Answer:
Explanation:
Let the velocity of firing be u at angle θ
At maximum height velocity will be equal to horizontal component of initial velocity or vcosθ
So , vtop = v cosθ
At height h/2
vertical component of velocity v₂
v₂² = (usinθ)² - 2 g . h/2
v₂² = u²sin²θ - gh
horizontal component of velocity at height h/2 = u cosθ
velocity at height h / 2
= √ ( u²sin²θ - gh + u² cos²θ)
Given
√ ( u²sin²θ - gh + u² cos²θ) = 2 vtop
u²sin²θ - gh + u² cos²θ = 4 v²top = 4 u² cos²θ
u²sin²θ - gh = 3 u² cos²θ
At height h , vertical component of velocity is zero
0 = u²sin²θ - 2gh
gh = u²sin²θ / 2
u²sin²θ - u²sin²θ / 2 = 3 u² cos²θ
u²sin²θ / 2 = 3 u² cos²θ
Tan²θ = 6
Tanθ = 2.45
θ = 68⁰ .
Answer:
2.7ohms
Explanation:
Given parameters:
Voltage of the battery = 12V
Current = 4.5A
Unknown:
Resistance of the resistor = ?
Solution:
From Ohm's law, we know that;
V = IR
V is the voltage
I is the current
R is the resistance
So;
R = = = 2.7ohms
It gets more power5ful case of the coil affect
a). is true. <span>All motion is relative to a frame of reference. That's a fancy
way of saying that whenever you talk about a distance, an acceleration,
or a speed, they're always compared to something. Many surprising
things come out of this:
-- There's no such thing as "how fast is it <u>really</u> moving".
-- There's no such thing as "how high is it <u>really</u>".
-- There's no such thing as "<u>really</u> moving" or "<u>really</u> at rest".
What about quantities where you use distance or motion to calculate them ?
Like . . . . .
Potential energy . . . . . (mass) x (gravity) x (<u>height</u>)
Kinetic energy . . . . . (1/2) (mass) (<u>speed</u>)²
Momentum . . . . . (mass) x (<u>speed</u>) .
Yep, that's right. Those things are all relative to a frame of reference too;
different observers can get different answers,and they're all correct ... for
their own frame of reference.
Simple example:
You're flying in a passenger jet to visit your grandma.
In the frame of reference of somebody on the ground, you're moving at
400 miles per hour.
In the frame of reference of the pilot or the person sitting next to you,
you are at rest, and you can listen to your pod or read a book ... (or
maybe you're <em>heavily</em> at rest and taking a nap :-).)
</span>
Answer: The height of the cloud = 394.55 m
Explanation:
The observer is 500m away from the spotlight.
Let x be the distance from the observer to the interception of the segment of the height, h with the floor. The equations are thus:
Tan 45° = h/x ... eq1
Tan 75° = h/(500- x ) ... eq2
From eq 1, Tan 45° = 1, therefore eq1 becomes:
h = x ... eq3
Put eq3 into eq2
Tan 75° = h/(500- h)
h = ( 500 - h ) Tan 75°
h = 500Tan 75° - hTan75°
h + h Tan 75° = 500 Tan 75°
h ( 1 + Tan 75° ) = 500 Tan75°
h = 500Tan75°/ (1 + Tan 75°)
h= 1866.02 / 4.73
h = 394.55m