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Vera_Pavlovna [14]

2 years ago

9

The Gulf Stream off the east coast of the United States can flow at a rapid 3.3 m/s to the north. A ship in this current has a c

ruising speed of 8.0 m/s . The captain would like to reach land at a point due west from the current position.
Part B) At this heading, what is the ship's speed with respect to land?

1 answer:

never [62]2 years ago

5 0

Answer:

2.29 m/s

Explanation:

given,

Speed of the stream to north,x = 3.3 m/s

speed of the ship,z = 8 m/s

Speed of the stream is in north direction, to move the ship in west direction ship should move in south direction.

To solve this a right angle triangle is constructed.

where speed of the stream is base and speed of cruise is hypotenuse.

now, speed of ship with respect to ground

z² = x² + y²

y² = z² - x²

y² = 8² - 3.3²

$y = \sqrt{53.11}$

y = 2.29 m/s

Hence, the speed of the ship with respect to ground is equal to 2.29 m/s.

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During the course of a demonstration the professor is called away. When he returns he finds a beaker of water that was at room t

Yanka [14]

Answer:

Either heat added or mechanical work done.

Explanation:

Since he found stirring rod on the desk and a cigarette lighter. This means that the beaker was probably either heated with the aid of fire from the lighter.

Also, the stirring rod could have been used to stir the water which will increase the kinetic energy which also means an increase in temperature.

Thus, it's either heat was added or mechanical work was done as a result of stirring.

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2 years ago

A truck traveled 400 meters north in 80 seconds and then it traveled 300 meters east in 70 seconds. The magnitude of the average

Varvara68 [4.7K]

Answer:

3.3m/s

Explanation:

You first get the total time (80 + 70 = 150s).

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d= 500m.

So now that you have displacement and time, you can find the velocity:

v=d/t

v=500/150

v=3.3

5 0

2 years ago

When NASA was communicating with astronauts on the Moon, the time from sending on the Earth to receiving on the moon was 1.33 s.

frez [133]

To solve this problem it is necessary to apply the concepts related to the kinematic equations of movement description, which determine the velocity, such as the displacement of a particle as a function of time, that is to say

$v = \frac{x}{t}\rightarrow x = v*t$

Where,

x = Displacement

v = Velocity

t = Time

Our values are given as,

$v=3*10^8m/s$

$t = 1.33 s$

Replacing we have that,

$x=v*t$

$x=(3*10^8)(1.33)$

$x = 399'000.000m$

Therefore the distance from Earth to the Moon is 399.000 km

5 0

3 years ago

Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea

Setler [38]

Answer:

a) M = 4,997 10²⁰ kg , b) T = 1.43 10³ s

Explanation:

a) This exercise should be solved in several parts, let's start by calculating the acceleration of gravity of this planet from kinematics

v = v₀ - a t

As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction

a = (v₀ - v) / t

a = (15 - (-15)) /9.00 = 30/9

a = 3.33 m / s²

Now we use Newton's second law where force is the force of universal attraction

F = m a

G m M / r² = m a

M = a r² / G

Let's calculate

M = 3.33 (1.00 10⁵)² / 6.67 10⁻¹¹

M = 4,997 10²⁰ kg

b) The period of the ship's orbit

In this case we have a centripetal acceleration

The radius of the orbit is the radius of the plant plus the height of the ship from the surface

R = $R_{m}$ + h

R = 1 10⁵ + 2.00 10⁴

R = 12 10⁴ m

F = m a

G m M / R² = m a

Centripetal acceleration is

a = v² / R

The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle

d = 2π R

v = d / t

v = 2π R / T

Let's replace

G m M / R² = m (2π R / T)² / R

G M = R³ 4π² / T²

T² = 4π² R³ / G M

T² = (4π² (12 10⁴)³ / (6.67 10⁻¹¹ 4,997 10²⁰)

T² = 6.82 10¹⁶ / 3.33 10¹⁰

T = √ (2,048 10⁶)

T = 1.43 10³ s

3 0

2 years ago

Twin A makes a one way trip at 0.6c to a star 12 light years away while twin B stays on Earth. Each twin sends the other a signa

hram777 [196]

Answer:

8 signals received by twin A during the trip.

Explanation:

Given that,

Distance = 12 light year

Speed = 0.6 c

Time = 1 year

We need to calculate the time by A

Using formula of time

$T=t\sqrt{\dfrac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}}}$

Put the value into the formula

$T=1\sqrt{\dfrac{1+0.6}{1-0.6}}$

$T=2\ years$

Similarly,

The expression for distance cover by A

$D=d\sqrt{1-\dfrac{v^2}{c^2}}$

$D=12\sqrt{1-(0.6)^2}$

$D=9.6\ ly$

We need to calculate the time

Using formula of time

$t=\dfrac{D}{v}$

$t=\dfrac{9.6}{0.6}$

$t=16\ years$

We need to calculate the signals received by twin A

Using formula for number of signals

$n=\dfrac{t}{T}$

Put the value into the formula

$n=\dfrac{16}{2}$

$n=8\ signals$

Hence, 8 signals received by twin A during the trip.

7 0

3 years ago