = 1000kg/m^3 \rightarow density of water at normal conditions

h = 7.3m

g = 9.8m/s^2

PART A) Replacing these values to find the total pressure difference we have to

P_1 = h_1 \rho g

P_1 = (7.3)(1000)(9.8)

P_1 = 71540Pa

In this way the pressure change would be subject to

\Delta = P_2-P_1

\Delta = 2.1*10^5Pa- 0.7154*10^5Pa

\Delta = 138460Pa

\Delta = 0.135Mpa

PART B) Considering the pressure gauge of the group as the ideal so that at a height H the water cannot flow even if it is open we have to,

P_2 = H\rho g

2.1*10^5 = H (1000)(9.8)

H = 21.42m

Therefore the high which could a faucet be before no water would flow from it is 21.42m

5 0
3 years ago
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