Which phrase desenbes an irregular galaxy ?
has a round shape
contains many young stars
has arms that extend from the center
Is larger than other types of galaxies

Answer:

Car A would have a better average speed

Explanation:

added weight to a object that is self propelled will be slower than a identical object with no added weight

Answer:82. Since you have a distance and a force, then the easiest principle to use is energy, i.e. work.

The work done by friction is F * d. This work cancels out the kinetic energy of the stone (1/2)mv^2

Fd = (1/2)mv^2

F = (1/2)mv^2/d.

Plug in m = 20 kg, v = 3 m/sec, d = 40 m.

83. With more mass, the kinetic energy is higher now. The work needed is higher. W = F * d and F is the same.

Explanation:Hope I helped :)

Answer:

Yes, geothermal energy is a renewable energy resource because the water can be heated by pumping it through the rocks.

Answer:

Acceleration = 18g

Explanation:

Given the following data;

Initial velocity, u = 26m/s

Final velocity, v = 0

Time = 0.15 secs

To find the acceleration;

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]

Substituting into the equation, we have;

[tex]a = \frac{0 - 26}{0.15}[/tex]

[tex]a = \frac{26}{0.15}[/tex]

Acceleration = 173.33m/s2

To express it in magnitude of g;

Acceleration = 173.33/9.8

Acceleration = 17.7 ≈ 18g

Acceleration = 18g

5 * 10² m

We are Given:

Frequency of the wave = 6 * 10⁵ Hz

Wavelength of the wave:

We know the relation:

c = λν  

[where λ (Lambda) is the wavelength ,ν (nu) is the frequency and c is the speed of light ]

3 * 10⁸  m/s= (6 * 10⁵ )* λ                          [replacing known values]

λ = [tex]\frac{3 * 10^{8}}{6 * 10^{5}}[/tex]                                                     [dividing both sides by 6 * 10⁵]

λ = 1/2 * 10³  

λ = 1/2 * 10 * 10²                                        [10³ can be rewritten as 10 * 10²]

λ = 5 * 10² m

Therefore, the wavelength of the wave is 5 * 10² m

Answer:

= 720000 [k]

Explanation:

The cost is equal to 5 [$/kW-h], kilowatt per hour, this value should be multiplied by the power, and then by the time.

[tex]5[\frac{k}{kw*h}]*200[w]*30[day]*24[\frac{h}{day} ][/tex]

= 720000 [k]

Answer:

The maximum speed of the car is 35 m/s

The total distance traveled by the car is 658.33 m

Explanation:

Given;

initial velocity of the car, u = 15 m/s

acceleration of the car, a = 2 m/s²

time of car motion, t = 10 s

(i)

Initial distance traveled by the car is given by;

d₁ = ut + ¹/₂at²

d₁ = (15 x 10) + ¹/₂(2)(10)²

d₁ = 150 + 100

d₁ = 250 m

The maximum speed of the car during this is given by;

v² = u² + 2ad₁

v² = (15)² + (2 x 2 x 250)

v² = 1225

v = √1225

v = 35 m/s

(ii)

The final distance cover by the car during the deceleration of 1.5 m/s².

Note: the final or maximum speed of the car becomes the initial velocity during deceleration.

v² = u² + 2ad₂

where;

v is the final speed of the car when it stops = 0

0 = u² + 2ad₂

0 = (35²) + (2 x - 1.5 x d₂)

0 = 1225 - 3d₂

3d₂ = 1225

d₂ = 1225 / 3

d₂ = 408.33 m

The total distance traveled by the car is given by;

d = d₁ + d₂

d = 250 m + 408.33 m

d = 658.33 m

Answer:

36 m/s

Explanation:

t = 3.6s

u = 0m/s

a = +g = 10m/s²

v = ?

using,

v = u + at

v = 0 + 10(3.6)

v = 36 m/s

Answer:

ou have I=200mA, E=40J, t=30s, and you want to find the voltage drop.

First, you should know that  P=V⋅I , so V=PI

Second, you have the amount of energy converted in a certain amount of time, so E=P⋅t

So, find the power and use it to find the voltage drop.

this works , but i thought energy was defined by W = P * t whitch would then be P = W/t

Answer:

Explanation:

F=kx

x=F/k

F=2000 kg

x=100 cm=9*10^-3

effective spring constant=k=F/x

k=2000/9*10^-3=2.2*10^-5

now frequency

f=1/2π√k/m

f=1/2*3.14√2.2*10^-5/310

f=1/6.28√7.097*10^-8

f=1/6.28*2.7*10^-4

f=0.16*2.7*10^-4

f=4.32*10^-5

The effective spring constant of the springs is 33755.55 N/m.

The frequency of the car's vibration is 2.07 Hz.

