Please answer my question

Please Answer My Question

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Answer 1

Answer:

Answer is (b) Mercury, venus and Mars.

Answer 2

Explanation:

i think b is correct!!

;-) :-) :-) :-)


Related Questions

"2.40 A pressure of 4 × 106N/m2 is applied to a body of water that initially filled a 4300 cm3 volume. Estimate its volume after the pressure is applied."

Answers

Answer:Final volume after pressure is applied=4,292cm3

Explanation:

Using the bulk modulus formulae

We have that The bulk modulus of waTer is given as  

K =-V dP/dV

Where  K, the bulk modulus of water = 2.15 x 10^9N/m^2

2.15 x 10^9N/m^2= - 4,300 x  4 × 106N/m2 / dV

dV = - 4,300 x  4 × 10^6N/m^2/ 2.15 x 10^9N/m^2

dV (change in volume)= -8.000cm^3

Final volume after pressure is applied,

V= V+ dV

V= 4300cm3 + (-8.000cm3)

=4300cm3 - 8.000cm3

Final Volume, V =4,292cm3

A spinning ice skater will slow down if she extends her arms away from her body. Which of the following statements explain this phenomenon
A) circular motion is always uniform
B) A centripetal force always points outward
C) Angular momentum is always conserved
D) Centripetal acceleration cannot change
Marking brainliest

Answers

The answer is B which is a centripetal force always points outwards

Answer:

B, which is why ice skaters often keep their arms close to their body when doing spins and jumps to minimize resistance.

What is the current in the wire now?

Answers

Answer:

220v

Explanation:

Sorry, the question is incomplete

Answer:

on the potential difference applied and on the resistance of the wire.

Explanation:

Ohms law state that the current through a conductor between two points is directly proportional to the potential difference across the two points. Imtroducing the comstant of proportionality, the resistance, one arrives at the usual athematical equation that describes this relationship: I = V/R.

Acceleration is sometimes expressed in multiples of g, where g = 9.8 m/s^2 is the magnitude of the acceleration due to the earth's gravity. In a test crash, a car's velocity goes from 26 m/s to 0 m/s in 0.15 s. How many g's would be experienced by a driver under the same conditions?

Answers

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

How much work would be done on a particle with 5.0 C of charge on it if it moved from an equipotential line at 5.5 volts to another equipotential line at 3.5 volts?

Answers

Answer:

10J

Explanation:

In this question we have the following information

The charge of the particle is q = 5 C

The equipotenetial level is V1 = 5.5 v

and also the

equipotenetial level is V2 = 3.5 v

So we calculate the

work done W=q x (v1-v2)

workdone = 5 x (5.5-3.5)

= 5x2

=10 J

Workdone = 10 J

So we conclude that the workdone on a particle with these information is 10j

A 5.3 kg block rests on a level surface. The coefficient of static friction is μ_s=0.67, and the coefficient of kinetic friction is μ_k= 0.48 A horizontal force, x is applied to the block. As x is increased, the block begins moving. Describe how the force of friction changes as x increases from the moment the block is at rest to when it begins moving. Show how you determined the force of friction at each of these times ― before the block starts moving, at the point it starts moving, and after it is moving. Show your work.

Answers

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

Magnitude of normal force ( object at rest );  n = 51.94 N Required magnitude of x before the movement of object ; x = 35 NMagnitude of x  after object start moving   x = 25 N

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

Learn more : https://brainly.com/question/21444366

If the particles were moving with a speed much less than c, the magnitude of the momentum of the second particle would be twice that of the first. However, what is the ratio of the magnitudes of momentum for these relativistic particles?

Answers

Answer:

p₂ / p₁ = 2 (v₁ / v₂)

Explanation:

The moment is a very useful concept, since it is one of the quantities that is conserved during shocks and explosions, for which it had to be redefined to be consistent with special relativity,

         p = m v / √[1+ (v/c)² ]

for the case of speeds much lower than the speed of light this expression is close to

         p = m v

 

In this exercise they indicate that the moment of the second particle is twice the moment of the first, when their velocities are small

        p₂ = 2 p₁

       p₂/p₁ = 2

in consecuense

       m v₂ = 2 m v₁

       v₂ = 2 v₁

consider particles of equal mass.

By the time their speeds increase they enter the relativistic regime

        p₂ = mv₂ /√(1 + v₂² /c²)

        p₁ = m v₁ /√(1 + v₁² / c²)

let's look for the relationship between these two moments

       p₂ / p₁ = mv₂ / mv₁   [√ (1+ v₁² / c²) /√ (1 + v₂² / c²)

       

from the initial statement

      p₂ / p₁ = 2 √(c² + v₁²) / (c² + v₂²)

we take c from the root

      p₂ / p₁ = 2 √ [(1+ v₁²) / (1 + v₂²)]

this is the exact result, to have an approximate shape suppose that the velocities are much greater than 1

      p₂ / p₁ = 2 √ [v₁² / v₂²] = 2 √ [(v₁ / v₂)²]

      p₂ / p₁ = 2 (v₁ / v₂)

