Answer:
See below
Explanation:
Vertical position is given by
df = do + vo t - 1/2 a t^2 df = final position = 0 (on the ground)
do =original position = 2 m
vo = original <u>VERTICAL</u> velocity = 0
a = acceleration of gravity = 9.81 m/s^2
THIS BECOMES
0 = 2 + 0 * t - 1/2 ( 9.81)t^2
to show t =<u> .639 seconds to hit the ground </u>
During this .639 seconds it flies horizontally at 10 m/s for a distance of
10 m/s * .639 s =<u> 6.39 m </u>
The current passing through a circuit consisting of a battery of 12 V and resistor of 2 ohms is 6 Ampere
.
Explanation:
- Assume the wires are ideal with zero resistance.
- The current passing through the circuit will be
I = V/R = 12/2 = 6.000 A.
There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.
You need to figure out t4 to know the tension in the string.
Since the whole thing is not moving t1 + t2 + t3 = t4.
torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)
t1 =3.2 * 44g
t2 = 7 * 49g
t3 = 3.5 * 24g
t4 = t1 + t2 + t3 = 5570,118
The t4 also is given by:
t4 = r * T * sin Ф
r = 7
Ф = 32°
T: tension in the string
T = t4 / (r * sinФ)
T = t4 / (7 * sin(32°))
T = 1501,6 N
<span>The velocity would be 54.2 m/s
We would use the equation 1/2mv^2top+mghtop = 1/2mv^2bottom+mghbottom where m is the mass of the bobsled(which can be ignored), vtop/bottom is the velocity of the bobsled at the top or bottom, g is gravity, and htop/bottom is the height of the bobsled at the top or bottom of the hill. Since the velocity of the bobsled at the top of the hill and height at the bottom of the hill are zero, 1/2mv^2top and mghbottom will equal zero. The equation will be mghtop=1/2mv^2bottom. Thus we would solve for v.</span>
Answer:
this is for the ppl who delete my answer
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Explanation:
