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2 years ago

How fast would you be going (in kmh) if you had a ship that accelerated at a constant 1g for 24 hours?

Physics

1 answer:

Nady [450]2 years ago

5 0

Answer:

Explanation:

1 g is 9.8 m/s^2 the problem wants the results in km/h so we'll fix that really quick.

9.8 m/s^2 (1 km/1000m)(60 sec/1 min)^2(60 min/1 hour)^2 = 127008 km/hour^2

Now, I'm assuming the ship is starting from rest, and hopefully you know your physics equations. We are going to use vf = vi + at. Everything is just given, or we can assume, so I'll just solve.

vf = vi + at

vf = 0 + 127008 km/hour^2 * 24 hours

vf = 3,048,192 km/hour

If there's anything that doesn't make sense let me know.

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What is the definition of displacement

Savatey [412]

the action of moving something from its place or position.

6 0

2 years ago

Read 2 more answers

A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north

Sedaia [141]

Answer:

Magnitude of displacement = 2.07 km

Magnitude of average velocity = 1.17 kmph

Explanation:

Let east represent positive x axis and north represent positive y axis.

A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north of west.

1.93 km due wast

s ₁ = 1.93 i km

1.03 km due south

s₂ = -1.03 j km

3.84 km in a direction 52.8 ° north of west

s₃ = -3.84 cos 52.8 i + 3.84 sin 52.8 j = -2.32 i + 3.06 j km

Total displacement

s = s ₁+ s₂+ s₃ = 1.93 i - 1.03 j -2.32 i + 3.06 j = -0.39 i + 2.03 j

Magnitude of displacement, $=\sqrt{(-0.39)^2+2.03^2}=2.07km$

Time taken = 1.771 hour

Magnitude of average velocity, $=\frac{2.07}{1.771}=1.17km/hr$

7 0

2 years ago

How many times higher could an astronaut jump on the Moon than on Earth if his takeoff speed is the same in both locations (grav

JulsSmile [24]

Answer:

maximum height on moon is 6 times more than the maximum height on Earth

Explanation:

Let the Astronaut has its maximum speed by which he can jump is "v"

now for the maximum height that it can jump is given as

$v_f^2 - v_i^2 = 2 aH$

now from above equation we will have

$0 - v^2 = 2(-g)H$

now we have

$H = \frac{v^2}{2g}$

now if Astronaut jump on the surface of moon with same speed

then we know that the acceleration of gravity on surface of moon is 1/6 times the gravity on earth

so at surface of moon we have

$0 - v^2 = 2(-g/6) H$

now we have

$H = \frac{6v^2}{2g}$

so maximum height on moon is 6 times more than the maximum height on Earth

8 0

2 years ago

Two billiard balls move toward each other on a table. The mass of the number three ball, m1, is 5 g with a velocity of 3 m/s. Th

Ymorist [56]

This question involves the concepts of the law of conservation of momentum and velocity.

The velocity of the eight ball is "5.7 m/s".

According to the law of conservation of momentum:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = mass of number three ball = 5 g

m₂ = mass of the eight ball = 6 g

u₁ = velocity of the number three ball = 3 m/s

u₂ = velocity of the eight ball = - 1 m/s (negative sign due to opposite direction)

v₁ = final velocity of the three number ball = - 5 m/s

v₂ = final velocity of the eight ball = ?

Therefore,

(5 g)(3 m/s) + (6 g)(- 1 m/s) = (5 g)(- 5 m/s) + (6 g)(v₂)

$v_2=\frac{34\ g.m/s}{6\ g}\\\\$

v₂ = 5.7 m/s

Learn more about the law of conservation of momentum here:

brainly.com/question/1113396?referrer=searchResults

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2 years ago

When finding the upper bound of the density, you put what number in the denominator?

lilavasa [31]

Answer: The result of "the upper bound of the density" does not go on the denominator.
So simplified, no. The answer is no.

3 0

2 years ago