Steve pushes a crate 20 m across a level floor at a constant speed with a force of 200 N, this time on a frictionless floor. The velocity of the crate is in the direction of the force Steve is applying to the crate. What is the net work done on the crate

Answer:

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Answer:

b is correct.

Explanation:

because of the question you have given

Answer:

The acceleration is a negative as the ball is now moving in the opposite direction.

The velocity would decrease as the ball moves upward

Acceleration remains constant and velocity is negative and decreasing as per the given scenario. The correct option is B.

In mechanics, acceleration is defined as the rate of change of an object's velocity with respect to time.

Vector quantities are accelerations. The orientation of an object's acceleration is determined by the orientation of its net force.

Velocity is the directional speed of a moving object as an indication of its rate of change in position as observed from a specific frame of reference and measured by a specific time standard.

The rate of displacement of an entity known as its velocity. It is measured in meters per second. The rate of change in velocity of an object is defined as acceleration.

According to the scenario, acceleration remains constant while velocity decreases and is negative.

Thus, the correct option is B.

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Your question seems incomplete, the missing options are:

Answer:

Mm, thats the answer trust me men

Answer: Hmmm im not sure but i'd go with 3.80x10^5

Explanation: Like i mentioned im not very good at physics...sorry if its wrong

Answer:

A.

sửa cho tôi nếu tôi sai

Answer:

the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

Explanation:

Given the data in the question;

initial velocity; u = 0 m/s

height; h = 2.5 m

we find the velocity of the ball just before it touches the foam.

using the equation of motion;

v² = u² + 2gh

we know that acceleration due gravity g = 9.81 m/s²

so we substitute

v² = ( 0 )² + ( 2 × 9.81 × 2.5 )

v² = 49.05

v = √49.05

v = 7.00357 m/s

Now as the ball touches the foam

final velocity v₀ = 0 m/s

compresses S = 3 cm = 0.03 m

so

v₀² = v² + 2as

we substitute

( 0 )² = 49.05 + 0.06a

0.06a = -49.05

a = -49.05 / 0.06

a = -817.5 m/s²

Therefore, the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

Answer:

below

Explanation:

Part A) This engine works  per cycle is 254.9 J.

Part B) The thermal efficiency is  23.42%

The thermal efficiency of any heat engine is represented in percentage of heat energy converted into work.

For isothermal expansion, work done is

W₁ =nRT₁ x ln(V₂/V₁)

W₁ = 0.1 x 8.314 x 800 x ln(5000/3000)

W₁ = 339.8 J =Q₁

For isochoric cooling ,

W₂ =0

Q₂ =nCvdT = 0.1 x 3R/2 x (T₂-T₁)

Q₂ = -748.3 J

For isothermal compression,

W₃ =nRT₂ ln (V₄/V₃)

W₃  = 0.1 x 8.314 x 200 x ln(3000/5000)

W₃ = -84.9J

For isochoric heating

W₄ =0

Q₄ =nCvdT = 0.1 x 3R/2 x (800-200)

Q₄  = -748.3 J

Total work done in all the process W = W₁ +W₂ +W₃ +W₄

W =254.9 J

Thus, the work done is 254.9 J

Thermal efficiency = Work done/Heat taken

η = W/ Q₁ +Q₄

η = [254.9 / 339.8 +748.3 ] x 100 %

η = 0.2342 x 100 %

η = 23.42%

Thus, the thermal efficiency is 23.42%

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Explanation:

1. Distance, d = 5 m

Time, t = 10 s

Speed = distance/time

[tex]v=\dfrac{5}{10}=0.5\ m/s[/tex]

2. Mass, m = 10 kg

Acceleration, a = 3 m/s³

Force, F = mass (m) × acceleration (a)

F = 10 × 3

= 20 N

3. Mass, m = 7 kg

Height, h = 4 m

Potential energy, E = mgh

E = 7 × 9.8 × 4

E = 274.4 J

Hence, this is the required solution.

