Describe the shape of the earth besides round and tell me how it rotates​

Answer:

reduce passive stiffness and increase range of movement during exercise.

Explanation:

stretching performed as part of a warm up prior to exercise is thought to reduce passive stiffness and increase range of movement during exercise. in general it appears that is static stretching is most beneficial for athletes requiring flexibility for their sports.

Conociendo la longitud y frecuencia de un péndulo, queremos encontrar la longitud de otro pendulo de tal forma que tenga otra frecuencia.

Veremos que la longitud del nuevo péndulo debe ser 6.25m

Sabemos que un péndulo de 4m de longitud tiene una frecuencia de 5Hz.

La frecuencia de un péndulo está dada por:

[tex]f = \frac{1}{2*\pi} *\frac{g}{l}[/tex]

Donde g es la aceleración gravitatoria y l es la longitud del péndulo, remplazando los datos que tenemos en esa ecuación obtenemos:

[tex]5 Hz = \frac{1}{2*3.14} *\sqrt{\frac{g}{4m} } \\\\(5Hz*2*3.14)^2*4m = g = 3,943.8 m/s^2[/tex]

Ahora debemos encontra la longitud de tal forma que la frecuencia sea 4Hz, entonces debemos resolver:

[tex]4Hz = \frac{1}{2*3.14} *\sqrt{ \frac{3943.8m/s^2}{l} }\\\\4hz*2*3.14 = \sqrt{ \frac{3943.8m/s^2}{l} }\\\\l =\frac{3943.8m/s^2}{ (4hz*2*3.14)^2} = 6.25m[/tex]

La longitud del nuevo péndulo deve ser 6.25m

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I just need the points

Explanation:

But just pick any and say something like "it stans out to me most/more" or "it sounds/looks more intristing to me"

Answer:I think Fossils show that some animals are extinct

Explanation:

Please mark as brainliest for me.Thanks

Answer:

A

Explanation:

all galaxies exploded in order to create the sun/stars

Hi there!

We know that:

Force due to gravity = Mgsinθ

Force due to friction = μMgcosθ

Let the positive direction be directed in the direction of the block's acceleration, which is downward.

Thus:

ΣF = Mgsinθ - μMgcosθ

Solving for acceleration requires diving all terms by the mass, so:

a = gsinθ - μgcosθ

Substitute in given values. (g = 9.8 m/s²)

a = 9.8sin(30) - 0.3(9.8)cos(30) = 2.354 m/s²

#Case -1

If Pulling force is less than frictional force the object won't move .

#Case-2

If Pulling force is greater than frictional force then object will be .

In order to calculate friction force you need Limiting friction first .

[tex]\\ \sf\longmapsto F_L=\mu sN[/tex]

u s is coefficient of static friction and N is normal reaction

Or

[tex]\\ \sf\longmapsto F_L=\mu smg[/tex]

Explanation:

Let's calculate the components of the football's velocity:

[tex]v_{0x} = (23.9\:\text{m/s})\cos{51.5°} = 14.9\:\text{m/s}[/tex]

[tex]v_{0y} = (23.9\:\text{m/s})\sin{51.5°} = 18.7\:\text{m/s}[/tex]

a) The time it takes for the football to travel 36.0 m horizontally is

[tex]t = \dfrac{x}{v_{0x}} = \dfrac{36.0\:\text{m}}{14.9\:\text{m/s}} = 2.4\:\text{s}[/tex]

During this time, the y-displacement of the football is

[tex]y = v_{0x}t - \frac{1}{2}gt^2[/tex]

[tex]\:\:\:\:= (18.7\:\text{m/s})(2.4\:\text{s}) - \frac{1}{2}(9.8\:\text{m/s}^2)(2.4\:\text{s})^2[/tex]

[tex]\:\:\:\:= 16.7\:\text{m}[/tex]

This means that the football cleared the crossbar by 16.7 m - 3.05 m = 13.7 m

b) To determine whether the football was rising or falling while clearing the crossbar, let's look at the y-component of its velocity after 2.4 s:

[tex]v_y = v_{0y} - gt = 18.7\:\text{m/s} - (9.8\:\text{m/s}^2)(2.4\:\text{s})[/tex]

[tex]\:\:\:\:\:\:= -4.82\:\text{s}[/tex]

Since its sign is negative, this means that the football was already on its way down.

