Answer:
See explanation.
Explanation:
If each runner was holding the pole, the runner in the water side of the pole would probably be behind the other runner. Since running in knee deep is hard and makes you slower, the pole would be slanted.
Color <span>is a physical property of all visible light determined by the light's frequency and visible to the human eye.</span>
Answer:
83%
Explanation:
On the surface, the weight is:
W = GMm / R²
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.
In orbit, the weight is:
w = GMm / (R+h)²
where h is the height of the shuttle above the surface of the Earth.
The ratio is:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
The shuttle in orbit retains 83% of its weight on Earth.
Answer:
The magnitude of the magnetic force exerted on the moving charge by the current in the wire is 2.18 x
N
The direction of the magnetic force exerted on the moving charge by the current in the wire is radially inward
Explanation:
given information:
current, I = 3 A
= +6.5 x
C
r = 0.05 m
v = 280 m/s
and direction of the magnetic force exerted on the moving charge by the current in the wire, we can use the following formula:
F = qvB sin θ
where
F = magnetic force (N)
q = electric charge (C)
v = velocity (m/s)
θ = the angle between the velocity and magnetic field
to find B we use
B = μ
I/2πr
μ
= 4π x
or 1.26 x
N/
, thus
B = 4π x
x 3 / 2π(0.05)
= 1.2 x
T
Now, we can calculate the magnitude force
F = qvB sin θ
θ = 90°, because the speed and magnetic are perpendicular
F = 6.5 x
x 280 x 1.2 x
sin 90°
= 2.18 x
N
Using the hand law, the magnetic direction is radially inward
Answer:
we have to find out the critical resolved shear stress. As it it given in the question
Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.
a) Slip will occur in the direction where cosФ cosλ are maximum. Cosine for all possible λ values are given as follows.
cos(62.4°) = 0.46
cos(72.0°) = 0.31
cos(81.1°) = 0.15
Thus, the slip direction is at the angle of 62.4° along the tensile axis.
b) now the critical resolved shear stress can be find out by the following equation.
τ
= σ
( cosФ cosλ)
now by putting values,
= (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23