Look at this model of an atom. Using a periodic table, which element does it represent?

30 m/s

We are given:

Initial Velocity (u) = 0 m/s

Acceleration of the Car (a) = 3 m/s²

Time Interval (t) = 10 seconds

Speed of the Car After 10 seconds:

From the First equation of motion:

v = u + at

replacing the given values

v = 0 + (3)(10)

v = 30 m/s

Hence, the car is moving at a velocity of 30 m/s after 10 seconds

Answer:

The value is  [tex]\Delta s = 8.537 *10^{25 } \ J/K[/tex]

Explanation:

From the we are told that

   The radius of the sphere is [tex]r = 6.32 *10^{8} \ m[/tex]

   The temperature is [tex]T_x = 5350 \ K[/tex]

    The average temperature of the rest of the universe is  [tex]T_r = 2.73 \ K[/tex]

Generally the change in entropy of the entire universe per second is mathematically represented as

         [tex]\Delta s = s_r - s_x[/tex]

Here [tex]s_r[/tex] is the entropy of the rest of the universe which is mathematically represented as

          [tex]s_r = \frac{Q}{T_r}[/tex]

Here Q is the quantity of heat radiated by the star which is mathematically represented as

           [tex]Q = 4 \pi * r^2 * \sigma * T^4_x[/tex]

Here [tex]\sigma[/tex] is the Stefan-Boltzmann constant with value  

           [tex]\sigma = 5.67 * 10^{-8 }W\cdot m^{-2} \cdot K^{-4}.[/tex]

=>         [tex]Q = 4 \pi * (6.32*10^{8})^2 * 5.67 * 10^{-8 } * 5350 ^4[/tex]

=>         [tex]Q = 2.332 *10^{26} \ J[/tex]

So

      [tex]s_r = \frac{2.332 *10^{26}}{2.73}[/tex]

=>   [tex]s_r = 8.5415 *10^{25}\ J/K[/tex]

Here [tex]s_x[/tex] is the entropy of the rest of the universe which is mathematically represented as

      [tex]s_x = \frac{Q}{T_x}[/tex]

=>   [tex]s_x = \frac{2.332 *10^{26} }{5350}[/tex]

=>   [tex]s_x = 4.359 *10^{22} \ J/K[/tex]

So

      [tex]\Delta s = 8.5415 *10^{25} - 4.359 *10^{22}[/tex]

=>   [tex]\Delta s = 8.537 *10^{25 } \ J/K[/tex]

This question involves the concepts of entropy and the thermal radiation

The entropy of the entire universe is increased by "8.41 x 10²⁵ J/k

".

The increase in entropy is given as follows:

[tex]\Delta s = s-s_T[/tex]

where,

Δs = increase in entropy = ?

σ = Stefan-Boltzman's constant = 5.67 x 10⁻⁸ W/m².k⁴

A = surface area = 4πr² = 4π(6.32 x 10⁸ m)² = 5.01 x 10¹⁸ m²

Tr = Absolute temperature of the star = 5350 K

T = absolute temperature of the rest of the universe = 2.73 k

Q = thermal radiation energy

Q = [tex]\sigma A T_r^4=(5.67\ x\ ^{-8}\ W/m^2.k^4)(5.01\ x\ ^{18}\ m^2)(5350\ k)^4=2.3\ x\ 10^{26}\ J[/tex]

s = entropy of the universe = [tex]\frac{Q}{T}=\frac{2.3\ x\ 10^{26}\ J}{2.73 k}=8.42\ x\ 10^{25}\ J/k[/tex]

[tex]s_T[/tex] = entropy of the star = [tex]\frac{Q}{T_r}=\frac{2.3\ x\ 10^{26}\ J}{5350\ k}=4.3\ x\ 10^{22}\ J/k[/tex]

Therefore,

Δs = 8.42 x 10²⁵ J/k - 4.3 x 10²² J/k

Δs = 8.41 x 10²⁵ J/k

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Answer:

I hope the above pic helps you mate

I will always help you understanding your assingments

have a great day

#Captainpower:)

Answer:

this is a homework helping website I don't know If I can help you with this problwm

Answer:

False

Explanation:

We still use the old way. It's easier to understand the old way, and its NOT outdated.

