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kondaur [170]
4 years ago
12

A sailboat is moving across the water at 3.0 m/s. A gust of wind fills its sails and it accelerates at a constant 2.0 m/s2. At t

he same instant, a motorboat at rest starts its engines and accelerates at 4.0 m/s2. After 3.0 seconds have elapsed, find the velocity of the sailboat.
Physics
1 answer:
Flauer [41]4 years ago
5 0

Answer:

v = 9 m/s

Explanation:

It is given that,

Initial speed of the sailboat, u = 3 m/s

Acceleration of the sailboat, a=2\ m/s^2

Initial speed of the motorboat, u = 0

Acceleration of the motorboat, a=4\ m/s^2

Time elapsed, t = 3 s

To find,

The velocity of the sailboat

Solve,

Let v is the velocity of the sailboat after 3 seconds. By using the equation of kinematics, it can be calculated.

v=u+at

v=3\ m/s+2\ m/s^2\times 3\ s

v = 9 m/s

Therefore, the velocity of the sailboat is 9 m/s.

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Given the acceleration, initial velocity, and initial position of a ball thrown vertically upward, how long does it take for it
kolbaska11 [484]

Answer:

t = \frac{v_{o} }{g}

ymax=  y₀ + v₀²/2g

Explanation:

The equations of uniformly accelerated rectilinear motion of upward (vertical) for the y axis are :

vfy = v₀y-gt Formula (1)

vfy² = v₀y²-2gΔy   Formula (2)

Where:  

t: time in any position (s)

Δy= y-y₀

y = vertical position in any time (m)

y₀ : initial vertical position in meters (m)  

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

v₀y = v₀  : total initial speed of the ball

y₀ : initial vertical position of the ball

g  :acceleration due to gravity

Time (t) calculation for the ball to reach maximum height

We apply the formula (1)

vfy =  v₀y-gt

When the ball reaches its maximum height (h), vy = 0:

0= v₀-gt

v₀ = gt

t = \frac{v_{o} }{g}

Calculation of the maximum hight that reaches the ball

When the ball reaches its maximum height (ymax), vy = 0:

We apply the formula (2)

vfy²= v₀y²-2gΔy

0= v₀²-2gΔy

2gΔy  = v₀²

Δy= v₀²/2g    ,Δy= ymax-y₀

ymax-y₀ = v₀²/2g

ymax=  y₀ + v₀²/2g

3 0
3 years ago
A ball is released from a vertical height of 40 cm. It rolls down a "perfectly frictionless" ramp and up similar ramp. What vert
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Well, first of all, if the ramps are perfectly frictionless, then the ball
won't roll ... it'll slide.  Kind of like a bowling ball does at the beginning
of a shiny, highly-waxed lane.  You need friction against the ball to roll it.

But in any case, no matter what the slopes of the ramps are, even
if they have different slopes, the ball will rise on the second ramp
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4 years ago
The horizontal blue line through the middle of the lens is called the optical axis. As the lamp is moved above or below the axis
lawyer [7]

Answer:

As the object moves parallel to principle axis the image of the object will continuously moves away from the principal axis so the horizontal distance will increase

Explanation:

As the object is moving towards the lens then in that case first the image will form between the focus and double the distance of focus.

So here the image is diminished

now when the object is at large distance from from lens then its image will form at focus and it is of very small size

Now as the object comes closer to the lens till the object distance is 2F the image size will increase but smaller then object size

Now when object is between F and 2F then the image size will be more than object size and it will increase to infinite magnification

So overall we can say that as the object moves parallel to principle axis the image of the object will continuously moves away from the principal axis so the horizontal distance will increase

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4 years ago
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When does gravitational lensing occur? High concentrations of dark matter cause length contraction of nearby objects. The gravit
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Answer:

A massive object (like a galaxy cluster) bends the light from an object (like a quasar) that lies behind it.

Explanation:

A massive object, like a galaxy cluster, is able to deform the space-time shape as a consequence of its own gravity, so the light that it is coming from a source that is behind it in the line of sight will be bend or distorts in a way that will be magnified, making small arcs around the cluster with the image of the background object.

This technique is useful for astronomers since they make research of faraway objects (at hight redshift) that otherwise will difficult to detect with a telescope.

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