The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

A spring balance can be used to calculate the Force. The Newton is the SI unit of force.

Weight of the four people: F = 310 × 9.80 N = 3038 Newton.

The additional compression of the spring: x = 0.90 cm = 0.90 × 10⁻² m.

Hence, the effective spring constant of the springs: k= force/compression

= 3038 N/0.90 × 10⁻² m

= 33755.55 N/m.

The frequency of the car's vibration is: f = 1/2π√(k/m)

=1/2π√(33755.55/2000)

= 2.07 Hz.

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Answer:

0.25 m.

Explanation:

mass of the block = 7.40 kg, height = 0.83 m, force constant of the spring = 9.50 x [tex]10^{2}[/tex] N/m.

The maximum compression on the spring can be determined by;

Potential energy stored in the spring = [tex]\frac{1}{2}[/tex] K[tex]x^{2}[/tex]

But, potential energy = mgh

So that,

mgh = [tex]\frac{1}{2}[/tex] K[tex]x^{2}[/tex]

7.4 x 9.8 x 0.83 = 9.50 x [tex]10^{2}[/tex] x [tex]x^{2}[/tex]

60.1916 = 9.50 x [tex]10^{2}[/tex] x [tex]x^{2}[/tex]

[tex]x^{2}[/tex]= [tex]\frac{60.1916}{9.50*10^{2} }[/tex]

  = 0.06336

x = 0.2517

x = 0.25 m

The maximum compression of the spring is 0.25 m.

Answer:

The displacement of the car after 6s is 43.2 m

Explanation:

Given;

velocity of the car, v = 12 m/s

acceleration of the car, a = -1.6 m/s² (backward acceleration)

time of motion, t = 6 s

The displacement of the car after 6s is given by the following kinematic equation;

d = ut + ¹/₂at²

d = (12 x 6) + ¹/₂(-1.6)(6)²

d = 72 - 28.8

d = 43.2 m

Therefore, the displacement of the car after 6s is 43.2 m

[tex] \LARGE{ \underline{ \tt{Required \: answer:}}}[/tex]

We have:

We need to find:

Solution:

According to Newton's 2nd law of motion, or quantitative measure of Force:

Using this,

➝ F = ma

➝ 9N = 3 kg × a

➝ a = 9/3 m/s²

➝ a = 3 m/s²

Hence,

⛱️ [tex] \large{ \blue{ \bf{FadedElla}}}[/tex]

Answer:

2

Explanation:

there are 2 significant figures in there

Answer:

Yeah it's ok I think. Also, I can't see the answer you gave so maybe updating the question would be nice.

Answer:

The resultant velocity has a magnitude of 38.95 m/s

Explanation:

Vector Addition

Given two vectors defined as:

[tex]\vec v_1=(x_1,y_1)[/tex]

[tex]\vec v_2=(x_2,y_2)[/tex]

The sum of the vectors is:

[tex]\vec v=(x_1+x_2,y_1+y_2)[/tex]

The magnitude of a vector can be calculated by

[tex]d=\sqrt{x^2+y^2}[/tex]

Where x and y are the rectangular components of the vector.

We have a plane flying due west at 34 m/s. Its velocity vector is:

[tex]\vec v_1=(-34,0)[/tex]

The wind blows at 19 m/s south, thus:

[tex]\vec v_2=(0,-19)[/tex]

The sum of both velocities gives the resultant velocity:

[tex]\vec v =(-34,-19)[/tex]

The magnitude of this velocity is:

[tex]d=\sqrt{(-34)^2+(-19)^2}[/tex]

[tex]d=\sqrt{1156+361}=\sqrt{1517}[/tex]

d = 38.95 m/s

The resultant velocity has a magnitude of 38.95 m/s

Answer:

A) 140 k

b ) 5.22 *10^3 J

c) 2910 Pa

Explanation:

Volume of Monatomic ideal gas = 1.20 m^3

heat added ( Q ) = 5.22*10^3 J

number of moles  (n)  = 3

A ) calculate the change in temp of the gas

since the volume of gas is constant no work is said to be done

heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT

make ΔT subject of the equation

ΔT = Q / n.(3/2).R

    = (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )

    = 140 K

B) Calculate the change in its internal energy

ΔU = Q  this is because no work is done

therefore the change in internal energy = 5.22 * 10^3 J

C ) calculate the change in pressure

applying ideal gas equation

P = nRT/V

therefore ; Δ P = ( n*R*ΔT/V )

                        = ( 3 * 8.3144 * 140 ) / 1.20

                        = 2910 Pa

A) The change in the temperature of the gas is; ΔT = 139.5 K

B) The change in internal energy of the gas is; ΔU = 5.22 × 10³ J

C) The change in pressure of the gas is; ΔP = 2899.5 Pa

We are given;

