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vampirchik [111]
3 years ago
14

A child pushes horizontally on a box of mass m which moves with constant speed v across a horizontal floor. The coefficient of f

riction between the box and the floor is μ. At what rate does the child do work on the box?
Physics
1 answer:
dexar [7]3 years ago
4 0

Answer:

Explanation:

Given

mass of box is m

speed of box is v

coefficient of friction \mu =\frac{1}{4}=0.25

Box is moving at a constant speed i.e. Net Force on box is zero

i.e. Frictional Force and Force Applied are Equal

F=f_r=\mu mg

The Rate at which power is generated is

P=F\cdot v

P=\mu mg\cdot v

P=0.25mgv

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A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle
xz_007 [3.2K]

Answer:

a) \Delta{t} = 5.39s

b) the motorcycle travels 155 m

Explanation:

Let t_2-t_1 = \Delta{t}, then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}

where:

v_{m2} is the speed of the motorcycle at time 2

v_{c} is the velocity of the car (constant)

v_{0} is the velocity of the car and the motorcycle at time 1

d is the distance between the car and the motorcycle at time 1

x is the distance traveled by the car between time 1 and time 2

Solving the system of equations:

\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]

v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933

3 0
3 years ago
Two equal magnitude electric charges are separated by a distance d. The electric potential at the midpoint between these two cha
Ann [662]

Answer:

The charges under study are of the same sign

The calculation of the electric field for each charge separately, there is no relationship between the charges

Explanation:

Let's start by writing the equation for the electric field

          E = k q / r²

where q is the charge under analysis and r the distance from this charge to a positive test charge.

When analyzing the statement the student has some problems.

* The charges under study are of the same sign, it does not matter if positive or negative.

* The calculation of the electric field for each charge separately, there is no relationship between the charges for the calculation of the electric field.

* What is added is the interaction of the electric field with the positive test charge, in this case each field has the opposite direction to the other, so the vector sum gives zero

8 0
3 years ago
Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawate
LenKa [72]

Answer:

12.64968 Hz

Explanation:

v = Velocity of sound in seawater = 1522 m/s

u = Velocity of dolphin = 7.2 m/s

f' = Actual frequency = 2674 Hz

From Doppler effect we get the relation

f=f'\frac{v-u}{v}\\\Rightarrow f=2674\frac{1522-7.2}{1522}\\\Rightarrow f=2661.35032\ Hz

The frequency that will be received is 2661.35032 Hz

The difference in the frequency will be

2674-2661.35032=12.64968\ Hz

6 0
4 years ago
Physically inactive (sitting down a lot during free time)<br>​
alekssr [168]
Yes, that’s physically inactive. You should be moving your arms and legs if you’re going to be sitting down.
6 0
3 years ago
Name THREE things that impact a collision.
Lapatulllka [165]

Answer:

The three types of impact that occur are those involving the vehicle, the body of the vehicle occupant, and the organs within the body of the occupant.

4 0
3 years ago
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