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olga_2 [115]
3 years ago
7

What effect will the north pole of one magnet have on the north pole of another magnet?

Physics
1 answer:
Y_Kistochka [10]3 years ago
3 0

Answer:

Repulsion

Explanation:

Every magnet has two poles: North and South. Magnets have a property that they attract or repel another magnet and iron-nickel objects. But whether two magnets will attract or repel each other depends on how they are brought together. If unlike poles are brought near the magnets will attract. If like poles are brought near then they will repel each other.

Here north pole of one magnet will repel north pole of another magnet.

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What are two ways in which all types of precipitation are alike
Digiron [165]
1. they all come from the clouds 2.they are  all formed by precipatation.                       hope this helps enjoyyyyyy!!!!!!!!!!!!!
7 0
2 years ago
a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro
finlep [7]

Answer:

\theta=53.13^o

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

V_x=V_{ox}=V_ocos\theta

V_y=V_{oy}-gt=V_osin\theta-gt

x=V_{ox}t

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}

\displaystyle y_{max}=\frac{V_{oy}^2}{2g}

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

x_{max}=3y_{max}

\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}

Using the formulas for V_{ox}, V_{oy}:

\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}

Simplifying

4cos\theta sin\theta=3sin^2\theta

Dividing by sin\theta

4cos\theta=3sin\theta

Rearranging

tan\theta=\frac{4}{3}

\theta=arctan\frac{4}{3}

\theta=53.13^o

4 0
3 years ago
You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height (h) from the starting
Eddi Din [679]

Answer:

h = 50.49 m

Explanation:

Data provided:

Speed of skier, u = 2.0 m/s

Maximum safe speed of the skier, v = 30.0 m/s

Mass of the skier, m = 85.0

Total work = 4000 J

Height from the starting gate = h

Now, from the law of conservation of energy

Total energy at the gate = total energy at the time maximum speed is reached

\frac{1}{2}mu^2+mgh=4000J+\frac{1}{2}mv^2

where, g is the acceleration due to the gravity

on substituting the values, we get

\frac{1}{2}\times85\times2.0^2+85\times9.81\times h=4000J+\frac{1}{2}\times85\times30^2

or

170 + 833.85 × h = 4000 + 38250

or

h = 50.49 m

7 0
3 years ago
Birds which are famous for performing a ritualized mating dance
dlinn [17]
<span>Birds which famous for performing a ritualized mating dance are: The Grebes

Grebes most commonly live near a fresh water source.

They conducted this ritual because When two Grebes with opposite sex dances near each other, they will release some sort of pheromones that make them attracted to one another. </span>
6 0
3 years ago
Read 2 more answers
A 0.01 kg spring toy is compressed 0.02 m and released vertically. The toy is measured to reach 0.25 m in the air. Determine the
Scorpion4ik [409]

Answer:

122.5 N/m

Explanation:

According to the law of conservation of energy, if there is no air resistance or frictional forces, the initial elastic potential energy of the spring toy is entirely converted into gravitational potential energy when the toy reaches the highest point.

Therefore, we can write:

\frac{1}{2}kx^2=mgh

where the term on the left is the initial elastic potential energy while the term on the right is the gravitational potential energy, and where

k is the spring constant

x = 0.02 m is the compression of the spring

m = 0.01 kg is the mass of the toy

h = 0.25 m is the height reached by the toy

g=9.8 m/s^2 is the acceleration due to gravity

Solving for k,

k=\frac{2mgh}{x^2}=\frac{2(0.01)(9.8)(0.25)}{(0.02)^2}=122.5 N/m

8 0
2 years ago
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