Which phrase describes an irregular galaxy O has a round shape O does not have a specific shape O has arms that extend from the center O is larger than other types of galaxies​

Answer:

here your answer

i am sorry if wrong

Answer:

wool maybe, I was a bit confused in silk and wool but wool

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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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.

Answer:

the ground pushes back on the player

Answer:

c

Explanation:

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

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

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

Answer:

whatttttttttttt

Explanation:

the ಠᴥಠಠᴥಠಠᴥಠಠᴥಠಠᴥಠಠᴥಠಠᴥಠಠᴥಠ

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.

Let's see

[tex]\\ \sf\longmapsto \overrightarrow{drag}[/tex]

[tex]\\ \sf\longmapsto \overleftarrow{Thrust}[/tex]

[tex]\\ \sf\longmapsto Lift\:Force\uparrow[/tex]

[tex]\\ \sf\longmapsto Acceleration\:due\:to\: gravity\downarrow[/tex]

Net force is 0

Answer:

Explanation:

Appears to be the vertexes of a triangle.

AB = √(-6 - (-2))² + (3 - 4)²) = √17

AC = √(-6 - (-8))² + (3 - 3)²) = 2

BC = √(-2 - (-8))² + (4 - 3)²) = √37

#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]

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

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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Answer:

A

Explanation:

all galaxies exploded in order to create the sun/stars

Answer:I think Fossils show that some animals are extinct

Explanation:

Please mark as brainliest for me.Thanks

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:

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.

Explanation:

this is actually not as simple as it sounds here.

quite some energy is lost in the deformation of the bodies of car and truck, and it also needs more energy to get a standing object going than to accelerate an already moving object.

but assuming the simple described circumstances, then the energy and impulse of the moving truck of 5000 kg is transferred to a new combined system of car and truck of now 5000 + 3000 = 8000 kg.

so, the 20m/s inertia energy of the truck is now distributed to the truck/car combination.

since the same energy has to move now more mass, it is clear that the combined speed will be lower.

20×5000 = x×8000

20×5 = x×8

x = 100/8 = 12.5 m/s

that is the resulting speed of the combined truck/car object.

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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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:

[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]

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:

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

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