Answer
688.32m and 277.44m
Explanation :
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The X and Y coordinates of the rocket relative of firing
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<u>The</u><u> </u><u>horizontal</u><u> </u><u>range</u><u> </u><u>of</u><u> </u><u>projectile</u><u> </u><u>at</u><u> </u><u>x</u><u>.</u><u> </u>
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The vertical position of projectile at y.
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<h3><u>Henceforth</u><u>,</u><u> </u><u>the</u><u> </u><u>distance</u><u> </u><u>at</u><u> </u><u>horizon</u><u> </u><u>is</u><u> </u><u>6</u><u>8</u><u>8</u><u>.</u><u>3</u><u>2</u><u>m</u><u> </u><u>and</u><u> </u><u>at</u><u> </u><u>vertical</u><u> </u><u>is</u><u> </u><u>2</u><u>7</u><u>7</u><u>.</u><u>4</u><u>4</u><u>m</u><u>.</u></h3>
Answer:
See the answers below
Explanation:
We can solve both problems using vector sum.
a)
Let's assume the forces that help the diver dive as positive downward, and the forces that oppose upward, as negative
![F_{resultant}=100+30-85+900\\F_{resultant}=845[N]](https://tex.z-dn.net/?f=F_%7Bresultant%7D%3D100%2B30-85%2B900%5C%5CF_%7Bresultant%7D%3D845%5BN%5D)
The drag force is horizontal d this way in the horizontal direction we will only have the drag force that produces the water stream.
![F_{drag}=50[N]](https://tex.z-dn.net/?f=F_%7Bdrag%7D%3D50%5BN%5D)
b)
Let's assume the forces that propel the rocket upwards as positive and forces like the weight of the rocket and other elements as negative forces.
![F_{resultant}=960+7080-7700\\F_{resultant}=340 [kN]](https://tex.z-dn.net/?f=F_%7Bresultant%7D%3D960%2B7080-7700%5C%5CF_%7Bresultant%7D%3D340%20%5BkN%5D)
Answer:
0.695s
Explanation:
From Hooke's law, the restoring force is given has
F = -ky .......1
Where F is the force, y is the spring displacement and k force constant of the spring.
Also recall,
F=mg ............ 2
Where m is the mass of object, g is the acceleration due to gravity.
Equating 1 and 2
Ky = mg
Given that g=9.8m/s2 , y is 3.4cm and g is 8g
K×3.4/100m =8/1000kg × 9.8m/s2
K= ( 0.008kg × 9.8m/s2 ) ÷ 0.034
K= 0.0784÷0.035
K=2.24N/m
Mass ofvthe second object is 25g =0.025kg
Period of oscillation T
T=2π√m/k
T=2×3.142√0.025/2.24
T=6.284√0.0111
T=0.659seconds
You oughtta be able to do this one.
The "efficiency" is just the portion of the input work that
comes out in a useful form.
If the efficiency is 70%, that tells you that however much work
you put INto the machine, the machine will do 70% of that much
work for you at the output side.
Put 20,000 J in ... out comes (0.70) x (20,000 J) = 14,000 J .
What happens to the other 30% of the work you put into it ?
It turns into HEAT. That's why machines always have to be
cooled somehow while they're running.