Answer:

ΔV = 2 10¹ V

Explanation:

The calculation of the uncertainty or error in an expression is given by

         ΔV = [tex]\frac{dV}{di}[/tex]  |Δi| + [tex]\frac{dV}{dR}[/tex]  |ΔR |

         V = i R

let's make the derivatives

        [tex]\frac{dV}{di}[/tex] = R

        [tex]\frac{dV}{dR}[/tex] = i

we substitute

         ΔV = R | Δi | + i | ΔR |

in the exercise give the values

         i = (5.9 ± 0.4) A

         R = (42.7 ± 0.6) Ω

we calculate

          ΔV = 42.7  0.4 + 5.9  0.6

          ΔV = 20.6 V

          ΔV = 2 10¹ V

the voltage is

         V = i R

         V = 5.9  42.7

          V = 251.9 V

the result is

         V = (25 ± 2) 10¹ V

Explanation:

Given that,

Two resistors of 10 ohms and 15 ohms are connected.

In series combination, the equivalent resistance is given by :

[tex]R_s=R_1+R_2\\\\R_s=10+15\\\\R_s=25\ \Omega[/tex]

In parallel combination, the equivalent resistance is given by :

[tex]\dfrac{1}{R_p}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\\dfrac{1}{R_p}=\dfrac{1}{10}+\dfrac{1}{15}\\\\R_p=6\ \Omega[/tex]

Hence, this is the required solution.

Answer:

The electric field is given by 4.5 N/C.

Explanation:

Charge density = 80 nC/m3

inner radius, r' = 1 mm

outer radius, r'' = 3 mm

distance,  r = 4 mm

The linear charge density is given by

[tex]\lambda =\rho \times\pi\times (r''^2 - r'^2)\\\\\lambda = 80\times 10^{-9}\times 3.14\times 10^{-6}\times(9-1)\\\\\lambda = 2\times 10^{-12}\\[/tex]

The electric field is given by

[tex]E = \frac{\lambda }{4\pi\varepsilon_or}\\E=\frac{9\times 10^9\times 2 \times 10^{-12}}{0.004}\\\\E=4.5 N/C[/tex]

Answer:

[tex]d=79.9m[/tex]

Explanation:

From the question we are told that:

coefficient of static friction [tex]\mu=0.38[/tex]

Velocity [tex]v=87.9=>24.41667m/s[/tex]

Generally the equation for Conservation of energy is mathematically given by

 [tex]\mu*mgd = 0.5 m v^2[/tex]

 [tex]d=\frac{0.5*24.42^2}{0.38*9.8}[/tex]

 [tex]d=79.9m[/tex]

Answer:

Option B

Explanation:

A gaseous, highly reactive non-metal

Answer:B

Explanation:I just took the test

It is Potential energy's

Answer:

because density of iron is more than that of water but less then that of Mercury

hope it's helpful

Answer:

wind

Explanation:

just a possible answer.

Answer:

-30 N/C

Explanation:

Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m

Now ΔV = final potential - initial potential = 1.2 V - 0.90 V = 0.30 V

Since E = -ΔV/Δx

substituting the values of the variables into the equation, we have

E = -ΔV/Δx

E = -0.30 V/0.01 m

E = -30 V/m

Since 1 V/m = 1 N/C.

E = -30 N/C

So, the average electric field is -30 N/C

Explanation:

The initial kinetic energy [tex]KE_0[/tex] for both blocks is zero. Let [tex]m_1= m[/tex] and [tex]m_2 =3m[/tex]. So using the conservation law of linear momentum, we can write

[tex]0 = m_1v_1 - m_2v_2[/tex]

or

[tex]v_2 = \dfrac{m_1}{m_2}v_1 = \dfrac{m}{3m}v_1 = \dfrac{1}{3}v_1[/tex]

The final kinetic energies for the two masses are

[tex]KE_1 = \frac{1}{2}m_1v_1^2 = \frac{1}{2}mv_1^2[/tex]

[tex]KE_2 = \frac{1}{2}m_2v_2^2 = \frac{1}{2}(3m)(\frac{1}{3}v_1)^2 = \frac{1}{2}m(\frac{1}{3}v_1^2)[/tex]

Therefore, the ratio of their kinetic energies is

[tex]\dfrac{\Delta KE_2}{\Delta KE_1} = \dfrac{\frac{1}{2}(\frac{1}{3}v_1^2)}{\frac{1}{2}v_1^2} = \dfrac{1}{3}[/tex]

Answer:

arge number of electrons free to move between the charged ions in the lattice.

