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3 years ago

14

A bird flies north 3 kilometers and then south 4 kilometers, what is the resultant displacement of the bird? Do not forget to in

clude the direction

1 answer:

Ber [7]3 years ago

8 0

Answer: let north be+ and south be-

therefore, -1 km is the resultant displacement of the bird

hope it helped u,

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A 120 W lightbulb and a 90 W lightbulb each operate at a voltage of 120 V. Part A Which bulb carries more current? Which bulb ca

uysha [10]

Answer:

120 W lightbulb

Explanation:

Let the two lightbulb be A and B respectively.

Given the following data;

Power A = 120W

Power B = 90W

Voltage = 120V

To find the current flowing through each lightbulb;

a. For lightbulb A

Power = current * voltage

120 = current * 120

Current = 120/120

Current = 1 Ampere.

b. For lightbulb B

Current = power/voltage

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Therefore, the lightbulb that carries more current is A with 1 Ampere.

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3 years ago

Here's a question from ~ [ AIEEE 2002 ]

lyudmila [28]

r=150m
coefficient of friction= $\mu$ =0.6

As car is avoid skidding

$\\ \sf\hookrightarrow \dfrac{mv^2}{r}=\mu mg$

Cancel m

$\\ \sf\hookrightarrow \dfrac{v^2}{r}=\mu g$

$\\ \sf\hookrightarrow v^2=\mu rg$

$\\ \sf\hookrightarrow v^2=0.6(10)(150)$

$\\ \sf\hookrightarrow v^2=60(150)$

$\\ \sf\hookrightarrow v^2=900$

$\\ \sf\hookrightarrow v=30ms^{-1}$

Done

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2 years ago

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6. A .25 kg arrow with a velocity of 12 m/s to the west strikes and pierces the center of a 6.8 kg target. a. What is the final

Alenkasestr [34]

Answer:

(a) the final velocity of the combined mass is 9.43 m/s

(b) the decrease in kinetic energy during the collision is 386.1 J

Explanation:

Given;

mass of arrow, m₁ = 25 kg

initial velocity of arrow, u₁ = 12 m/s

mass of target, m₂ = 6.8 kg

initial velocity of the target, u₂ = 0

Part (a)

From the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁+m₂)

where;

v is the final velocity of the combined mass

25 x 12 + 0 = v(25 + 6.8)

300 = v(31.8)

v = 300/31.8

v = 9.43 m/s

Part(b)

Kinetic Energy, K.E = ¹/₂mv²

Initial kinetic energy = ¹/₂m₁u₁² + ¹/₂m₂u₂² = ¹/₂ x 25 x (12)² + 0 = 1800 J

Final kinetic energy = ¹/₂m₁v² + ¹/₂m₂v² = ¹/₂v²(m₁ + m₂)

= ¹/₂ x (9.43)²(25+6.8)

= 1413.91 J

Decrease in kinetic energy = Initial K.E - Final K.E

Decrease in kinetic energy = 1800J - 1413.91 J = 386.1 J

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A. answer is the correct answer

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