Volume of Monatomic ideal gas; V = 1.2 m³

Amount of heat added; Q = 5.22 × 10³ J

number of moles; n = 3

A) To calculate the change in temperature of the monatomic idea gas, we will use formula;

Q = ³/₂nRΔT

Where R is a constant = 8.314 J/mol.K

ΔT is the change in temperature

Making ΔT the subject of the formula;

ΔT = ²/₃(Q/(nR))

ΔT = ²/₃(5.22 × 10³)/(3 × 8.314)

ΔT = 139.5 K

B) Due to the fact that no work was done, then from first law of thermodynamics, we can say that;  

ΔU = Q

Thus;

change in internal energy; ΔU = 5.22 × 10³ J

C) The change in pressure will be calculated from the formula;

ΔP = (n*R*ΔT)/V

ΔP = (3 * 8.314 * 139.5)/1.2

ΔP = 2899.5 Pa

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Answer:

Explanation:

m1v1=m2v2

m1=70 kg

m2=10 g=0.01 kg

v2=500 m/s

m1v1=m2v2

v1=m2v2/m1

v1=0.01*500/70

v1=0.07

Answer:

[tex]P=2.57\times 10^{-7}\ N/m^2[/tex]

Explanation:

Given that,

A radio wave transmits 38.5 W/m² of power per unit area.

A flat surface of area A is perpendicular to the direction of propagation of the wave.

We need to find the radiation pressure on it. It is given by the formula as follows :

[tex]P=\dfrac{2I}{c}[/tex]

Where

c is speed of light

Putting all the values, we get :

[tex]P=\dfrac{2\times 38.5}{3\times 10^8}\\\\=2.57\times 10^{-7}\ N/m^2[/tex]

So, the radiation pressure is [tex]2.57\times 10^{-7}\ N/m^2[/tex].

Answer:

Approximately [tex]35.2\; \rm m[/tex].

Explanation:

Given:

Initial velocity: [tex]u = 13\; \rm m \cdot s^{-1}[/tex].

Acceleration: [tex]a = -2.40\; \rm m \cdot s^{-2}[/tex] (negative because the car is slowing down.)

Implied:

Final velocity: [tex]v = 0\; \rm m \cdot s^{-1}[/tex] (because the car would come to a stop.)

Required:

Displacement, [tex]x[/tex].

Not required:

Time taken, [tex]t[/tex].

Because the time taken for this car to come to a full stop is not required, apply the SUVAT equation that does not involve time:

[tex]\begin{aligned} x &= \frac{v^2 - u^2}{2\, a} \\ &= \frac{{\left(0\; \rm m \cdot s^{-1}\right)}^2 - {\left(13\; \rm m \cdot s^{-1}\right)}^2}{2\times \left(-2.40\; \rm m\cdot s^{-2}\right)} \approx 35.2\; \rm m \end{aligned}[/tex].

In other words, this car would travel approximately [tex]35.2\; \rm m[/tex] before coming to a stop.

Answer:

metals like iron or nickel

Explanation:

Answer:

A. 24.25 m/s

Explanation:

velocity = [tex]\sqrt{2 * g * d}[/tex]

velocity = sqr 2 * 9.8 * 30 = sqr 588 = 24.25 m/s

The velocity of the watermelon as it smashes into the ground will be 24.2 m/s

The third equation of motion is -

v² - u² = 2aS

Given David Wetterman drops a 5 kg watermelon from the top of a 30 m building.

Height of building [S] = 30 m

Mass of watermelon [M] = 5 Kg

Initial velocity [v] = 0 m/s

acceleration [g] = 9.8 m/s²

Using the third equation of motion -

v² - u² = 2aS

v² = 2aS

v² = 2 x 9.8 x 30

v² = 588

v = 24.2 m/s

Therefore, the velocity of the watermelon as it smashes into the ground will be 24.2 m/s.

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Answer:half-life (T1/2) of this isotope =

Explanation:

The number of nuclei of any radioactive substance at a given time is expressed by

Nt = N0e⁻kt

Nt=decay of material  at a time t, =3110 decays per minute

N=decays at t=0, 8255 decays per minute

k=constant

Nt=N0e−kt

3110= 8255 e⁻k(4.50)

3110/ 8255=e−k(4.50)

0.3767 =  e−k(4.50)

In 0.3767  =   -k (4.50)

0.976=-4.5k

k=0.976/4.5

=0.2159

Also we know that t 1/2= time that it takes half the original material to decay.it is  related to the rate constant by

T₁/₂=ln  2 / k

Therefore half-life (T1/2) of this isotope

T₁/₂=ln  2/0.2159

T₁/₂=3.12 days

Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is  [tex]X = 38.3 \ m[/tex]