The third law of motion defined the action and reaction force that the heavy ball should be fall at the same kind of the speed.

It defined that at the time when the two bodies should be interacted so here the forces should be applied to one another that should be equivalent to the magnitude also the direction should be opposite.

Also, we can called as the action law and the reaction.

Therefore, we can conclude that the third option is correct.

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Answer:d

Explanation: oil would form droplets but only tiny ones because it’s surface tension is lower than that of water

Both water and vegetable oil would made droplets but Oil would form  only very tiny droplets because its surface tension is lower than that of water.

Surface tension is the propensity for liquid surfaces that are at rest to condense into the smallest surface area. Razor blades and insects (like water striders), which have a higher density than water, can float on the surface of the water without even becoming partially buried because to surface tension.

Surface tension at liquid-air contacts originates from the liquid molecules' stronger attraction to one another due to cohesion than to the air molecules (due to adhesion).

As the surface tension of the oil is lower than that of water, it forms droplets will smaller size.

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#SPJ5

Answer:

miami dolphins.

Explanation:

......

Answer:

miami :(

Explanation:

Answer:

pressure=force/area

p=60N/(2m×1m)

=35N/M

Answer:

25m

Explanation

Let the horizontal displacement formula be expressed as;

Dx = Dcos theta

theta is the angles subtended by the displacement

Given

D = 50m

theta = 60°

Required

horizontal displacement of the chimpanzee

Substitute into the formula

Recall that: Dx = Dcos theta

Dx = 50cos60°

Dx = 50(0.5)

Dx = 25m

Hence the horizontal displacement of the chimpanzee is 25m

Answer:

85.4 N

Explanation:

Weight of the ball, W = 60 N

W = mg, m is mass

m = W/g

m = 60/9.8 = 6.12 kg

Length of a rope, r = 6 m

Speed, v = 5 m/s

We need to find the tension in the rope at this point. Tension is equal to the centripetal force It is given by :

[tex]F=\dfrac{mv^2}{r}+mg\\\\F=\dfrac{6.12\times (5)^2}{6}+60\\\\=85.5\ N[/tex]

So, the correct option is (b) " 85.4 N".

Answer:

The value is  [tex]v = 2 \ m/s[/tex]

Explanation:

From the question we are told that

   The velocity of the each of the three cars is [tex]u_1 = u_2 = u_3 = 2 \ m/s[/tex]

    The velocity of the fourth car is  [tex]u_4 = 4 \ m/s[/tex]

    The initial velocity of the fifth car [tex]u_5 = 0 \ m/s[/tex]

Generally from the law of momentum conservation we have that

    [tex]m_1 u_1 + m_2 u_2 + m_3 u_3 +m_4u_4 + m_5u_5 = [m_1 + m_2 + m_3 +m_4+ m_5]v[/tex]

Given that the cars are identical then their mass will be the same

i.e

    [tex]m_1 =m_2 = m_3 = m_4 = m_5 = m[/tex]

=>   [tex][u_1 + u_2 + u_3 +u_4 + u_5]m = 5mv[/tex]

=>   [tex]2+ 2 + 2 +4 + 0 = 5v[/tex]

= >   [tex]v = 2 \ m/s[/tex]

Answer:

The minimum coefficient of static friction required, µ = 0.10

Note. The question is incomplete. The complete question is given below:

While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now the driver court date is approaching.

The log has a mass of m = 929 kg; the truck has a mass of M = 8850 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 23.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction μs needed to keep the log in the back of the truck.

