answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
VikaD [51]
2 years ago
14

A wagon full of manure accidentally rolls down a driveway for 5.0m while a person pushes against the wagon with a force of 420 N

to bring it to a stop. What is the change in kinetic energy for the wagon?
Physics
2 answers:
Cerrena [4.2K]2 years ago
8 0

Answer:

2100 J

Explanation:

Parameters given:

Force acting on the object, F = 420 N

Distance moved by object, d = 5m

The change in kinetic energy of an object is equal to the work done by a force acting on the object:

W = F * d

∆KE = F * d

∆KE = 420 * 5

∆KE = 2100 J

Studentka2010 [4]2 years ago
6 0

Answer:

∆KE = 2100 J

Explanation:

Given:

Force of person pushes against the wagon=  F = 420 N

Distance = d = 5.0 m

When you accelerate an object, you are doing work against inertia. It is proven that work equals the change in kinetic energy of the object.

therefore, W = ΔKE

where,

W is the work done against the resistance of inertia

ΔKE is the change in kinetic energy (Δ is Greek letter capital delta)

Also, W = F x d

∴∆KE = F x d

∆KE = 420 x 5

∆KE = 2100 J

The change in kinetic energy for the wagon is 2100 J

You might be interested in
One nucleus contains 31 protons and 40 neutrons another nucleus contains 31 protons and 41 neutrons what can you conclude about
Pavel [41]

They are isotopes of the same element.

5 0
2 years ago
Read 2 more answers
A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c
netineya [11]

Answer:

E=\frac{\lambda}{2\pi r\epsilon_0}

Explanation:

We are given that

Linear charge density of wire=\lambda

Radius of hollow cylinder=R

Net linear charge density of cylinder=2\lambda

We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R

By Gauss theorem

\oint E.dS=\frac{q}{\epsilon_0}

q=\lambda L

E(2\pi rL)=\frac{L\lambda}{\epsilon_0}

Where surface area of cylinder=2\pi rL

E=\frac{\lambda}{2\pi r\epsilon_0}

8 0
2 years ago
A wave travels through a medium at 251 m/s and has a wavelength of 5.10 cm. What is its frequency? What is its angular frequency
allochka39001 [22]

Explanation:

It is given that,

Speed of a wave, v = 251 m/s

Wavelength of the wave, λ = 5.1 cm = 0.051 m

(1) The frequency of the wave is given by :

\nu=\dfrac{v}{\lambda}

\nu=\dfrac{251\ m/s}{0.051\ m}

\nu=4921.56\ Hz

(2) Angular frequency of the wave is given by :

\omega=2\pi\nu

\omega=2\pi\times 4921.56\ Hz

\omega=30923.07\ rad/s

(3) The period of oscillation is given by T as :

T=\dfrac{1}{\nu}

T=\dfrac{1}{4921.56}

T = 0.000203 seconds

or

T = 0.203 milliseconds

Hence, this is the required solution.

5 0
2 years ago
A 94-ft3/s water jet is moving in the positive x-direction at 18 ft/s. The stream hits a stationary splitter, such that half of
vitfil [10]

Answer:

FR<em>x  </em>= 960.37 lbf   (←)

FR<em>z </em>= 0 lbf

Explanation:

Given:

Q = 94 ft³/s

vx = 18 ft/s

ρ = 62.4 lbm/ft³

∅ = 45°

<em>Assumptions: </em>

1. The flow is steady and incompressible.

2 . The water jet is exposed to the atmosphere, and thus the  pressure of the water jet before and after the split is the  atmospheric pressure which is disregarded since it acts on all  surfaces.

3. The gravitational effects are disregarded.

4. The  flow is nearly uniform at all cross sections, and thus the effect  of the momentum-flux correction factor is negligible, β ≅ 1.

<em>Properties:</em> We take the density of water to be ρ = 62.4 lbm/ft³

Analysis: The mass flow rate of water jet is

M = ρ*Q = (62.4 lbm/ft³ )(94 ft³/s) = 5865.6 lbm/s

We take the splitting section of water jet, including the splitter as the control volume, and designate the entrance by 1 and  the outlet of either arm by 2 (both arms have the same velocity and mass flow rate <em>M</em>). We also designate the horizontal  coordinate by x with the direction of flow as being the positive direction and the vertical coordinate by z.

The momentum equation for steady flow is

∑ F = ∑ (β*M*v) <em>out</em> - ∑ (β*M*v) <em>in</em>

We let the x- and y- components of the  anchoring force of the splitter be FR<em>x</em> and FR<em>z,  </em>and assume them to be in the positive directions. Noting that

v₂ = v₁ = v  and  M₂ = (1/2) M, the momentum equations along the x and z axes become

FR<em>x </em>= 2*(1/2) M*v₂*Cos ∅ - M*v₁ = M*v*(Cos ∅ - 1)

FR<em>z </em>= (1/2) M*(v₂*Sin ∅) + (1/2) M*(-v₂*Sin ∅) = 0

Substituting the given values,

FR<em>x </em>= (5865.6 lbm/s)*(18 ft/s)*(Cos (45°) - 1)(1 lbf / 32.2 lbm*ft/s²)

⇒  FR<em>x  </em>= - 960.37 lbf

FR<em>z </em>= 0 lbf

The negative value for FR<em>x</em> indicates the assumed direction is wrong, and should be reversed. Therefore, a force of 960.37 lbf  must be applied to the splitter in the opposite direction to flow to hold it in place. No holding force is necessary in the  vertical direction. This can also be concluded from the symmetry.

In reality, the gravitational effects will cause the upper stream to slow down and the lower stream to speed  up after the split. But for short distances, these effects are negligible.  

3 0
2 years ago
The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational
s344n2d4d5 [400]

Answer:

v = √2G M_{earth} / R

Explanation:

For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)

        Eo = K + U = ½ m1 v² - G m1 m2 / r1

        Ef = - G m1 m2 / r2

When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf

       Eo = Ef

       ½ m1v² - G m1 M_{earth} / R = - G m1 M_{earth} / R

      v² = 2G M_{earth} (1 / R - 1 / Rinf)

If we do Rinf = infinity     1 / Rinf = 0

       v = √2G M_{earth} / R

      Ef = = - G m1 m2 / R

The mechanical energy is conserved  

 

      Em = -G m1  M_{earth} / R

      Em = - G m1  M_{earth} / R

     R = int        ⇒  Em = 0

6 0
1 year ago
Other questions:
  • A floating balloon can be formed when the substance helium is released from a compressed container into a flat rubber balloon. T
    9·2 answers
  • Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele
    13·1 answer
  • How many neutrons are contained in 2 kg? Mass of one neutron is 1.67x10-27 kg.
    5·1 answer
  • If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed?
    12·1 answer
  • Assume that the cart is free to roll without friction and that the coefficient of static friction between the block and the cart
    15·2 answers
  • A solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of the cylinder is 3.2 m and its rad
    13·1 answer
  • The researcher requires the force on the wire to point upward with a magnitude of 4.2x10-4 N. The length of the wire that is in
    12·1 answer
  • 16) A wheel of moment of inertia of 5.00 kg-m2 starts from rest and accelerates under a constant torque of 3.00 N-m for 8.00 s.
    15·2 answers
  • The grooved pulley of mass m is acted on by a constant force F through a cable which is wrapped securely around the exterior of
    13·1 answer
  • A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!