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kykrilka [37]
2 years ago
6

In a position vs. time graph depicting the motion of two different objects, the point at which the lines intersect is where the

objects:
a.
have the same position.
c.
collide.
b.
have the same speed.
d.
none of the above.
Physics
1 answer:
myrzilka [38]2 years ago
3 0
I think the correct answer from the choices listed above is option A. In a position vs. time graph depicting the motion of two different objects, the point at which the lines intersect is where the objects <span>have the same position. Hope this answers the question. Have a nice day.</span>
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Alex787 [66]
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3 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
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Answer:

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3 0
3 years ago
A mass of 3.6 kg oscillate on a horizontal spring with a spring constant of 160 N/m.
Darya [45]

Answer:

48.7 J

Explanation:

For a mass-spring system, there is a continuous conversion of energy between elastic potential energy and kinetic energy.

In particular:

- The elastic potential energy is maximum when the system is at its maximum displacement

- The kinetic energy is maximum when the system passes through the equilibrium position

Therefore, the maximum kinetic energy of the system is given by:

KE=\frac{1}{2}mv^2

where

m is the mass

v is the speed at equilibrium position

In this problem:

m = 3.6 kg

v = 5.2 m/s

Therefore, the maximum kinetic energy is:

KE=\frac{1}{2}(3.6)(5.2)^2=48.7 J

6 0
3 years ago
What is the magnetic flux density (B-field) at a distance of 0.36 m from a long, straight wire carrying a current of 3.8 A in ai
olga nikolaevna [1]

Answer:

The magnetic flux density is 2.11\times10^{-6}\ T

Explanation:

Given that,

Distance = 0.36 m

Current = 3.8 A

We need to calculate the magnetic flux density

Using formula of magnetic field

B =\dfrac{\mu_{0}I}{2r}

Where,

r = radius

I = current

Put the value into the formula

B =\dfrac{4\pi\times10^{-7}\times3.8}{2\times\pi\times0.36}

B=2.11\times10^{-6}\ T

Hence, The magnetic flux density is 2.11\times10^{-6}\ T

3 0
3 years ago
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