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Flura [38]
2 years ago
5

An object of mass 100kg is raised 2m above the ground using an Inclined Plane of length 10m calculate the effort parallel to inc

lined plane (g=9.8m/s2
Physics
1 answer:
MArishka [77]2 years ago
8 0

Answer:

Slope = 2 m / 10 m = 1/5

For every  5 m of effort the object will be raised 1 m

W = work done on object = M g h      increase in PE of object

E S = W      where E is effort and S the distance thru which the effort acts

E S = M g H

E = 100 kg * 9.8 m/s^2 * 2 m / 10 m = 196 kg m / s^2 = 196 N

Check: total work = 2 * 9.8 * 100 = 1960 J

Force Needed = 1960 J / 2 m = 980 Newtons

Mechanical advantage = 980 / 196 = 5   as one would expect since the object is raised 1 m for every 5 m of force input

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Two point sources are vibrating in phase producing two-dimensional water wave interference. The first antinodal line on either s
Viktor [21]

For all antinodal positions we know that

\Delta \phi = 2N\pi

now we also know the relation between phase difference and path difference

\Delta \phi = \frac{2\pi}{\lambda}\Delta x

now we will have

2N\pi = \frac{2\pi}{\lambda}\Delta x

now from above equation we will have

\Delta x = N\lambda

now for the first anti node position

N = 1

\Delta x = \lambda

Option D is correct

4 0
3 years ago
A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross
Sav [38]

Answer:

The energy that the truck lose to air resistance per hour is 87.47MJ

Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

F_D=\frac{1}{2}\rhoC_dAV^2

Our values are:

V=30.1m/s

C_d=0.45

A=3.3m^2

\rho=1.2kg/m^3

Replacing,

F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2

F_D=807.25N

We need calculate now the energy lost through a time T, then,

W = F_D d

But we know that d is equal to

d=vt

Where

v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

W=(807.25N)(30.1m/s)(3600s/1hr)

W=87.47*10^6J (per hour)

Therefore the energy that the truck lose to air resistance per hour is 87.47MJ

4 0
3 years ago
6. A baseball player hits a 0.155-kilogram fastball traveling at -44.0 m/s into center field at a speed of 50.0
Zolol [24]

Answer:

3,237.78N

Explanation:

According to Newton's second law of motion

F = mass * acceleration

Since a = v-u/t

F = m(v-u)/t

Given

Mass m = 0.155kg

v = 50.0m/s

u = -44.0m/s

Time t = 0.00450secs

Substitute

F = 0.155(50-(-44))/0.00450

F = 0.155(50+44)/0.00450

F = 0.155(94)/0.00450

F = 14.57/0.00450

F = 3,237.78N

hence he hit the baseball with a force of 3,237.78N

5 0
3 years ago
Here is a graph of speed vs time. If the object is moving to the east, which BEST describes the speed and velocity of the graph?
san4es73 [151]

Answer: A JUST DID IT ON USA TESTPREP

Explanation:

5 0
3 years ago
I need answers for this wave unit
Svetllana [295]

Answer:

a

Explanation:

6 0
3 years ago
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