Answer:

A.It’s not moving.

Explanation:

Position-Time graphs display the motion of a object by showing the changes of velocity with respect to time.

The motion of a car on a position-time graph that is represented with a horizontal line indicates that the car has stopped moving.

A straight line with a positive slope indicates that the car is moving at a constant velocity, and thus the slope is constant. On the other hand, a curve with a changing slope, shows that the velocity is changing.

Answer:

What Um Sorry Where's The Answer?....

Answer:

[tex]-1.99\:\mathrm{m/s}[/tex]

Explanation:

Assuming that the equation is intended to be [tex]\displaystyle x=\cos\left(\frac{2\pi}{3}t\right)[/tex], we can find the velocity vs. time equation by taking the first derivative with respect to [tex]t[/tex]:

[tex]\displaystyle \frac{dx}{dt}=\frac{d}{dt}\left(\cos\left(\frac{2\pi}{3}t\right)\right)[/tex]

Recall the chain rule:

[tex]\displaystyle f(g(x))'=f'(g(x))\cdot g'(x)[/tex]

Therefore,

[tex]\displaystyle \frac{d}{dt}\left(\cos\left(\frac{2\pi}{3}t\right)\right)=-\sin\left(\frac{2\pi}{3}t\right)\cdot \frac{2\pi}{3}[/tex]

Therefore, the velocity vs. time equation of the object is [tex]\displaystyle v=-\sin\left(\frac{2\pi}{3}t\right)\cdot \frac{2\pi}{3}[/tex].

Substitute [tex]t=0.6\text{ s}[/tex] into this equation to find the velocity at that given time:

[tex]\displaystyle v=-\sin\left(\frac{2\pi}{3}(0.6)\right)\cdot \frac{2\pi}{3}\approx \boxed{-1.99\text{ m/s}}[/tex]

The angular acceleration of the larger camshaft is 995.72 rad/s².

The given parameters;

The angular acceleration of the 3.75 cm camshaft is calculated as follows;

[tex]\omega _f = \omega _i + \alpha t\\\\\omega _f =0 + \alpha t\\\\\omega _f = \alpha t\\\\(1250 \ \frac{rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1\min}{60 \ s} ) = 0.032 \alpha \\\\130.92 = 0.032\alpha \\\\\alpha = \frac{130.92}{0.032} = 4091.25 \ rad/s^2[/tex]

The angular momentum of the camshaft is calculated as follows;

[tex]I_1 \alpha _1 = I_2 \alpha_2 \\\\\frac{1}{2} mr_1^2 \alpha _1 = \frac{1}{2}m R^2 \alpha_2\\\\r_1^2 \alpha _1 = R^2 \alpha_2\\\\\alpha_2 = \frac{r_1^2 \alpha _1 }{R^2} \\\\\alpha_2 =\frac{(0.037)^2 \times (4091.25)}{(0.075)^2} \\\\\alpha _2 = 995.72 \ rad/s^2[/tex]

Thus, the angular acceleration of the larger camshaft is 995.72 rad/s².

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The conservation of mechanical energy allows finding the result for the speed of the pendulum when it is at 30º is:

The conservation of mechanical energy is a theorem of greater importance in physics and ordinary life, it states that if there is no friction force the total mechanistic energy remains constant at all points.

Mechanical energy is the sum of kinetic energy plus all potential energies. In the attachment we see a diagram of the pendulum's movement at the two points of interest.

They indicate that the pendulum is released from an initial angle of θ₁ = 60º, let's find the mechanical energy at that point.

