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Luda [366]
1 year ago
11

A 50.0-kg person steps on a scale in an elevator. The scale reads 5000 N. What is the magnitude of the acceleration of the eleva

tor
Physics
1 answer:
zaharov [31]1 year ago
5 0

The magnitude of the acceleration of the elevator is 90m/s^2

Due to Newton's Law ∑ Forces in direction of motion is equal to mass

multiplied by the acceleration

We have here two forces 5000 N in direction of motion and the weight of the person in opposite direction of motion

The weight of the person is his mass multiplied by the acceleration of gravity

So ,

W = mg , where m is the mass and g is the acceleration of gravity

m = 50 kg and g = 9.8 m/s²

Substitute these values in the rule above

W = 50 × 9.8 = 490 N

The scale reads 5000 N

F = 5000 N , W = 490 N , m = 50kg

F - W = ma

5000 - 490 = 50 a

4510 = 50 a

Divide both sides by 50

a = 90.2 m/s²

Hence the acceleration is 90m/s^2

Learn more about Acceleration here

brainly.com/question/14344386

#SPJ4

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Can someone please illustrate how the refracted ray will look like?
andrew11 [14]

Answer

As the angle of incidence increases in Figure 2.8, a point is finally reached where the refracted ray does not emerge at the second layer but lie along the interface. This particular angle of incidence at which the angle of refraction is 90° and the refracted ray lies along the interface is known as the critical angle. At and beyond the critical angle, there is no transmitted ray and therefore a very high reflected ray will be recorded .

Therefore,

sinθisin90=Vp1Vp2

But, sin 90 = 1.

At critical angle,

sinθcritical=Vp1Vp2

A critical refracted wave travels along the interface between layers and is refracted back into the upper layer at the critical angle. The waves refracted back into the upper layer are called head waves or first-break refractions because at certain distances from a source, they are the first arriving energy. Recorded first-break refraction is shown in Figure 2.10.

Note that these first-break refractions can give us important information about the shallow velocities on land seismic data.

Note also that seismic data are acquired in such a way that reflections from horizons of interest are in the pre-critical region, even at the farthest offset in the data.

In reality, part of the seismic energy arriving at an interface is transmitted and refracted, and another part of the energy is reflected at that same interface. Given that there are many reflectors in the subsurface, there will be many paths from source to receiver, each of them with a different travel time. The proportion of energy reflected depends on the material properties of the two bounding layers and on the angle of incidence

8 0
3 years ago
The relative density of oak wood is 0.64 .find the density in the cgs system
Zepler [3.9K]
Relative density is defined as:
 dr = ds / dw
 where:
 dr: relative density
 ds: density of the substance.
 dw: density of water.
 In this case we have the relative density of oak wood:
 dr = 0.64.
 We want to find the density of the substance: ds
 Therefore we need to know the density of water in the cgs unit system:
 dw = 1g / c ^ 3
 Finally:
 ds = dr * dw
 ds = 0.64 * 1 g / cm ^ 3
 ds = 0.64 g / c ^ 3
 The density of the oak wood in cgs is 0.64 g / cm ^ 3
6 0
3 years ago
What is the potential energy of the system composed of the three charges q1, q3, and q4, when q1 is at point R? Define the poten
Ad libitum [116K]

Answer:

Incomplete question check attachment for complete question

Also it is given that

q1=-0.7uC

q4=-1.7uC

q3=-1.7uC

Also the distance are given as

a=2.2cm=0.022m

d2=3.6cm=0.036m

Explanation:

The potential energy due to point R is given as

The potential energy due to charge q1 and q3 plus the potential energy due to charge q4 and q1 plus the potential energy due to charge q3 and q4

So, let take it one after the other

Potential energy is give as

P.E=kq1q2/r

Therefore,

Potential energy due to charge q1 and q3

U¹³=kq1q3/r

To get the distance between charge q1 and q3, we will apply Pythagoras theorem

r=√(d2²+a²)

r=√(0.036²+0.022²)

r=0.0422m

k is a constant =9×10^9Nm²/C²

Then,

U¹³=kq1q3/r

U¹³=9×10^9×0.7×10^-6×1.7×10^-6/0.0422

U¹³=0.254J

Potential energy due to charge q1 and q4

U¹⁴=kq1q4/r

To get the distance between charge q1 and q4, we will apply Pythagoras theorem

r=√(d2²+a²)

r=√(0.036²+0.022²)

r=0.0422m

k is a constant =9×10^9Nm²/C²

Then,

U¹⁴=kq1q4/r

U¹⁴=9×10^9×0.7×10^-6×1.7×10^-6/0.0422

U¹⁴=0.254J

Potential energy due to charge q3 and q4

U³⁴=kq3q4/r

r=2a=2×0.022=0.044m

k is a constant =9×10^9Nm²/C²

Then,

U³⁴=kq3q4/r

U³⁴=9×10^9×1.7×10^-6×1.7×10^-6/0.044

U¹⁴=0.591J

Then, the total energy is

U= U¹³+ U¹⁴ + U³⁴

U=0.254+0.254+0.591

U=1.099J

Then also, the potential energy is zero because at infinity both U¹³ and U¹⁴ will have infinite potential because their distance apart will be infinite.

6 0
3 years ago
Which of the following accurately describes the way in which a muscle moves?
Vaselesa [24]
<h3><u>Answer</u>;</h3>

B. When actin filaments are pulled toward the center of the sarcomere, the fiber shortens.

<h3><u>Explanation;</u></h3>
  • <em><u>The events of muscle fiber shortening occurs with in the sacromeres in the fibers. </u></em>
  • <em><u>Contraction of striated muscle fibers takes place as the sacromeres shorten as myosin heads pull on the actin filaments.</u></em>
  • <em><u>Filament movement starts at the region or zone where thin and thick filaments overlap. </u></em>
  • <em><u>Myofibril contains many sacromeres along its length and thuse myofibrils and muscle cells contract as the sacromeres contract.</u></em>
7 0
2 years ago
Hurricanes are considered____ because they lose power over cool waters or land A. Short-lived B. Heat engines C. Weak
klasskru [66]
My best guess would be heat engines
7 0
3 years ago
Read 2 more answers
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