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3 years ago

What was the kickball was originally named

Physics

2 answers:

Misha Larkins [42]3 years ago

5 0

Answer:

kick baseball

Explanation:

The sport of kickball, originally called "kick baseball," has roots going back to the early part of the 20th century in Cincinnati, Ohio. In the 1920's physical education teachers used kickball to teach the basics of baseball

Serggg [28]3 years ago

5 0

When it was originally invented, it was known as “kick baseball”. A few years later, kickball was adopted into the curriculum of education by physical education teachers nationwide, as an introduction to playing the most American sport of them all – baseball.

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You are watching an object that is moving in shm. when the object is displaced 0.600 m to the right of its equilibrium position,

larisa [96]

It should take at least 4.50

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3 years ago

two horses pull against a rope with forces of 100 newtons in opposite directions. this is an example of

statuscvo [17]

It is an example of balanced force.

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3 years ago

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Does kinetic energy stay the same at all heights? pls help I have 12 minutes to finish the project and the teacher won't help me

Alex Ar [27]

Answer:

Kinetic energy does not stay the same at all heights

Explanation:

Well as the height and wind increase so does the kinetic energy it's like when you fall as you are about to hit the floor you speed increases

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7 0

3 years ago

PLS ANSWER ASAP!!

Molodets [167]

b) between poles M1 and M2

Explanation:

From the expression, we can deduce that r is the distance between two magnetic poles M1 and M2.

The law of attraction between two magnetic poles states that:

<em> the force of attraction or repulsion between two magnetic poles is a function of the product of the strength of the magnetic poles and the square of the distance between the pole</em>s

Mathematically:

FM = K $\frac{M1 M2}{r^{2} }$

here r is the distance between the poles

FM is the magnetic force between the poles

M1 is the strength of the first magnetic pole

M2 is the strength of the second pole

K is the magnetic field constant

learn more:

magnetic pole brainly.com/question/2191993

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8 0

3 years ago

A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 50.9 J and a maximum d

Llana [10]

(a) 2446 N/m

When the spring is at its maximum displacement, the elastic potential energy of the system is equal to the total mechanical energy:

$E=U=\frac{1}{2}kA^2$

where

U is the elastic potential energy

k is the spring constant

A is the maximum displacement (the amplitude)

Here we have

U = E = 50.9 J

A = 0.204 m

Substituting and solving the formula for k,

$k=\frac{2E}{A^2}=\frac{2(50.9)}{(0.204)^2}=2446 N/m$

(b) 50.9 J

The total mechanical energy of the system at any time during the motion is given by:

E = K + U

where

K is the kinetic energy

U is the elastic potential energy

We know that the total mechanical energy is constant: E = 50.9 J

We also know that at the equilibrium point, the elastic potential energy is zero:

$U=\frac{1}{2}kx^2=0$ because x (the displacement) is zero

Therefore the kinetic energy at the equilibrium point is simply equal to the total mechanical energy:

$K=E=50.9 J$

(c) 8.55 kg

The maximum speed of the block is v = 3.45 m/s, and it occurs when the kinetic energy is maximum, so when

K = 50.9 J (at the equilibrium position)

Kinetic energy can be written as

$K=\frac{1}{2}mv^2$

where m is the mass

Solving the equation for m, we find the mass:

$m=\frac{2K}{v^2}=\frac{2(50.9)}{(3.45)^2}=8.55 kg$

(d) 2.14 m/s

When the displacement is

x = 0.160 m

The elastic potential energy is

$U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J$

So the kinetic energy is

$K=E-U=50.9 J-31.3 J=19.6 J$

And so we can find the speed through the formula of the kinetic energy:

$K=\frac{1}{2}mv^2 \rightarrow v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(19.6)}{8.55}}=2.14 m/s$

(e) 19.6 J

The elastic potential energy when the displacement is x = 0.160 m is given by

$U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J$

And since the total mechanical energy E is constant:

E = 50.9 J

the kinetic energy of the block at this point is

$K=E-U=50.9 J-31.3 J=19.6 J$

(f) 31.3 J

The elastic potential energy stored in the spring at any time is

$U=\frac{1}{2}kx^2$

where

k = 2446 N/m is the spring constant

x is the displacement

Substituting

x = 0.160 m

we find the elastic potential energy:

$U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J$

(g) x = 0

The postion at that instant is x = 0, since it is given that at that instant the system passes the equilibrium position, which is zero.

4 0

3 years ago