Explanation:

First, velocity in mph is converted to m/s

1 mph = 0.447 m/s

55 mph ≈ 24.6 m/s

The acceleration of the empty truck is a = v/t = 24.6 / 23 = 1.07 m/s²

Force that can be generated by the truck, F = ma

F = 8850kg * 1.07 m/s² = 9469.5 N

However, with the added mass of the log on it, the acceleration of the truck will become;

a = F / m = 9469.5 N / (8550+929)kg = 0.97 m/s²

Frictional force between the log and the truck = 0.97 m/s² * 929 kg = 901.13 N

Normal reaction on the truck due to the weight of the log, R = mg

R = 929 kg * 9.8m/s² = 9104.2 N

Coefficient of static friction, µ = F/R

µ = 901.13/9104.2

µ = 0.098 ≈ 0.10

Therefore, the minimum static friction required is µ = 0.10

Answer:

hello your question is incomplete attached below is the complete question

answer :

a) I1 = I2

b) J1 > J2

c) E 1 > E2

d) ( vd1 ) > ( vd2 )

Explanation:

a) The currents in the two segments are the same  i.e. I1 = I2  and this is because the segments are connected in series

b) Comparing the current densities J1 and J2 in the two segments

note : current density ∝ 1 / area

The area of the second segment is > the area of first segment  therefore

J1 > J2

J1 ( current density of first segment )

J2 ( current density of second segment )

c) Comparing the electric field strengths E1 and E2

 note : electric field strength ∝ current density

since current density of first segment is > current density of second segment  and conductivity of the materials are the same hence

E 1 > E2

d) Comparing the drift speeds Vd1 and Vd2

( vd1 ) > ( vd2 )

this because  ; vd ∝ current density

Currents in the two segments are equal. Current densities, electric field strengths and drift speeds of the first segment are greater than the second segment.

The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.

The wire in consists of two segments of different diameters but made from the same metal.

The current in segment 1 is [tex]I_1[/tex] and the current in segment 2 is [tex]I_2[/tex]. As, the segments are connected in the series connection. Thus, the value of both the currents will be equal.

[tex]I=I_1=I_2[/tex]

The current density is total electric current which is flowing in a unit cross sectional area. Current density is inversely p[proportional to the cross sectional area.

The cross-sectional area of the second segment is greater than the first one as shown in the image attached below.

Thus, the current densities of first segment will be J1 greater than the current density of second segment J2.

[tex]J_1 > J_2[/tex]

The current density of the first segment is greater than the second one. Thus, the electric field of the first segment will also be greater than the second one for the same material.

[tex]E_1 > E_2[/tex]

Current density is directly proportional to the drift velocity. As the current densities of first segment is greater than the current density of second segment.Therefore, there drift speed,

[tex]v_{d1} > v_{d2}[/tex]

Hence, the currents in the two segments are equal. Current densities, electric field strengths and drift speeds of the first segment are greater than the second segment.

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i think correct answer is b

Answer:

Explanation:

Answer:

see attached file

Explanation:

Answer:

If gravity on Earth is increased, this gravitational tugging would have influenced the moon's rotation rate. If it was spinning more than once per orbit, Earth would pull at a slight angle against the moon's direction of rotation, slowing its spin. If the moon was spinning less than once per orbit, Earth would have pulled the other way, speeding its rotation.

Answer:

e

Explanation:

Answer: E B

Explanation:

Answer:

What is the question though?

Explanation:

Potential energy of a particle is directly proportional its position, thus, potential energy of particle increases with increase in the height of the particle and decreases with a decrease in the height of the particle.

The potential energy of a particle is the energy possessed by the particle by virtue of its position. This potential energy is form of mechanical energy and it is expressed mathematically as the product of the mass of the particle, acceleration due to gravity and the particles displacement or position.

P.E = mgh

where;

P.E is the potential energy of the particle

m is the mass of the particle

g is the acceleration due to gravity

h is the position of the particle.

Potential energy of a particle is directly proportional its position, thus, potential energy of particle increases with increase in the height of the particle and decreases with a decrease in the height of the particle.

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Answer:

m1 ÷ m2 = 1 ÷ 2

Explanation:

The ratio of their masses is as follows:

As it is given that the mass m1 is twice of mass m2

So the equation could be written as

[tex]\frac{1}{2} m1 v1 v1 = 2 \times \frac{1}{2} m2 v2 v2\\\\\frac{m1}{m2} = \frac{2 v2 v2}{v1 v1} ..........(1)[/tex]

Here we used the  conservation of momentum

Prior to the explosion, the object is in rest so the momentum would be zero but after the explosion the total momentum would be  m1v1 + m2v2,  that is also zero.

