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almond37 [142]
3 years ago
10

Ship A is located 4.1 km north and 2.3 km east of ship B. Ship A has a velocity of 22 km/h toward the south and Ship B has a vel

ocity of 40 km/h in a direction 38° north of east. What are the (a) x-component and (b) y-component of the velocity of A relative to B?
Physics
1 answer:
castortr0y [4]3 years ago
6 0

Answer:

a) x component  = -31.25 km/hr

b) y component = 46.64 km/hr

Explanation:

Given data:

A position is 4km north and 2.5 km east to B

Ship A velocity = 22 km/hr

ship B velocity = 40 km/hr

A velocity wrt to velocity of B

\vec{V_{AB}} =\vec{V_A} - \vec{V_B}

\vec{V_A} = 22 km/hr

\vec{V_B} = 40 cos38\hat{i} + 40sin 38 \hat{j}

                 = 31.52\hat{i} + 24.62 \hat{j}

putting respective value to get velocity of  A with respect to B

\vec{V_{AB}} = -22 \hat {j} - (31.52\hat{i} + 24.62 \hat{j})

\vec{V_{AB}} = -31.52\hat{i} - 46.62\hat{j}

a) x component  = -31.25 km/hr

b) y component = 46.64 km/hr

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Answer:

8\mu C

Explanation:

t = Time taken = 2\mu s

i = Current = 3 A

q(0) = Initial charge = 2\mu C

Charge is given by

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The magnitude of the net electric charge of the capacitor is 8\mu C

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3 years ago
Calculate the heat flux (in W/m^2) through a sheet of a metal 14 mm thick if the temperatures at the two faces are 350 and 140°C
ExtremeBDS [4]

Answer:

Explanation:

The rate of conductive heat transfer in watts is:

q = (k/s) A ΔT

where k is the heat conductivity, s is the thickness, A is the area, and ΔT is the temperature difference.

a)

Given k = 52.4 W/m/K, s = 0.014 m, and ΔT = 350-140 = 210 K, we can find q/A:

q/A = (52.4 / 0.014) (210)

q/A = 786,000 W/m²

b)

Given that A = 0.42 m², we can find q:

q = (0.42 m²) (786,000 W/m²)

q = 330,120 W

A watt is a Joule per second.  Convert to Joules per hour:

q = 330,120 J/s * 3600 s/hr

q = 1.19×10⁹ J/hr

c)

If we change k to 1.8 W/m/K:

q = (k/s) A ΔT

q = (1.8 / 0.014) (0.42) (210)

q = 11,340 J/s

q = 4.08×10⁷ J/hr

d)

If k is 52.4 W/m/K and s is 0.024 m:

q = (k/s) A ΔT

q = (52.4 / 0.024) (0.42) (210)

q = 192,570 J/s

q = 6.93×10⁸ J/hr

5 0
3 years ago
A 34n force is applied to a 213kg mass how much does the mass accelerate
motikmotik
We Know, F = m*a
Here, F = 34 N
m = 213 Kg

Substitute their values in the equation,
34 = 213 * a
a = 34/213
a = 0.159 m/s²

So, your final answer & the acceleration of the object would be 0.159 m/s²

Hope this helps!
3 0
3 years ago
What you filling your heart with <br>oxygen and blood
svlad2 [7]

Answer:

Explanation:

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5 0
2 years ago
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On a bet, you try to remove water from a glass by blowing across the top of a vertical straw immersed in the water. What is the
MArishka [77]

Answer:

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Explanation:

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The pressure at the two points is the same because they are open to the atmosphere, if we assume that the surface of the vessel is much sea that the area of ​​the straw the velocity of the surface of the vessel is almost zero v₁ = 0

The difference in height between the level of the glass and the straw is constant and equal to 1.6 cm = 1.6 10⁻² m

We substitute in the equation

         P_{atm} + ρ g y₁ = P_{atm} + ½ ρ v₂² + ρ g y₂

         ½ v₂² = g (y₂-y₁)

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Let's calculate

        v₂ = √ (2 9.8 1.6 10⁻²)

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