Explanation:

The metallic bond occurs when an atom with few electrons is united in its last level, therefore the best way to decrease the total energy of the system is to lose all its electrons to remain with the configuration of a noble gas. The electrons that it loses cannot be acquired by other atoms since they all have few electrons, thus leaving a large number of electrons free to move between the charged ions in the lattice.

Some important characteristics emerge from this description of the metallic bond:

* It has many free electrons therefore its electrical conductivity is high

* As the charged ions are fixed, the material can be malleable, bent without breaking since the free electrons create the bond that keeps the system stable.

* As the electrons are free when heating a part of the material, these electrons acquire energy and rapidly propagate it to the other side, giving a high thermal conductivity

* As the temperature increases, the electrons acquire more kinetic energy, which is why there are more collisions between them and consequently the resistivity of the material increases.

Answer:

B

Explanation:

The liquid evaporates in the solid is left in the dish..

Answer:

Explanation:

Here's what we know and in which dimension:

y dimension:

[tex]v_0=30[/tex] m/s

v = 0 (I'll get to that injust a second)

a = -9.8 m/s/s

The final velocity of 0 is important because that's the velocity of the ball right at the very top of its travels. If we knew how long it takes to get to that max height, we can also use that to find out how long it will take to hit the ground. Therefore, we will find the time it takes to reach its max height and pick up with the investigation of what this means after.

x dimension:

Δx = 70 m

v = ??

Velocity is our unknown.

Solving for the time in the y dimension:

[tex]v=v_0+at[/tex] and filling in:

0 = 30 + (-9.8)t and

-30 = -9.8t so

t = 3.1 seconds

We know it takes 3.1 seconds to get to its max height. In order to determine how long it will take to hit the ground, just double the time. Therefore, it will take 6.2 seconds for the ball to come back to the ground, which is where the persom trying to catch the ball comes in. We will use that time in our x dimension now.

In the x dimension, the equation we need is just a glorified d = rt equation since the acceleration in this dimension is 0.

Δx = vt and

70 = v(6.2) so

v = 11.3 m/s

Answer: The statement [tex]T_{1} > T_{2}[/tex] must be true.

Explanation:

As it is given that heat is being transferred from a substance with temperature [tex]T_{1}[/tex] to a substance with temperature [tex]T_{2}[/tex].

It is known that heat will always being transferred from a higher temperature to a lower temperature. Because at higher temperature the molecules of a substance acquire more energy and when they lose energy then a decrease in temperature occurs.

Hence, in the given situation [tex]T_{1} > T_{2}[/tex].

Thus, we can conclude that the statement [tex]T_{1} > T_{2}[/tex] must be true.

Answer:

(a) 81.54 N

(b) 570.75 J

(c) - 570.75 J

(d) 0 J, 0 J

(e) 0 J  

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

(c) Work done by the friction

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J  

Answer:

B

Explanation:

B

Answer:

4.60 × 10⁻⁸

Explanation:

From the given information;

Assuming that q charges are transferred, then:

[tex]F = \dfrac{kq^2}{d^2}[/tex]

where;

k = 9 ×10⁹

[tex]900000 = \dfrac{9*10^9 \times q^2}{1.2^2}[/tex]

[tex]q = \sqrt{\dfrac{900000\times 1.2^2 }{9*10^9}}[/tex]

q = 0.012 C

No of the electrons transferred is:

[tex]= \dfrac{0.012}{1.6\times 10^{-19}} C[/tex]

[tex]= 7.5 \times 10^{16} \ C[/tex]

Initial number of electrons =  N × 47 × no  of moles

here;

[tex]\text{ no of moles }= \dfrac{6.2}{107.87}[/tex]

no of moles = 0.0575 mol

∴

Initial number of electrons =  [tex]6.023\times 10^{23} \times 47 \times 0.0575 mol[/tex]

= 1.63 × 10²⁴

The fraction of electrons transferred  [tex]=\dfrac{7.5\times 10^{16} }{1.6 3\times 10^{24}}[/tex]

= 4.60 × 10⁻⁸

Answer:

Loudness of the second sound is more than the first one.

Explanation:

There are two sounds, the second sound is identical to first but the loudness of second is more than the first one.

As the frequency is same so the itch is same for both the sounds.

As the loudness depends on the amplitude of the sound so the loudness of the second sound is more than the first sound.