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rusak2 [61]
3 years ago
8

Please help! I have another part to this too.

Physics
1 answer:
JulsSmile [24]3 years ago
5 0

first one is true, there's no net force acting on it thats greater than another or making it unbalanced, if there was the object would be in some kind of motion

All scientist use meters, that way scientist can share information across country without needing to convert the data.

3. is air resistance

4. The large rock

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What is the main cause of twice-daily tides around the world?<br><br><br> Help me.
liraira [26]

<em>The main cause of twice-daily tides around the world are;</em>

Gravitational pull of the moon

<u>Tides are the rise and fall of water caused by gravitational forces of the moon and sun on the oceans of the earth. </u>

<u>The moons gravity is large enough to actually pull water out of the ocean from space. Just enough to create motion of the tides.</u>

4 0
3 years ago
A skier has an acceleration of 2.5 m/s2. How long does it take her to come to a complete stop from a speed of 18 m/s?
aleksandr82 [10.1K]
(Assuming acceleration is -2.5m/s2) 7.2 seconds.
4 0
3 years ago
Read 2 more answers
A metal ring 4.30 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend
CaHeK987 [17]

Answer:

A)0.00966 N/C

B) counterclockwise direction

Explanation:

We are given;

Diameter of the metal ring; d = 4.3 cm

Radius;r = 2.15 cm = 0.021- m

Initial magnetic field, B = 1.12 T

Rate of decrease of the magnetic field;dB/dt = 0.23 T/s

Now, as a result of change in magnetic field, an emf will be induced in it. Thus, , electric field is induced and given by the formula :

∫E•dr = d/dt∫B.A •dA

This gives;

E(2πr) = dB/dt(πr²)

Gives;. 2E = dB/dt(r)

E = dB/dt × 2r

We are given;

E = 0.23 × 2(0.021)

E = 0.00966 N/C

The magnitude of the electric field induced in the ring has a magnitude of 0.00966 N/C

B) The direction of electric field will be in a counterclock wise direction when viewed by someone on the south pole of the magnet

6 0
3 years ago
Consider a 100 g object dropped from a height of 1 m. Assuming no air friction (drag), when will the object hit the ground and a
Katyanochek1 [597]

Answer:

speed and time are Vf = 4.43 m/s and  t = 0.45 s

Explanation:

This is a problem of free fall, we have the equations of kinematics

      Vf² = Vo² + 2g x

As the object is released the initial velocity is zero, let's look at the final velocity with the equation

      Vf = √( 2 g X)

      Vf = √(2 9.8  1)

      Vf = 4.43 m/s

This is the speed with which it reaches the ground

 

Having the final speed we can find the time

      Vf = Vo + g t

       t = Vf / g

       t = 4.43 / 9.8

       t = 0.45 s

This is the time of fall of the body to touch the ground

3 0
4 years ago
Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up
Westkost [7]

Answer:

24 N

Explanation:

m = mass of the cube = 6.0 kg

Consider the three cubes together as one.

M = mass of the three cubes together = 3 m = 3 (6.0) = 18 kg

a = acceleration of the combination = 2 ms⁻²

F = Force applied on the combination

Using Newton's second law

F = ma = (18) (2) = 36 N

F_{L} = Force by the left cube on the middle cube

Consider the forces acting on left cube, from the force diagram, we have

F - F_{L} = ma \\36 - F_{L} = (6) (2)\\F_{L} = 24 N

4 0
3 years ago
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