A spring with a mass attached to it is stretched from the rest position of 20 cm to the position of 87 cm. If the spring constant of this spring is 1729 N/m, what is the potential energy stored in this spring?

561.29 J
120.97 J
388.07 J
254.79 J

Answer:

e

Explanation:

Answer: E B

Explanation:

i think correct answer is b

Answer:

Explanation:

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:

see attached file

Explanation:

Answer:

miami dolphins.

Explanation:

......

Answer:

miami :(

Explanation:

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:

What is the question though?

Explanation:

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:

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

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:

ertdhrsg

Explanation:

gesrgsg

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:

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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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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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: 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:

False

Explanation:

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 It's easier to understand the old way, and its NOT outdated.

☃ 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

Explanation:

wavelength =200cm=2m

frequency is reciprocal of wavelength

f=1/2

f=0.5Hz or 1/s

The frequency of the wave with a Wavelength of 200 cm should be considered as the f=0.5Hz or 1/s.

Since the wavelength is 200 cm or 2m

Also we know that

frequency is reciprocal of wavelength

So,

f=1/2

f=0.5Hz

Hence, The frequency of the wave with a Wavelength of 200 cm should be considered as the f=0.5Hz or 1/s.

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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: ⇒  Answer is 3

"A small force may produce a large change in momentum by acting on a very massive object".

THEY HAVEN'T GIVEN US THE TIME PERIOD NOR THE DISTANCE TRAVELED. THEREFORE WE CANNOT ACTUALLY DECIDE IF THE FORCE IS KEPT FOR A LONG TIME OR SHORT TIME. ANYWAYS SINCE THE MASS IS GIVEN AS MASSIVE , THE  MOMENTUM SHOULD BE DEFINITELY HIGH.

WHY I SAY OTHERS ARE WRONG:

1) For a small force to give a large change in momentum, it should act for a long time interval.

2) By applying a large force for a short time interval, the change of momentum should be large.

3) Correct answer.

4) Acting over a short distance can be the same as acting over a short period of time.Therefore the distance should be large in order for a larger momentum.

I HOPE IT HELPS!

The true statement is required.

The true statement is A large force the change in momentum may be small acting over a short time interval.

Force is given by

[tex]F=ma[/tex]

m = Mass of object

a = Acceleration = [tex]\dfrac{v-u}{t}[/tex]

t = Time

So, force is

[tex]F=m\dfrac{v-u}{t}\\\Rightarrow F=\dfrac{mv-mu}{t}\\\Rightarrow F=\dfrac{\Delta p}{t}[/tex]

[tex]\Delta p[/tex] = Change in momentum.

If the time taken is small then the force will be large.

The change in momentum can also be small.

So, for a large force the change in momentum may be small acting over a short time interval.

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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:

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:

I hope the above pic helps you mate

I will always help you understanding your assingments

have a great day

#Captainpower:)

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:

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:

pressure=force/area

p=60N/(2m×1m)

=35N/M