First thing plot a right angle with angle of elevation of 35degrees as shown in figure then use SOHCAHTOA.
Ay/Helicopter = sin 35
Ay = 86 sin 35 = 36.8237
Ax/Helicopter = cos 35
Ax = 86 cos 35 = 77.7175
Note: I ignore the negative signs coz they signify direction
Answer:
(A) = 3.57 m
Explanation:
from the question we are given the following:
diameter (d) = 3.2 m
mass (m) == 42 kg
angular speed (ω) = 4.27 rad/s
from the conservation of energy
mgh = 0.5 mv^{2} + 0.5Iω^{2} ...equation 1
where
Inertia (I) = 0.5mr^{2}
ω = \frac{v}{r}
equation 1 now becomes
mgh = 0.5 mv^{2} + 0.5(0.5mr^{2})(\frac{v}{r})^{2}
gh = 0.5 v^{2} + 0.5(0.5)(v)^{2}
4gh = 2v^{2} + v^{2}
h = 3v^{2} ÷ 4 g .... equation 2
from ω = \frac{v}{r}
v = ωr = 4.27 x (3.2 ÷ 2)
v = 6.8 m/s
now substituting the value of v into equation 2
h = 3v^{2} ÷ 4 g
h = 3 x (6.8)^{2} ÷ (4 x 9.8)
h = 3.57 m
Answer:
People can capture geothermal energy through: Geothermal power plants, which use heat from deep inside the Earth to generate steam to make electricity. Geothermal heat pumps, which tap into heat close to the Earth's surface to heat water or provide heat for buildings
When the weather is cold, the water or refrigerant heats up as it travels through the part of the loop that's buried underground. Once it gets back above ground, the warmed water or refrigerant transfers heat into the building. The water or refrigerant cools down after its heat is transferred.
Answer:
1.6675×10^-16N
Explanation:
The force of gravity that the space shuttle experiences is expressed as;
g = GM/r²
G is the gravitational constant
M is the mass = 1.0 x 10^5 kg
r is the altitude = 200km = 200,000m
Substitute into the formula
g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²
g = 6.67×10^-6/4×10^10
g = 1.6675×10^{-6-10}
g = 1.6675×10^-16N
Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N
The electrical potential energy of a charge q located at a point at potential V is given by
Therefore, if the charge must move between two points at potential V1 and V2, the difference in potential energy of the charge will be
In our problem, the electron (charge e) must travel across a potential difference V. So the energy it will lose traveling from the metal to the detector will be equal to
Therefore, if we want the electron to reach the detector, the minimum energy the electron must have is exactly equal to the energy it loses moving from the metal to the detector:
