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ahrayia [7]
2 years ago
10

Any help????????????

Physics
2 answers:
kolezko [41]2 years ago
5 0
(-2,2) is the answer
pashok25 [27]2 years ago
3 0
The answer should be D
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A 590-kg rocket is at rest on the launch pad. what upward thrust force is needed to accelerate the rocket uniformly to an upward
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V(final)=V₀+a(t), V(final)=28, V₀=0, t=3.3, solve for a. a=28/3.3=8.5m/s^2. F=ma, F=?, m=590kg, a=8.5m/s^2, F=5015N. 
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Electron can only lose energy by transition from one allowed orbit to another ______ electromagnetic radiation
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A ball is launched with an initial velocity of 3m/s, at an angle of 40 degrees above the horizontal, and from a height of 0.5m.
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Answer:

The ball travels <u>1.31 m horizontally</u> before hitting the ground.

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Given:

Initial velocity of the ball is, u=3\ m/s

Angle of projection is, \theta=40\ degree

Initial height of the ball is, y_0=0.5\ m

Final height of the ball is, y=0\ m

Now, horizontal  and vertical components of initial velocity are given as:

u_x=u\cos\theta\\\\u_y=u\sin\theta

Plug in the given values and find u_x\ and\ u_y. This gives,

u_x=3\ m/s\times \cos(40)\\\\u_x=2.3\ m/s\\\\u_y=3\ m/s\times \sin(40)\\\\u_y=1.9\ m/s

Now, there is acceleration due to gravity acting in the vertical direction. The direction is downward. So, g=-9.8\ m/s^2.

Now, using equation of motion in the vertical direction, we have:

y-y_0=u_yt+\frac{1}{2}a_y t^2

Plug in the given values and solve for time 't'. This gives,

0-0.5=1.9t-0.5\times 9.8\times t^2\\\\-0.5=1.9t-4.9t^2\\\\4.9t^2-1.9t-0.5=0

On solving the above quadratic equation, we get:

t=0.57\ s\ or\ t=-0.18\ s

Ignoring the negative result as time can't be negative. So, time taken by the ball to reach the ground is 0.57 s.

Now, as the ball moves, there is no force acting on it in the horizontal direction. So, the acceleration in the horizontal direction is 0.

Now, we know that, when acceleration is zero, the body moves with a constant speed.

Hence, distance traveled in the horizontal direction is given as:

Horizontal distance = Horizontal component of initial velocity × Time

R=u_x\times t

Plug in the given values and solve for 'R'. This gives,

R=2.3\times 0.57\\\\R=1.31\ m

So, the ball travels 1.31 m horizontally before hitting the ground.

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