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zvonat [6]
3 years ago
7

Both the Moon and the Sun influence the tides on Earth. The moon has a much greater influence though. Why is that? A) because th

e Sun is much less dense B) because the Sun is gaseous, not solid C) because the Sun is so far away from Earth D) because the Sun is less massive than the Moon
Physics
2 answers:
uysha [10]3 years ago
8 0
Because the sun is much farther from Earth than the moon
Temka [501]3 years ago
3 0
C. The sun is 400 times farther from Earth than the moon is.
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water in a cup and a kettle can have the same temperature even though the quantities are different . give reasons​
jekas [21]

Answer:

The reason is because both are exposed to a virtually infinite heat sink, due to the virtually infinite mass  and of the surrounding environment, compared to the sizes of either the cup or the kettle such that the equilibrium temperature, T_{(equilibrium)} reached is the same for both the cup and the kettle as given by the relation;

\infty M_{(environ)} \times  c_{(environ)} \times (T_2 - T_1) = m_{1} \times  c_{(water)} \times (T_3 - T_2) + m_{2} \times  c_{(water)} \times (T_4 - T_2)

Due to the large heat sink, T₂ - T₁ ≈ 0 such that the temperature of the kettle and that of the cup will both cool to the temperature of the environment

Explanation:

4 0
3 years ago
A bird is traveling west with a velocity of 8 m/s and momentum of 40 mg*m/s what's the birds mass?
VikaD [51]

Answer:

m = 5 [mg]

Explanation:

We must remember that the definition of linear momemtum is defined as the product of mass by distance.

P = m*v

P = momentum = 40 [mg*m/s]

m = mass [mg]

v = velocity = 8 [m/s]

Now clearing m:

m = P/v

m = 40/8

m = 5 [mg]

3 0
3 years ago
1500 kg wrecking ball traveling at a speed of 3.5 m/s hits a wall that does not crumble but is pushed back 75 cm. If the wreckin
Rudiy27

Answer:

The size of the force that pushes the wall is 12,250 N.

Explanation:

Given;

mass of the wrecking ball, m = 1500 kg

speed of the wrecking ball, v = 3.5 m/s

distance the ball moved the wall, d = 75 cm = 0.75 m

Apply the principle of work-energy theorem;

Kinetic energy of the wrecking ball = work done by the ball on the wall

¹/₂mv² = F x d

where;

F is the size of the force that pushes the wall

¹/₂mv² = F x d

¹/₂ x 1500 x 3.5² = F x 0.75

9187.5 = 0.75F

F = 9187.5 / 0.75

F = 12,250 N

Therefore, the size of the force that pushes the wall is 12,250 N.

7 0
3 years ago
The fast French train known as the TGV (Train à Grande Vitesse) has a scheduled average speed of 216 km/h. (a) If the train goes
faust18 [17]

Answer:

a)  r = 6122 m and b) v = 32.5 m / s

Explanation:

a) The train in the curve is subject to centripetal acceleration

         a = v2 / r

Where v is The speed and r the radius of the curve

They indicate that the maximum acceleration of the person is 0.060g,

        a = 0.060 g

        a = 0.060 9.8

        a = 0.588 m /s²

Let's calculate the radius

        v = 216 km / h (1000m / 1km) (1 h / 3600 s =

        v = 60 m / s

        r = v² / a

        r = 60² /0.588

        r = 6122 m

b) Let's calculate the speed, for a radius curve 1.80 km = 1800 m

        v = √a r

        v = √( 0.588 1800)

        v = 32.5 m / s

6 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
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