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4 years ago

If the spring constant is doubled, what value does the period have for a mass on a spring?

2 answers:

7 0

If spring constant is doubled, the mass on spring will be doubled as well. according to this formula, F=ke
k stands for spring constant and e stands for the length extended

Nat2105 [25]4 years ago

5 0

Answer:

T' = T / sqrt 2

Explanation:

According to the formula of time period of a loaded spring

$T = 2\pi \sqrt{\frac{m}{k}}$ .... (1)

where, m is the mass on the spring and k be the spring constant.

Here we observe that the time period is inversely proportional to the square root of the spring constant.

If the spring constant is doubled then the time period is

$T' = 2\pi \sqrt{\frac{m}{2k}}$ ....(2)

Dividing equation (2) by equation (1), we get

T' = T / sqrt 2

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How much carbon dioxide (co2) does natural gas emit compared with other fossil fuel

amid [387]

Natural gas is a fossil fuel, though the global warming emissions from its combustion are much lower than those from coal or oil. Natural gas emits 50 to 60 percent less carbon dioxide (CO2) when combusted in a new, efficient natural gaspower plant compared with emissions from a typical new coal plant.

5 0

4 years ago

A bicyclist notes that the pedal sprocket has a radius of rp = 9.5 cm while the wheel sprocket has a radius of rw = 4.5 cm. The

ANTONII [103]

Answer:

Explanation:

a) ωp = 2π radians / 1.7 s = 3.7 rad/s

b) ωs = 3.7 rad/s(9.5 cm / 4.5 cm) = 7.8 rad/s

v = (ωs)R = 7.8(65) = 507 cm/s or 5.1 m/s

c) ωs = 3.5 m/s / 0.65 m = 5.38 rad/s

ωp = 5.38(4.5 cm / 9.5 cm) = 2.55 rad/s

t = θ/ω = 2π / 2.55 = 2.463... 2.5 s

4 0

3 years ago

A force of 100. newtons is used to move an object a distance of 15 meters with a power of 25 watts. Find the time it takes to do

Flauer [41]

<h3>It takes 60 seconds to do the work</h3>

Solution:

Given that,

Force = 100 newtons

Distance = 15 meters

Power = 25 watts

To find: time it takes to do the work

Find the work done:

$work = force \times displacement\\\\work = 100\ newtons \times 15\ meters\\\\work = 1500\ joule$

Find the time taken

$power = \frac{work}{time}\\\\25\ watts = \frac{1500\ joule}{time}\\\\time = \frac{1500\ joule}{25\ watt}\\\\time = 60\ second$

Thus it takes 60 seconds to do the work

3 0

3 years ago

The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1

telo118 [61]

<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life $h$ of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.

In this case, we are given the half life of two elements:

beryllium-13: $h_{B-13}=5(10)^{-10}s=0.0000000005s$

beryllium-15: $h_{B-15}=2(10)^{-7}s=0.0000002s$

As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?

We can find it out by the following expression:

$h_{B-15}=X.h_{B-13}$

Where $X$ is the amount we want to find:

$X=\frac{h_{B-15}}{h_{B-13}}$

$X=\frac{2(10)^{-7}s}{5(10)^{-10}s}$

Finally:

$X=400$

Therefore:

The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.

8 0

3 years ago

If a scuba diver fills his lungs to full capacity of 5.5 L when 10 m below the surface, to what volume would his lungs expand if

umka21 [38]

Answer:

The volume at the surface is 10.97 L.

Explanation:

Given that,

Volume = 5.5 L

Height = 10 m

Density of sea water= 1025 kg/m³

We need to calculate the pressure at that point

Using formula of pressure

$P'=P+\rho gh$

Put the value into the formula

$P'=1.01\times10^{5}+1025\times9.8\times10$

$P'=201450\ Pa$

We need to calculate the volume at the surface

Using equation of ideal gas

$PV= RT$

So, for both condition

$PV=P'V'$

Put the value into the formula

$V=\dfrac{201450\times5.5}{1.01\times10^{5}}$

$V=10.97\ L$

Hence, The volume at the surface is 10.97 L.

3 0

3 years ago

Read 2 more answers