      Em₀ = U = m g h

Where the height is measured from the lowest point of the movement.

      h = L - L cos tea1 = L (1 cos tea1)

The second point of interest occurs for θ₂ = 30º.

At this point part of the energy is indica and part gravitational potential.  

     [tex]Em_f[/tex]  = K + U₂

      [tex]Em_f[/tex] = ½ m v² + m g h ’

There is no friction in the system, therefore mechanical energy is conserved.

       Em₀ = Em₀_f

       mg L (1 - cos θ₁) = ½ m v² + m g L (1 - cos θ₂)

       v² = 2g L (cos θ₂ - cos θ₁)

Let's calculate.

       v² = 2 9.8 2.1 (cos 30 - cos 60)

       v² = 41.16 0.366

       v = 3.88 m / s

In conclusion using the conservation of mechanical energy we can find the result for the speed of the pendulum when it is at 30º is:

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Answer:

3.0 J

Explanation:

Just took the test

Answer:

19

Explanation:

The mass of an atom is found in the nucleus: number of protons + number of neutrons; 9 + 10 = 19

The mass number of fluorine is 19

Answer:

720 J

Explanation:

Kinetic Energy = 1/2 * m *(v*v)

M= 90 kg

V=4 m/s

KE=1/2 * 90 * (4*4)

KE= 45*16= 720 J

[tex]\frac{m}{s}[/tex]Answer: D) 720 J

Explanation:

Given : m=90kg

Total Energy = K. E. + P. E.  =[tex]\frac{1}{2}[/tex][tex]mv^{2}[/tex]+mgh

Since bike is present over a surface of Earth (top of hill) hence h=0m.

Therefore,

Total Energy=  [tex]\frac{1}{2}[/tex]*90kg*[tex](4m/s)^{2}[/tex]+(90kg)*(9.81[tex]\frac{m}{s}^2[/tex])x(0)

= 720 J

Hence total energy of bike is equal to 720 J

If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.

Given the data in the question;

Since the hockey puck was initially in the referee's hands

First we will find the velocity of the Puck when it hits the ground

From the Third Equation of Motion:

[tex]v^2 = u^2 + 2as[/tex]

Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.

Since the pluck is under gravity, we will have:

[tex]v^2 = u^2 + 2gh[/tex]

We substitute in our value and find "v"

[tex]v^2 = 0 + (2 \ *\ 9.8m/s^2\ *\ 2.5m )\\\\v^2 = 47.04m^2/s^2\\\\v= \sqrt{47.04m^2/s^2}\\\\v = 6.85857m/s[/tex]

Now, Velocity of the hock puck after it hits the ground and bounce back;

We know that; Coefficient of restitution [tex]= \frac{Relative\ velocity\ after\ collision}{Relative\ velocity\ before\ collision}[/tex]

Hence, Relative Velocity after collision = Coefficient of restitution × Relative Velocity before collision

we substitute in our values;

Relative Velocity after collision [tex]= 0.35 \ *\ 6.85857m/s[/tex]

Relative Velocity after collision [tex]= 2.4 m/s[/tex]

Now, to determine how high should the puck bounced back

We use the Third Equation of Motion:

[tex]v^2 = u^2 + 2as[/tex]

Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.

Since the pluck is under gravity, we will have:

[tex]v^2 = u^2 + 2gh[/tex]

Now, since the hockey puck bounces back, it is experiencing a negative acceleration

Hence, the equation becomes

[tex]v^2 = u^2 - 2gh[/tex]

We substitute our values into the equation and find "h"

[tex](0m/s)^2 = (2.4m/s)^2 - ( 2*9.8m/s^2*h)\\\\0 = 5.76m^2/s^2 - (19.6m/s^2*h)\\\\(19.6m/s*h) = 5.76m^2/s^2 \\\\h= \frac{ 5.76m^2/s^2 }{19.6m/s^2}\\\\h = 0.3m[/tex]

Therefore, If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.

