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3 years ago

Help I'm desperate and need help ASAP I will give brainliest to whoever gets it correct.

Physics

2 answers:

Gennadij [26K]3 years ago

6 0

Answer:

Explanation:

Two or more identical waves moving in opposite directions

frez [133]3 years ago

6 0

the answer is C

Thanks for correcting me!

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A hollow plastic sphere is held below the surface of a fresh-water lake by a cord anchored to the bottom of the lake. The sphere

Naya [18.7K]

Answer: a) B = 6811N

b) m = 603.2kg

c) 86.8%

Explanation: Buoyant force is a force a fluid exerts on a submerged object.

It can be calculated as:

$B=\rho_{fluid}.V_{obj}.g$

where:

$\rho_{fluid}$ is density of the fluid the object is in;

$V_{obj}$ is volume of the object;

g is acceleration due to gravity, is constant and equals 9.8m/s²

a) For the hollow plastic sphere, density of water is 1000kg/m³:

$B=10^{3}.0.695.9.8$

B = 6811N

b) Anchored to the bottom, the forces acting on the sphere are Buoyant, Tension and Force due to gravity:

B = T + $F_{g}$

B = T + mg

mg = B - T

$m=\frac{B - T}{g}$

Calculating:

$m=\frac{6811 - 900}{9.8}$

m = 603.2kg

c) When the shpere comes to rest on the surface of the water, there are only buoyant and gravity acting on it:

B = m.g

$\rho_{w}.V_{sub}.g=m.g$

$V_{sub}=\frac{m}{\rho_{w}}$

$V_{sub}=\frac{603.2}{1000}$

$V_{sub}$ = 0.6032m³

Fraction of the submerged volume is:

$\frac{V_{sub}}{V_{obj}}$ = $\frac{0.6932}{0.695}$ = 0.868

3 0

3 years ago

When does the air parcel stop losing energy?

djverab [1.8K]

Answer:

If there is no cloud (liquid water) in the parcel

Explanation:

5 0

3 years ago

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A straight wire of length 0.56 m carries a conventional current of 0.4 amperes. What is the magnitude of the magnetic field made

djverab [1.8K]

Answer:

The magnitude of the magnetic field made by current in the wire is 3.064 x 10⁻⁶ T.

Explanation:

Given;

length of the straight wire, L = 0.56 m

conventional current, I = 0.4 A

distance of magnetic field from the wire, r = 2.6 cm = 0.026 m

To determine magnitude of magnetic field made by current in the wire, we will apply Bio-Savart Law;

$B = \frac{\mu_o}{4\pi r} \frac{LI}{\sqrt{r^2 +(L/2)^2} } \\\\B = \frac{4\pi *10^{-7}}{4\pi *0.026} \frac{0.56*0.4}{\sqrt{(0.026)^2 +(0.56/2)^2} }\\\\B = 3.846*10^{-6}(0.7966)\\\\B = 3.064*10^{-6} \ T$

Therefore, the magnitude of the magnetic field made by current in the wire is 3.064 x 10⁻⁶ T.

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3 years ago

A young man heaves a 6.5 kg rock at a velocity of 6.9 m/s. What is the kinetic energy of the rock?

earnstyle [38]

Answer:

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2 years ago

Given the following lens combination:

natulia [17]

Given:

Lens.........diameter ...fl#

eyepiece...2cm............5

objective...40cm........15

focal length of eyepiece = 2*5 = 10cm

focal length of objective = 40*15 = 600cm

magnification = FL obj / FL eyp = 600/10 = 60x

7 0

3 years ago

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