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PIT_PIT [208]
3 years ago
6

Relative velocity.....Please help....​

Physics
1 answer:
Romashka [77]3 years ago
5 0

Answer:

3. Usinθ

Explanation:

The vertical component of velocity of any object reaching maximum height is 0 and the horizontal component of velocity of any object in flight remains constant.

Therefore the horizontal component of velocity is Ucosθ for both objects at any given time.

At maximum height A is having 0 vertical velocity but B which just starting its flight is having Usinθ as its vetical component of velocity.

Velocity of B relative to A = Velocity of B - Velocity of A

As velocity can be resolved to components,( to simplify the sum )

Horizontal component of velocity of B relative to A = Horizontal component of velocity of B - Horizontal component of velocity of A

which is 3. <u>Usinθ</u>

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Inside a 30.2 cm internal diameter stainless steel pan on a gas stove water is being boiled at 1 atm pressure. If the water leve
dybincka [34]

Answer:

Q = 20.22 x 10³ W = 20.22 KW

Explanation:

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Therefore,

V = π(0.151 m)²(0.0145 m)

V = 1.038 x 10⁻³ m³

Now, we find the mass of the water that is vaporized.

m = ρV

where,

m = mass = ?

ρ = density of water = 1000 kg/m³

Therefore,

m = (1000 kg/m³)(1.038 x 10⁻³ m³)

m = 1.038 kg

Now, we calculate the heat required to vaporize this amount of water.

q = mH

where,

H = Heat of vaporization of water = 22.6 x 10⁵ J/kg

Therefore,

q = (1.038 kg)(22.6 x 10⁵ J/kg)

q = 23.46 x 10⁵ J

Now, for the rate of heat transfer:

Rate of Heat Transfer = Q = q/t

where,

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Therefore,

Q = (23.46 x 10⁵ J)/1116 s

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3 years ago
Enter the expression 2gΔym−−−−−√, where Δ is the uppercase Greek letter Delta.
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One long wire lies along an x axis and carries a current of 43 A in the positive x direction. A second long wire is perpendicula
qwelly [4]

Answer:

Bnet=1.006*10^-6T

Explanation:

One long wire lies along an x axis and carries a current of 43 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 5.9 m, 0), and carries a current of 41 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0, 1.7 m, 0)?

the magnetic field Bnet=\sqrt{b1^2+b2^2}

the magnetic field due this long wire is given by

B1=∨I1/(2\pi *R1)..............................1

B2=∨I2/(2\pi *R2)............................2

Bnet=\sqrt{(vI1/2*pi*R1)^2+(vI2/2*pi*R2)^2}.......................3

Bnet=v/2*pi\sqrt{(I1/R1)^2+(i2/R2)^2}

Bnet=4*pi*10^-7/(2\pi)\sqrt{(43/1.7)^2+(41/29.5)^2}

Bnet=0.0000002*(641.72)^.5

Bnet=1.006*10^-6T

8 0
4 years ago
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