we see the value of the moment depends on the speed of the particles

In the winter sport of curling, players give a 20 kg stone a push across a sheet of ice. The Slone moves approximately 40 m before coming to rest. The final position of the stone, in principle, onlyndepends on the initial speed at which it is launched and the force of friction between the ice and the stone, but team members can use brooms to sweep the ice in front of the stone to adjust its speed and trajectory a bit; they must do this without touching the stone. Judicious sweeping can lengthen the travel of the stone by 3 m.1. A curler pushes a stone to a speed of 3.0 m/s over a time of 2.0 s. Ignoring the force of friction, how much force must the curler apply to the stone to bring it op to speed?A. 3.0 NB. 15 NC. 30 N
D. 150 N2The sweepers in a curling competition adjust the trajectory of the slope byA. Decreasing the coefficient of friction between the stone and the ice.
B. Increasing the coefficient of friction between the stone and the ice.C. Changing friction from kinetic to static.D. Changing friction from static to kinetic.3. Suppose the stone is launched with a speed of 3 m/s and travel s 40 m before coming to rest. What is the approximate magnitude of the friction force on the stone?A. 0 NB. 2 NC. 20 ND. 200 N4. Suppose the stone's mass is increased to 40 kg, but it is launched at the same 3 m/s. Which one of the following is true?A. The stone would now travel a longer distance before coming to rest.B. The stone would now travel a shorter distance before coming to rest.C. The coefficient of friction would now be greater.D. The force of friction would now be greater.

Answers

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 :)

8. A rectangle is measured to be 6.4 +0.2 cm by 8.3 $0.2 cm.

a) Calculate its perimeter in cm

b) Calculate the uncertainty in its perimeter.

Answers

Answer:

a) The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty in its perimeter is 0.8 centimeters.

Explanation:

a) From Geometry we remember that the perimeter of the rectangle ([tex]p[/tex]), measured in centimeters, is represented by the following formula:

[tex]p = 2\cdot (w+l)[/tex] (1)

Where:

[tex]w[/tex] - Width, measured in centimeters.

[tex]l[/tex] - Length, measured in centimeters.

If we know that [tex]w = 6.4\,cm[/tex] and [tex]l = 8.3\,cm[/tex], then the perimeter of the rectangle is:

[tex]p = 2\cdot (6.4\,cm+8.3\,cm)[/tex]

[tex]p = 29.4\,cm[/tex]

The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty of the perimeter ([tex]\Delta p[/tex]), measured in centimeters, is estimated by differences. That is:

[tex]\Delta p = 2\cdot (\Delta w + \Delta l)[/tex]  (2)

Where:

[tex]\Delta w[/tex] - Uncertainty in width, measured in centimeters.

[tex]\Delta l[/tex] - Uncertainty in length, measured in centimeters.

If we know that [tex]\Delta w = 0.2\,cm[/tex] and [tex]\Delta l = 0.2\,cm[/tex], then the uncertainty in perimeter is:

[tex]\Delta p = 2\cdot (0.2\,cm+0.2\,cm)[/tex]

[tex]\Delta p = 0.8\,cm[/tex]

The uncertainty in its perimeter is 0.8 centimeters.

A man walks south at a speed of 2.00 m/s for 60.0 minutes. He then turns around and walks north a distance 3000 m in 25.0 minutes. What is the average velocity of the man during his entire motion?

Answers

Answer:

v = 0.823 m/s

Explanation:

A man walks south at a speed of 2.00 m/s for 60.0 minutes.

The distance covered in South = 60 × 60 × 2 = 7200 m

He then turns around and walks north a distance 3000 m in 25.0 minutes.

As they moved in opposite direction, net displacement will be : 7200 - 3000 = 4200 m

Average velocity of the man = net displacement/time

[tex]v=\dfrac{4200\ m}{(60+25)\times 60}\\\\=0.823\ m/s[/tex]

So, the average velocity of the man is 0.823 m/s.

While riding a multispeed bicycle, the rider can select the radius of the rear sprocket that is fixed to the rear axle. The front sprocket of a bicycle has radius 12.0 cm. If the angular speed of the front sprocket is 0.600 rev/s, what is the radius of the rear sprocket for which the tangential speed of a point on the rim of the rear wheel will be 5.00 m/s?

Answers

Answer:

2.9 cm

Explanation:

Assuming that the rear wheel has a radius of 0.330 m

Given that

r(a) = 12 cm -> 0.12 m

w(a) = 0.6 rev/s -> 3.77 rad/s

v = 5 m/s

r(w) = 0.330 m

The speed on any point on the rim at the sprocket in the front is

v(a) = w(a).r(a) = 3.77 * 0.12 = 0.4524 m/s

Also,

v(a) = speed at any point on the chain

v(b) = speed at any point on the rim of the rear sprocket

v(a) = v(b)

where v(b) = w(b).r(b)

Recall that the speed at any point on the rear wheel is v, where

v = w(b).r(w)

5 = w(b) * 0.330

w(b) = 5/0.330

w(b) = 15.15 rad/s

On substituting this in the equation, we have

v(b) = w(b).r(b).

Remember also, that v(a) = v(b), so

0.4524 = 15.15 * r(b)

r(b) = 0.4524 / 15.15

r(b) = 0.029 m -> 2.9 cm

Therefore, the radius of the rear sprocket needed is 2.9 cm

during a baseball game you are running home and slide into home plate. However you come up short and you are tagged out. Which force stops you from sliding all the way home? a friction b gravity c pull d push

Answers

Answer:1 because

Explanation: it’s pointing to the earth and gravity

Pulls things down to earth

A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?

Answers

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

A baseball is thrown across the field. The ____________is measured from where the ball is thrown to where landed was 75 feet.

motion
direction
distance
reference point

Answers

Answer:

distance i think

Explanation:

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