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Karo-lina-s [1.5K]
3 years ago
10

Which factor is MOST responsible for the ocean tides?

Physics
2 answers:
12345 [234]3 years ago
8 0
Most responsible? I'm pretty sure it's the moon's gravity but I maybe wrong....I hoped I *sorta* helped
xenn [34]3 years ago
4 0
C is the answer ocean tides

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Two blocks joined by a string have masses of 6 and 9 kg. They rest on a frictionless horizontal surface. A 2nd string, attached
Tom [10]

Answer:

12N

Explanation:

Suppose the string mass is negligible, the total mass of the 2 block system is 6 + 9 = 15 kg

So the acceleration of the system when subjected to 30N force is

a = F / M = 30 / 15 = 2 m/s2

So both blocks would have the same acceleration, however, the force acting on the 6kg block would have a magnitude of

f = am = 2 * 6 = 12N

This is the tension in the string between the blocks

5 0
3 years ago
a concave lens creates a virtual image at -47.0 cm and a magnification of +1.75. what is the focal length?
Lilit [14]

The focal length of given concave lens will be -26.85 cm

The height of an image to the height of an object is the ratio that is used to determine a lens' magnification. Additionally, it is provided in terms of object and image distance. It is equivalent to the object distance to image distance ratio.

Given  concave lens creates a virtual image at -47.0 cm and a magnification of +1.75.

We have to find focal length

The focal length can be found out by following way:

Magnification = m = +1.75

m = hi/h

hi = -47 cm

1.75 = -47/h

h = -26.85 cm

So the focal length of given concave lens will be -26.85 cm

Learn more about magnification factor here:

brainly.com/question/6947486

#SPJ10

8 0
1 year ago
Define wheel and axle in science terms
topjm [15]

Answer:

A simple machine consisting of an axle to which a wheel is fastened so that torque applied to the wheel winds a rope or chain onto the axle, yielding a mechanical advantage equal to the ratio of the diameter of the wheel to that of the axle.

5 0
3 years ago
A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff
postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

F_c=\frac{mv^2}{R}

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

f=\mu N

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

N=mg

Therefore, frictional force, f=\mu mg

Now, frictional force = centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0
3 years ago
After coming down a slope, a 60-kg skier is coasting northward on a level, snowy surface at a constant 15 m&gt;s. Her 5.0-kg cat
Vinil7 [7]

To solve this exercise, it is necessary to apply the concepts of conservation of the moment especially in objects that experience an inelastic colposition.

They are expressed as,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where,

m_1= mass of the skier

m_2= mass of the cat

v_1 = initial velocity of skier

v_2 = initial velocity of cat

v_f= final velocity of both

Re-arrange to find V_f we have,

V_f = \frac{m_1v_1+m_2v_2}{(m_1+m_2)}

V_f = \frac{(60)(15)+(5)(-3.8)}{(60+5)}

V_f = 13.55m/s

Once the final velocity is found it is possible to calculate the change in kinetic energy, so

\Delta KE = KE_i-KE_f

\Delta KE = \frac{1}{2}(m_1v_1^2+m_2v^2_2)-\frac{1}{2}(m_1+m_2)v_f^2

\Delta KE = \frac{1}{2}((60)(15)^2+(5)(-3.8)^2)-\frac{1}{2}(60+5)(13.55)^2

\Delta KE = 819.1J

Therefore the amount of kinetic energy converted in to internal energy is 819J

3 0
3 years ago
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