A) 0.189 N
The weight of the person on the asteroid is equal to the gravitational force exerted by the asteroid on the person, at a location on the surface of the asteroid:
where
G is the gravitational constant
8.7×10^13 kg is the mass of the asteroid
m = 130 kg is the mass of the man
R = 2.0 km = 2000 m is the radius of the asteroid
Substituting into the equation, we find
B) 2.41 m/s
In order to orbit just above the surface of the asteroid (r=R), the centripetal force that keeps the astronaut in orbit must be equal to the gravitational force acting on the astronaut:
where
v is the speed of the astronaut
Solving the formula for v, we find the minimum speed at which the astronaut should launch himself and then orbit the asteroid just above the surface:
Answer:
I. 0 m/s
II. 20 m/s
III. Part BC
Explanation:
I. Determination of the initial velocity.
From the diagram given above,
The motion of the car starts from the origin. This implies that the car start from rest and as such, the initial velocity of the car is 0 m/s
II. Determination of the maximum velocity attained.
From the diagram given above, we can see clearly that the maximum velocity is 20 m/s.
III. Determination of the part of the graph that represents zero acceleration.
It important that we know the meaning of zero acceleration.
Zero acceleration simply means the car is not accelerating. This can only be true when the car is moving with a constant velocity.
From the graph given above, the car has a constant velocity between B and C.
Therefore, part BC illustrates zero acceleration.
Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:
The potential at a distance can be represented as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D)
₀
V(r) = ![-[\frac{qr^2}{8 \pi E_0R^3 }]](https://tex.z-dn.net/?f=-%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D)
₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:
= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]^R](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D%5ER)
V(r) = 
V(r) = 
V(r) 
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V
Answer:
a) a = - 0.106 m/s^2 (←)
b) T = 12215.1064 N
Explanation:
If
F₁ = 9*1350 N = 12150 N (→)
F₂ = 9*1365 N = 12285 N (←)
∑Fx = M*a = (M₁ +M₂)*a (→)
F₁ - F₂ = (M₁ +M₂)*a
→ a = (F₁ - F₂) / (M₁ +M₂ ) = (12150-12285)N/(9*68+9*73)Kg
→ a = - 0.106 m/s^2 (←)
(b) What is the tension in the section of rope between the teams?
If we apply ∑Fx = M*a for the team 1
F₁ - T = - M₁*a ⇒ T = F₁ + M₁*a
⇒ T = 12150 N + (9 * 68 Kg)*(0.106 m/s^2)
⇒ T = 12215.1064 N
If we choose the team 2 we get
- F₂ + T = - M₂*a ⇒ T = F₂ - M₂*a
⇒ T = 12285 N - (9 * 73 Kg)*(0.106 m/s^2)
⇒ T = 12215.1064 N
Answer:
hello your question lacks some data and required diagram
G = 77 GPa, т all = 80 MPa
answer : required diameter = 252.65 * 10-^3 m
Explanation:
Given data :
force ( P ) = 660 -N force
displacement = 15 mm
G = 77 GPa
т all = 80 MPa
i) Determine the required diameter of shaft BC
considering the vertical displacement ( looking at handle DC from free body diagram )
D' = 0.3 sin∅ , where D = 0.015
hence ∅ = 2.8659°
calculate the torque acting at angle ∅ of CD on the shaft BC
Torque = 660 * 0.3 cos∅
= 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N
hello attached is the remaining part of the solution
