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zavuch27 [327]
3 years ago
7

An example of a single displacement reaction is

Physics
1 answer:
Elden [556K]3 years ago
4 0
Single replacement would be represented by a single element being replaced.

This is shown in answer choice B
Where the positions of A and B are swapped
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The mass of an electron is ____. A higher than the mass of the proton or the neutron
Harrizon [31]
Mass of an electron = 9.110 x 10⁻³¹ kg.
Mass of a proton = 1.6727 x 10⁻²⁷ kg

∴ mass of a proton/mass of an electron = 1.6727 x 10⁻²⁷ kg/9.110 x 10⁻³¹ kg.
                                                               ~1836

∴ mass of a proton = 1836 x mass of an electron.
∴ mass of an electron is insignificant to the mass of an atom.

∴mass of an atom = mass of protons + mass of neutrons
5 0
3 years ago
Read 2 more answers
Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels
True [87]

Answer:

No.

Explanation:

Given that Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels. Note that the speedometer in a car is calibrated based on the tire's diameter and on the distance the tire covers in each revolution. (a) Will the reading of the speedometer change ?

Considering the formula

V = wr

Where

V = linear speed

W = angular speed

r = radius of the wheel.

But W = 2πrf

Where the the 2 and pi are constant. The radius of the first wheel will be small but counter balance with the larger frequency.

While the radius of the second wheel may be large but it will be of a small frequency.

We can therefore conclude that the reading on the speedometer will not change. Because speedometer will read the linear speed V.

7 0
2 years ago
Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig
HACTEHA [7]

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

y=-\frac{1}{2} gt^2+v_0t+y_0

With the given values in the question the equation has one unknown v₀:

v_0=\frac{y-y_0}{t}+\frac{1}{2}gt

Solving for t=1:

1) v_0=y-y_0+\frac{g}{2}

To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

2) E=\frac{1}{2}m{v_0}^2+mgy_0

If h is the height of the tower, the energy on top of the tower:

3) E=mgh

Combining equation 2 and 3 and solving for h:

4) h=\frac{{v_0}^2}{2g}+y_0

Combining equation 1 and 4:

h=\frac{{(y-y_0+\frac{g}{2}})^2}{2g}+y_0

4 0
3 years ago
If you live in a very cold area, you may have seen the depth of a bank of snow shrink even though temperatures remain below the
Aleks04 [339]
<span>To answer the question above, if the day sky is clear it collects radiation. If air is dry snow sublimates faster. If both cases overlap it disappears faster where ever coldest. Thank you for posting your question here. I hope my answer helps. </span>
7 0
3 years ago
To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(
DerKrebs [107]

Answer:

0.0549 m

Explanation:

Given that

equation y(x,t)=Acos(kx−ωt)

speed  v = 8.5 m/s

amplitude A = 5.5*10^−2 m

wavelength λ   = 0.5 m

transverse displacement = ?

v = angular frequency / wave number

and

wave number = 2π/ λ

wave number =  2 * 3.142 / 0.5

wave number = 12.568

angular frequency = v k

angular frequency = 8.5 * 12.568

angular frequency = 106.828 rad/sec ~= 107 rad/sec

so

equation y(x,t)=Acos(kx−ωt)

y(x,t)= 5.5*10^−2 cos(12.568 x−107t)

when x =0 and and t = 0

maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))

maximum y(x,t)= 5.5*10^−2  m

and when x =  x = 1.52 m and t = 0.150 s

y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )

y(x,t)= 5.5*10^−2 × (0.9986)

y(x,t) = 0.0549 m

so the transverse displacement is  0.0549 m

5 0
3 years ago
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