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nika2105 [10]
3 years ago
9

1. calculate agni's average speed during the race.

Physics
1 answer:
MakcuM [25]3 years ago
7 0
The formula to find the speed/velocity is : 
v = Δx/Δt = (\frac{ x_{final} - x_{initial} }{ t_{travel} }

B.
1. From the graph given, we can get an information such as the distance travelled, and the time given.
2. If the graph are horizontal, it means that the Agni's speed is zero because Agni doesn't move.
3. If the question ask the velocity during all time, we can say that the velocity is :
.v =  \frac{x}{t} =  \frac{50 km}{2.5 hour} = 20 \frac{km}{h}
4. So, the final velocity can be found by looking the third part of the graph. So,..
v =  \frac{x}{t} =  \frac{(50-35) km }{0.5 hour} = 30 \frac{km}{h}

Hope this will help you :)
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60 meters

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3 years ago
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Which type of muscle cell can have multiple nuclei
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In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the
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Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, u_0= 10 m/s.

Angle of launch with the horizontal, \theta = 53 ^{\circ}

So, the vertical component of the initial velocity,

u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i).

The horizontal component of the initial velocity,

u_0\cos\theta=10 \cos 53 ^{\circ}

Let, t be the time of flight, to the horizontal distance covered

D=10 \cos (53 ^{\circ})t\cdots(ii).

Not, applying the equation of motion in the vertical direction.

s= ut +\frac 1 2 at^2

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, u =10 \sin 53 ^{\circ} (from equation (i), s=0 (as the final height is same as the launch height) and a = -9.81 m/s^2 (negative sign is due to the downward direction).

\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2

\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63 seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m.

Now, for the maximum height, H, applying the equation of motion as

v^2=u^2+2as

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, 0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H

\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}

\Rightarrow H = 3.25 m.

Hence, the maximum height covered is 3.25 m.

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