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Answer:

the ground pushes back on the player

Answer:

c

Explanation:

force is every where so anything that goes up must come down

Answer:

I think it's B

Explanation:

I think its trying to tell you that no matter who you are you could still do regular fitness but I don't know‍♀️

Answer:

According to Newton's first law of motion, an object maintains its state unless a force acts on it. Therefore, a moving car does not change its direction and keeps its speed unless a force acts on it.

Answer:

Explanation:

It's a velocity•time chart. As distance = vt, the area under the curve between limits is the distance traveled.

Pic is fuzzy so I will ASSUME the horizontal axis is seconds and vertical is meters per second.

0 s ≤ t ≤ 5 s = ½(5 - 0)(30 - 0) = 75 m

5 s ≤ t ≤ 10 s = (10 - 5)(30 - 0) = 150 m

10 s ≤ t ≤ 15 s = ½(30 + 20)(15 - 10) = 75 m

0 s ≤ t ≤ 25 s = 75 + 150 + 75 + 20(20 - 15) + ½(20)(25 - 20) = 450 m

Explanation:

S=(V^2-U^2)/2a a=g (gravity) a=10

=(0^2-25^2/2*(-10)

=625/20

=31.25m

Newton's second law allows us to find the results for the acceleration of the blocks are:

1) The acceleration is a = 559 m / s²

2) The acceleration is a = 7.88 m / s²

Newton's second law states that the net force is equal to the product of the mass and the acceleration of the body.

       ∑ F = m a

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the body.

The reference system is a coordinate system with respect to which the decomposition of the forces is carried out, in the attached we have a free body diagram of the system.

1) They indicate that the body mass is 25 kg.

y-axis

       N - W = 0

       N = W = m g

x-axis

        F -fr = ma

The friction force is a macroscopic force that results from the sum of all the microscopic interactions between the two surfaces, it has the formula

        fr = μ N

Where fr is the friction force, N the normal and very the friction coefficient.

This friction coefficient has two values:

We substitute.

         F - μ m g = m a

a) The system moves which is the acceleration.

Suppose that the force that star to move the system keeps constant, just before the system begins to move the coefficient of friction is static, let's find the applied force.

         F = μ m g

         F = 0.73 25 9.8

         F = 178.85 N

The block begins to move and the friction coefficient decreases to the dynamic value, we look for the acceleration.

         a = [tex]\frac{F - \mu \ m g}{m}[/tex]  

         a = [tex]\frac{178.85 - 0.16 \ 25 \ 9.8 }{25}[/tex]  

         a = 5.59 m / s²

2) In this case the mass of the block is 79 kg and the applied force is

        F = 909 N

We look for acceleration.        

        a = [tex]\frac{909 - 0.37 \ 79 \ 9.8 }{79}[/tex]  

        a = 7.88 m / S²

In conclusion using Newton's second law we can find the results for the acceleration of the blocks are:

         1) The acceleration is a = 559 m / s²

         2) The acceleration is a = 7.88 m / s²

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Explanation:

Since its accelerating, the velocity vs time graph is linear

For displacement we need initial velocity (which is zero because it starts from rest) and final velocity (which is calculatee thro acceleration formula

A= (vf - vi)/t

a= vf-0/t

1.25=vf / 7

1.25*7=vf

8.75 = vf

Now for displacement plug all the values in

X = 1/2(vf-vi)/t formula

The displacement (x) is 30.625 m

For part 3, we know new displacement that is 22m , the final and initial velocities are the same so just plug in the values for same formula above

The answer is t = 5.02

Im pretty sure all the answers are correct

The Two examples of contact forces are:

The two examples of non contact forces are:

Contact forces happens due to the contact between two objects

Non Contact forces happens because there is no contact between two objects. There is no attraction.

The amount of work done by the gas in the given expansion is 1800 J.

The given parameters;

The amount of work done by the gas in the given expansion is calculated as follows;

W = PΔV

where;

Substitute the given parameters and solve for the work done;

W = 1000(2.4  -  0.6

W = 1800 J

Thus, the amount of work done by the gas in the given expansion is 1800 J.

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