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Viktor [21]
2 years ago
13

if a person with a mass of 70kg is standing on scale in an elevator when the cable snaps, what will the reading on the scale be

during free-fall?
Physics
1 answer:
dybincka [34]2 years ago
3 0

Answer:

The reading will be the same.

Explanation:

Mass does not depend upon anything and it remains the same anywhere. What changes is the weight of the body because it depends upon gravity and is different at different places.

Giving me the brainest will be helpful.

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Problem 1: Two sources emit waves that are coherent, in phase, and have wavelengths of 26.0 m. Do the waves interfere constructi
Anton [14]

1) Destructive interference

The condition for constructive interference to occur is:

\delta = m\lambda (1)

where

\delta =|d_1 -d_2| is the path difference, with

d_1 is the distance of the point from the first source

d_2 is the distance of the point from the second source

m is an integer number

\lambda is the wavelength

In this problem, we have

d_1 = 78.0 m\\d_2 = 143 m\\\lambda=26.0 m

So let's use eq.(1) to see if the resulting m is an integer

\delta =|78.0 m-143 m|=65 m\\m=\frac{\delta }{\lambda}=\frac{65 m}{26.0 m}=2.5

It is not an integer so constructive interference does not occur.

Let's now analyze the condition for destructive interference:

\delta = (m+\frac{1}{2})\lambda (2)

If we apply the same procedure to eq.(2), we find

m=\frac{\delta}{\lambda}-\frac{1}{2}=\frac{65.0 m}{26.0 m}-0.5=2

which is an integer: so, this point is a point of destructive interference.

2) Constructive interference

In this case we have

d_1 = 91.0 m\\d_2 =221.0 m

So the path difference is

\delta =|91.0 m-221.0 m|=130.0 m

Using the condition for constructive interference:

m=\frac{\delta }{\lambda}=\frac{130.0 m}{26.0 m}=5

Which is an integer, so this is a point of constructive interference.

3) Destructive interference

In this case we have

d_1 = 44.0 m\\d_2 =135.0 m

So the path difference is

\delta =|44.0 m-135.0 m|=91.0 m

Using the condition for constructive interference:

m=\frac{\delta }{\lambda}=\frac{91.0 m}{26.0 m}=3.5

This is not an integer, so this is not a point of constructive interference.

So let's use now the condition for destructive interference:

m=\frac{\delta}{\lambda}-\frac{1}{2}=\frac{91.0 m}{26.0 m}-0.5=3

which is an integer: so, this point is a point of destructive interference.

3 0
3 years ago
Lava that cools so quickly that ions do not have time to arrange themselves into crystals will form igneous rocks with a _____ t
Alborosie
Lava cools so quickly that ions do not have time to arrange themselves into crystals will form igneous rocks with a glassy texture. Lava is the substance that flows from the volcano's. 
8 0
3 years ago
Hydrogen is the second most abundant gas in the atmosphere? True or false?
Alenkinab [10]
False as oxygen is the second most abundant and nitrogen is the most abundant at 78%.
7 0
3 years ago
Why is the answer B?
djyliett [7]

Answer:

Explanation:

The center of mass lies on a line that joins position 4 of one start with position 4 of the other star.  The shortest distance between these two points will produce the largest velocity. You are using F = m v^2/R

Small R = large force.

Large Force = increased speed.

The masses don't have any effect on the outcome: they remain constant.

7 0
3 years ago
In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.
vaieri [72.5K]

Answer:

(A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

Explanation:

Given that,

A nucleus undergoes a nuclear decay.

(A). In alpha decay,

We know that,

When the nucleus emit alpha particle then atomic mass of particle reduce by 4 and atomic number reduce by 2.

We need to calculate the resulting nucleus

Using given data

^{227}_{89}Ac\Rightarrow ^{227-4}_{89-2}X

^{227}_{89}Ac\Rightarrow ^{223}_{87}Fr

The resulting nucleus is Fr.

(B). In beta-minus decay,

We know that,

When the nucleus emit beta- minus particle then atomic mass of particle is same and atomic number increase by 1.

We need to calculate the resulting nucleus

Using given data

^{211}_{83}Bi\Rightarrow ^{211}_{83+1}X

^{211}_{83}Bi\Rightarrow ^{211}_{84}Po

The resulting nucleus is Po.

(C). In beta-plus decay,

We know that,

When the nucleus emit beta- plus particle then atomic mass of particle is same and atomic number decrease by 1.

We need to calculate the resulting nucleus

Using given data

^{22}_{11}Na\Rightarrow ^{22}_{11-1}X

^{22}_{11}Na\Rightarrow ^{22}_{10}Ne

The resulting nucleus is Ne.

(D). In gamma decay,

We know that,

When the nucleus emit gamma particle then atomic mass and atomic number of particle is same.

We need to calculate the resulting nucleus

Using given data

^{98}_{43}Tc\Rightarrow ^{98}_{43}Tc

The resulting nucleus is Tc.

Hence, (A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

4 0
3 years ago
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