Explanation:

From the question we are told that

The acceleration along the x axis is  [tex]a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)[/tex]

  The position of the object at t = 0 is  x = -14.0 m

  The velocity at t = 0 s is  [tex]v_{0}x = 7.10 m/s[/tex]

Generally from the equation for acceleration along x axis we have that

     [tex]a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)[/tex]

=>   [tex]\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt[/tex]

=>   [tex]V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]

At  t =0  s   and  [tex]v_{0}x = 7.10 m/s[/tex]

=>   [tex]7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1[/tex]

=>   [tex]K_1 = 7.10[/tex]      

So  

      [tex]\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]

=>  [tex]\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}[/tex]

=>  [tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2[/tex]

At  t =0  s   and   x = -14.0 m

  [tex]-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2[/tex]

=>   [tex]K_2 = -14[/tex]

So

     [tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14[/tex]

At  t = 10.0 s

      [tex]X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14[/tex]

=>   [tex]X = 38.3 \ m[/tex]

As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.

Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.

The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:

n + (-w) = 0

n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N

The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be

f = 0.67 (51.94 N) ≈ 35 N

so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.

The block starts moving as soon as x = 35 N, at which point f = 35 N.

At any point after the block starts moving, we have

f = 0.48 (51.94 N) ≈ 25 N

so that x = 25 N is the required force to keep the block moving at a constant speed.

As x  is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force

Given data :

mass of block at rest ( m ) = 5.3 kg

Coefficient of static friction ( μ_s ) =0.67

Coefficient of kinetic friction is ( μ_k ) = 0.48

Horizontal force applied to block = x  

First step : magnitude of normal force ( n ) when object is at rest

n = w            where w = m*g

n - w = 0

n - ( 5.3 * 9.81 ) = 0     ∴  n = 51.94 N

Second step : Required magnitude of x before the movement of object

F =  μ_s * n

F = 0.67 * 51.94  = 34.79 N  ≈ 35 N

∴ The object will start moving once F and x = 35 N

Final step : Magnitude of x  after object start moving

F = μ_k  * n

  = 0.48 * 51.94 = 24.93 N  ≈ 25 N

∴ object will continue to accelerate at a constant speed once F and x = 25N

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Answer:

b. matter

Explanation:

Waves disturb matter but do not transmit it.

Waves are disturbances that transmit energy from one point to another. Although they cause disturbances, they do not transfer the matters in the medium.

Answer: 1.14 kg*m/s

Explanation:

The first person explained everything right, they just forgot to convert the angular acceleration to rads/sec^2 from revs/sec^2. Once that is converted, your answer should come out right.

Another small thing, the answer there has an extra unnecessary step. It tells you to find the square root of w^2 to find w but that is unnecessary since the final equation asked for w^2. Hope this helps! :)

The moment of inertia I of the flywheel about its rotation axis is

Given

Angular displacement,

[tex]\theta = 30rev \\\\\theta = (30) * 2\pi rad \\\\\theta = 188.495rad[/tex]

Therefore, Final angular velocity (w) will be:

[tex]w^2 = 2\alpha\theta\\\\w^2 = 2 * (0.200 * 2\pi) * (188.49)\\\\w^2 = 473.73\\\\w = 21.76 rad/s[/tex]

Therefore,

moment of inertia

[tex]I = 2 * K / w^2[/tex]

[tex]I = 2 * 330 / 473.73[/tex]

[tex]I = 1.39kgm^2[/tex]

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Complete Question

A rolling ball moves from [tex]x_1 = 8.0 \ cm[/tex] to [tex]x_2 = - 4.1 \ cm[/tex] during the time from [tex]t_1 = 2.9 s[/tex]  to  [tex]t_2 = 6.0s[/tex]

What is its average velocity over this time interval?

Answer:

The velocity is  [tex]v = 3.903 \ m/s[/tex]

Explanation:

From the question we are told that

    The first position of the ball is  [tex]x_1 = 8.0 \ cm[/tex]

    The second position of the ball is  [tex]x_2 = - 4.1 \ cm[/tex]

Generally the average velocity is mathematically represented as

       [tex]v = \frac{ x_1 - x_2}{t_2 - t_1}[/tex]

=>    [tex]v = \frac{ 8 - -4.1 }{ 6 - 2.9 }[/tex]

=>    [tex]v = 3.903 \ m/s[/tex]

Answer:

option 3

Explanation:

can i get brainliest

Answer:

option c is correct

Explanation:

we know that

2as=vf^2-vi^2

vf=24 m/s

vi= 0 m/s

a=g= 9.8 m/s^2

s=vf^2-vi^2/2a

s=(24)²-(0)²/2*9.8

s=576/19.6

s=29.4 m

therefore option c is correct