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Furkat [3]
2 years ago
14

With detailed explaniation

Physics
1 answer:
belka [17]2 years ago
4 0
  • Ø=37°
  • Initial velocity=u=20m/s
  • g=10m/s²

#A

\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}

\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}

\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}

\\ \rm\Rrightarrow H_{max}=20sin^237

\\ \rm\Rrightarrow H_{max}=7.2m

#B

\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}

\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}

\\ \rm\Rrightarrow R=40sin74

\\ \rm\Rrightarrow R=38.5m

#C

\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}

\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}

\\ \rm\Rrightarrow T=4sin37

\\ \rm\Rrightarrow T=2.4s

Now

\\ \rm\Rrightarrow v=u-gt

\\ \rm\Rrightarrow v=20-10(2.4)

\\ \rm\Rrightarrow v=20-24

\\ \rm\Rrightarrow v=-4m/s

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Why does an air filled balloon get brust when it reaches its great height​
baherus [9]

Answer:

because there is external pressure is less in the height.

hope it helps.

4 0
3 years ago
If the PLATE SEPARATION of an isolated charged parallel-plate capacitor is doubled: A. the electric field is doubled
mash [69]

Explanation:

The electric field of an isolated charged parallel-plate capacitor is given by :

E=\dfrac{q}{A\epsilon_o}........(1)

Where

q is the electric charge

A is the area of cross section of parallel plate

It is clear from equation (1) that the electric field of a parallel plate capacitor is directly proportional to the charge on the plate and inversely proportional to the area of cross section of a plate.

So, the correct option is (E) i.e. "none of the above".

5 0
3 years ago
An unknown substance has a mass of 15 g and a volume of 4 cm3. Calculate the density of the unknown substance.
myrzilka [38]

Answer:

Density is 3.75g/cm3

Explanation:

Density = mass ÷ volume

= 15g ÷ 4cm3

= 3.75g/cm3

5 0
3 years ago
You are given four resistors, 2 ohms, 3 ohms, 5 ohms, and 10 ohms. Your friend say you can connect them so you obtain an equival
AlexFokin [52]

If we will connect the resistors 2ohms, 3ohms, 5ohms in series and the 10ohms resistance parallel then we get equivalent resistance of 5 ohms.

The equivalent circuit is,

R equivalent for the series connection is,

\begin{gathered} Req(S)=2+3+5=10ohms \\ Now,\text{ } \\ Req\text{ }for\text{ }10\text{ }ohms\text{ }and\text{ }10\text{ }ohms\text{ }is, \\ \frac{1}{Req}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}=\frac{1}{5} \\ So, \\ Req=5\text{ ohms} \end{gathered}

The equivalent resistance is 5 ohms.

So your friend is saying true.

6 0
10 months ago
It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an
Lemur [1.5K]

Answer:

q=1.7346×10⁻⁶C

Explanation:

Since the electric field is perpendicular to the bottom and top of the cube,the total flux is equals the flux over the top of surface plus the flex over the lower surface

Ф(total)=Ф₃₀₀+Ф₂₃₀

But the flux is given by  Ф=E.A=EACos(θ) where θ is the angle between Area vector and electric field

So

Ф(total)=E₃₀₀A Cos(180)+E₂₃₀ACos(0)

Ф(total)=A(E₃₀₀ - E₂₃₀)

The total flux is given by Gauss Law as:

Ф(total)=q/ε₀

q=ε₀Ф(total)

q=ε₀(A(E₃₀₀ - E₂₃₀))

Substitute the given values

q=(8.85×10⁻¹²){(70²)(100 - 60)}

q=1.7346×10⁻⁶C

7 0
3 years ago
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