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Svetach [21]
2 years ago
7

The launching velocity of a missile is 20.0 m/s, and it is shot at 53º above the horizontal. Which of the velocity components (n

eglecting air resistance) remains constant throughout the flight?
Physics
2 answers:
DochEvi [55]2 years ago
6 0

Answer:

Horizontal component of velocity will remains conserved which is v = 12 m/s

Explanation:

Two components of velocity of the projectile is given as

\vec v = 20 cos53 \hat i + 20 sin53 \hat j

\vec v = 12\hat i + 16\hat j

now we know that the acceleration is only due to gravity as we are considering the force is only vertically downwards due to gravity and there is no air resistance during whole motion.

So we will have acceleration as

\vec a = 0\hat i - g\hat j

now we can say that here we do not have any acceleration along horizontal direction so this component of velocity will remain conserved.

So the constant component of velocity is along horizontal component and it is given by v = 12 m/s

Gekata [30.6K]2 years ago
5 0
Neglecting air resistance, the horizontal component remains constant. The angle doesn't matter.
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(Double points) A machine receives electricity that enables it to deliver a total of 8,542 N of force for the completion of its
storchak [24]

Answer: machine's efficiency = 82.2%

Explanation:

Efficiency of a machine is the capability of a machine to convert input to output without waste.

It can be expressed as

Efficiency = output/ input × 100%

Output = 7,023N

Imput = 8,542N

Efficiency = 7,023N/8,542N × 100%

Efficiency = 82.2%

4 0
2 years ago
A spaceship which is 50,000 kilometers from the center of Earth has a mass of 3,000 kilograms. What is the magnitude of the forc
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Fg=Gx(M1M2/r^2)
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6 0
2 years ago
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Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the
astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

B = \frac{\mu_oI}{2\pi R}

Where \mu_o is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

B_1 = \frac{\mu_oI_1}{2\pi R}

B_2 = \frac{\mu_oI_2}{2\pi R}

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T =  \frac{\mu_oI_1}{2\pi R} - \frac{\mu_oI_2}{2\pi R}

or

4.0×10⁻⁶T =  \frac{\mu_o}{2\pi R}\times (I_1-I_2 )

also

\frac{I_1}{I_2} = \frac{3}{1}

⇒I_1 = 3\times I_2

substituting the values in the above equation we get

4.0×10⁻⁶T =  \frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)

⇒I_2 = 1A

also

I_1 = 3\times I_2

⇒I_1 = 3\times 1A

⇒I_1 = 3A

Hence, the larger of the two currents is 3A

3 0
2 years ago
For which of the following problems would a scientist most likely use carbon-14?
spayn [35]

Answer:

To calculate the age of a piece of bone

Explanation:

Carbon 14 is an isotope of carbon that is unstable and decays into Nitrogen 14 by emitting an electron. The decay rate of radioactive material is  normally expressed in terms of its "half-life" (the time required by half the radioactive nuclei of a sample to undergo radioactive decay). The nice thing about carbon 14 is that its "half-life" is about 5730 years, which gives a nice reference to measure the age of fossils that are some thousand years old.

Carbon 14 dating is used to determine the age of objects that have been living organisms long ago. They measure how much carbon 14 is left in the object after years of decaying without having exchange with the ambient via respiration, ingestion, absorption, etc. and therefore having renewed the normal amount of carbon 14 that is in the ambient.

A rock is not a living organism, so its age cannot be determined by carbon 14 dating.

3 0
2 years ago
A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c
netineya [11]

Answer:

E=\frac{\lambda}{2\pi r\epsilon_0}

Explanation:

We are given that

Linear charge density of wire=\lambda

Radius of hollow cylinder=R

Net linear charge density of cylinder=2\lambda

We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R

By Gauss theorem

\oint E.dS=\frac{q}{\epsilon_0}

q=\lambda L

E(2\pi rL)=\frac{L\lambda}{\epsilon_0}

Where surface area of cylinder=2\pi rL

E=\frac{\lambda}{2\pi r\epsilon_0}

8 0
2 years ago
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