So,

[tex]m1 v1 + m2 v2 = 0\\\\m1 v1 = - m2 v2\\\\\frac{m1}{m2} = \frac{- v2}{v1}[/tex]

Now squaring to the both sides

m1 m1 ÷ (m2 m2) = v2 v2 ÷ (v1 v1) ........(2 )

Solved both the equations

After solving it, the ratio is

m1 ÷ m2 = 1 ÷ 2

Answer:

Impulse = 0.1 Kgm/s

Explanation:

Given the following data;

Force, F = 10N

Time, t = 0.01 seconds

To find impulse

An impulse can be defined as the net force acting an object for a very short period of time.

Mathematically, impulse is given by the formula;

Impulse = force * time

Substituting into the equation, we have;

Impulse = 10 * 0.01

Impulse = 0.1 kgm/s

Therefore, the impulse of the hammer is 0.1 kilogram meter per seconds.

The impulse of the hammer is 0.1 Ns.

To solve the given problem we need to use the formula for calculating impulse.

Impulse: This can be defined as the product of force and time on a body.

The formula of impulse is given below.

I = Ft....................... Equation 1

Where :

From the question,

Given:

Substitute these values into equation 1

Hence, The impulse of the hammer is 0.1 Ns.

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Answer:

The two forces acting on rockets at the moment of launch are the thrust upwards and the weight downwards. Weight is the force due to gravity and is calculated (at the Earth’s surface) by multiplying the mass (kilograms) by 9.8.The resultant force on each rocket is calculated using the equation resultant force = thrust – weight.

Hopefully, this answer helps you! :)

Answer: net primary productivity

Explanation: i just took my test and this was correct!

This question is incomplete, the complete question is;

A high-resistance material is used as an insulator between the conductors of a length of coaxial cable. The resistance material, which forms a hollow tube, has an inner radius a and an outer radius b, and the insulator provides a resistance R between the conductors. If a second insulator, made of the same material and having the same length, is made with double both the inner radius and the outer radius of the first, what resistance would it provide between the conductors

a) (In2)R

b) 4R

c) R/(In2)

d) 2R

e) R

Answer: Option e) R is the correct answer.

Explanation:

Given that;

Inner radius = a

Outer radius = b

Conical Cylinder

∫dR = ∫(edr/2πrL)

R = e/2πL In e |ᵇₐ

R = e/2πL In(b/a) ------------- let this be equation 1

Taking a look at the second cone

a' = 2a

b' = 2a

R' = e/2πL In(2b/2a)

{L = L'}

R' = e/2πL In(b/a) -------let this be equation 2

now lets compare the two equation

R = e/2πL In(b/a)

R' = e/2πL In(b/a)

so R' = R

Therefore Option e) R is the correct answer.

Answer:

ertdhrsg

Explanation:

gesrgsg

☃ First of all we have to calculate frictional force by using below equation :

→ F = μR

→ F = 0.2 × 5 × 10

→ F = 10 N

☃ But we know that frictional force is acting on opposite direction of the motion :

⇏Net force = 15 - 10

⇏Net force = 5 N

☃ By using Newton's second law we get :

➝ Force = mass × acceleration

➝ 5 = 5 × a

➝ a = 5 ÷ 5

➝ a = 1 m/s²

Answer:

MA = 2.5

Explanation:

Given that,

Effort force applied to a screwdriver to pry the lid off of a can= 40 N

The screwdriver applies 100n of force to the lid = 100 N

We need to find the mechanical advantage of the screwdriver. Mechanical advantage of a machine is given by the ratio of load to the effort.

m = load/effort

= 100/40

= 2.5

Hence, the MA of the screwdriver is 2.5

Answer: mass remains the same

Explanation:

My teacher gave me the answer

As an astronaut travels from the surface of the earth to a position that is four times  as far away from the center of the earth, the astronaut's:

According to the given question, we can see that an astronaut is on a journey to a new location which is four times further from the center of the earth. We need to find out if there is a change to the mass of the astronaut.

As a result of this, we can see that the mass of the astronaut remains the same as he travels to a position which is four times further from the center of the earth, because the gravitational forces are the same.

Therefore, the correct answer is option B

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