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Snezhnost [94]
2 years ago
12

At what angle North of East must the ship travel to reach its destination? Let East be 0◦ and North 90◦.

Physics
1 answer:
soldi70 [24.7K]2 years ago
3 0
That kind of depends on where the ship is now, and where it wants to go.
We learn these kinds of things in navigation school, and I'll just bet that
it's the kind of information that's right there in the part of the question
that you didn't bother copying.
You might be interested in
In a 400-m relay race the anchorman (the person who runs the last 100 m) for team A can run 100 m in 9.8 s. His rival, the ancho
melisa1 [442]

Answer:

largest lead = 3 m

Explanation:

Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.

So, first we need to calculate the velocities of both the anchorman  

given data:

Distance = d = 100 m

Time arrival for A = 9.8 s

Time arrival for B = 10.1 s

Velocity of anchorman A = D / Time arrival for A

=100/ 9.8 = 10.2 m/s

Velocity of anchorman B = D / Time arrival for B

=100/10.1 = 9.9 m/s

As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use

d = vt

= 9.9 x 9.8 = 97 m  

So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.

largest lead = 100 - 97 = 3 m

So if his lead no more than 3 m anchorman A win the race.

5 0
2 years ago
You throw a ball upward with a speed of 14m/s. What is the acceleration of the ball after it leaves your hand? Ignore air resist
omeli [17]

The acceleration of the ball after leaving the hand is 9.8 m/s^2 downward

Explanation:

In order to find the acceleration of the ball during its motion, we have to study which forces are acting on it.

After the ball leaves the hand, if we neglect air resistance, there is only one force acting on the ball: the force of gravity, whose magnitude is

F=mg

where m is the mass of the ball and g is the acceleration of gravity (g=9.8 m/s^2), acting in the downward direction.

According to Newton's second law, the acceleration of the ball is given by

a=\frac{\sum F}{m}

where

\sum F is the net force acting on the ball

After the ball leaves the hand, the only force acting on it is the force of gravity, so we can substitute (mg) into the previous equation:

a=\frac{mg}{m}=g=9.8 m/s^2

This means that the acceleration of the ball remains 9.8 m/s^2 downward for the entire motion, after leaving the hand.

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0
2 years ago
Weather balloons are filled with only a small amount of helium because the ____________________ of the balloon will increase as
Shkiper50 [21]
Weather balloons are filled with only a small amount of helium because the __Volume__. of the balloon will increase as the air pressure decreases at higher altitudes.
7 0
3 years ago
A bird is about 6.26.2 in.​ long, with a​ thin, dark bill and a​ wide, white wing stripe. If the bird can fly 9292 mi with the w
Trava [24]

Answer:

209 mph

Explanation:

V = Speed of bird in still air

v = Speed of wind = 44 mph

Consider the motion of the bird with the wind

D_{1} = distance traveled with the wind = 9292 mi

t_{1} = time taken to travel the distance with wind

Time taken to travel the distance with wind is given as

t_{1} = \frac{D_{1}}{V + v}

t_{1} = \frac{9292}{V + 44}                              eq-1

Consider the motion of the bird with the wind

D_{2} = distance traveled against the wind = 6060 mi

t_{2} = time taken to travel the distance against wind

Time taken to travel the distance against wind is given as

t_{2} = \frac{D_{2}}{V + v}

t_{2} = \frac{6060}{V - 44}                              eq-2

As per the question,

Time taken with the wind = Time taken against the wind

t_{1} = t_{2}

\frac{9292}{V + 44} = \frac{6060}{V - 44}

(9292) (V - 44) = (6060) (V + 44)

9292V - 408848 = 6060V + 266640

3232V = 675488

V = 209 mph

5 0
2 years ago
In the diagram, R1 = 40.0 ,
Nostrana [21]

Answer:

51 Ω.

Explanation:

We'll begin by calculating the equivalent resistance of R₁ and R₃. This can be obtained as follow:

Resistor 1 (R₁) = 40 Ω

Resistor 3 (R₃) = 70.8 Ω

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) =?

Since the two resistors are in parallel connection, their equivalent can be obtained as follow:

R₁ₙ₃ = R₁ × R₃ / R₁ + R₃

R₁ₙ₃ = 40 × 70.8 / 40 + 70.8

R₁ₙ₃ = 2832 / 110.8

R₁ₙ₃ = 25.6 Ω

Finally, we shall determine the equivalent resistance of the group. This can be obtained as follow:

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) = 25.6 Ω

Resistor 2 (R₂) = 25.4 Ω

Equivalent Resistance (Rₑq) =?

Rₑq = R₁ₙ₃ + R₂ (series connection)

Rₑq = 25.6 + 25.4

Rₑq = 51 Ω

Therefore, the equivalent resistance of the group is 51 Ω.

4 0
2 years ago
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