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2 years ago

URGENT

Physics

1 answer:

olga55 [171]2 years ago

3 0

If the boat's speed is s, and the river's speed is r, and the boat is traveling east (0 degrees),

(0,r) + (s cos297,s sin297) = (6,0)

now just solve for r and s.

Pls mark me as brainliest

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How do fossils provide evidence of continental drift?

Alex777 [14]

C explanation: There are many examples of fossils found on separate continents and nowhere else, suggesting the continents were once joined. If Continental Drift had not occurred, the alternative explanations would be: They swam to the other continent/s in breeding pairs to establish a second population. ...

8 0

2 years ago

An electric field of 280000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -

nordsb [41]

<h2>The magnitude of the force that acts on a charge of -7.9C at this spot is 2.21 x 10⁶ N.</h2>

Explanation:

Electric field is the ratio of force and charge.

Electric field, E = 280000 N/C

Charge, q = -7.9 C

We have

$E=\frac{F}{q}\\\\280000=\frac{F}{7.9}\\\\F=280000\times 7.9\\\\F=2.21\times 10^6N$

The magnitude of the force that acts on a charge of -7.9C at this spot is 2.21 x 10⁶ N.

4 0

3 years ago

When dissolved into water, a molecule of potassium chloride ___________.

avanturin [10]

Answer:

ttrttrv vt vrtttv

Explanation:

4 0

2 years ago

Read 2 more answers

A 50.0 kg object is moving at 18.2 m/s when a 200 N force

gayaneshka [121]

Answer:

distance = 21.56 m

Explanation:

given data

mass = 50 kg

initial velocity = 18.2 m/s

force = -200 N ( here force applied to opposite direction )

final velocity = 12.6 m/s

solution

we know here acceleration will be as

acceleration a = force ÷ mass

a = $\frac{-200}{50}$ = -4 m/s²

we get here now required time that is

required time = $\frac{V_{(final)} - V_{(initial)}}{a}$ ...............1

put here value

required time = $\frac{12.6-18.2}{-4}$

so distance will be

distance = $\frac{V_{(final)}^2 - V_{(initial)}^2}{2a}$ ........2

distance = $\frac{12.6}^2 -{18.2}^2}{2\times (-4)}$

distance = 21.56 m

7 0

2 years ago

A lightning strike can transfer as much as electrons from the cloud to the ground. if the strike lasts 2ms , what is the average

malfutka [58]

The average electric current in the lightning will be 8 × $10^{-17}$ A

<h3>Why Lightning Conductors on top of a tower ?</h3>

The lightning conductors are long metal strips running from the spike end of a conductor on the top of a building to the earth. They are used to prevent buildings from destruction when struck by thunder or lightning.

Given that a lightning strike can transfer as much as electrons from the cloud to the ground. if the strike lasts 2ms, to calculate the average electric current in the lightning, we will first consider the charge released.

one charge = 1.6 × $10^{-19}$ C

Average current I = Q/t

Where

Q = charge = 1.6 × $10^{-19}$ C

t = time = 2ms = 2 × $10^{-3}$ s

I = current = ?

Substitute all the parameters into the formula

I = 1.6 × $10^{-19}$ C ÷ 2 × $10^{-3}$

I = 8 × $10^{-17}$ A

Therefore, the average electric current in the lightning will be 8 × $10^{-17}$ A

Learn more about Lightning here: brainly.com/question/3183045

#SPJ1

5 0

2 years ago