Which is not expecting acceleration?
1 answer:
#1
A point on the outside of a spinning top whose rotational speed is constant.
Since the point at the top is at rest so it will not accelerate at all
A skydiver whose air resistance is equal to that of her weight.
Since weight is counter balanced by air resistance so net force on it is zero so acceleration will be zero
A car on the freeway experiencing a net force of -120 N.
Since net force is not zero so it will accelerate here
A submerged beach ball whose buoyant force is eight times the force of gravity on it.
Since buoyant force is more than the weight so it will not at rest and move upwards with some acceleration
#2
A frictionless spinning merry go round.
Since spinning wheel accelerating due to centripetal force so it is non inertial frame
A falling rock.
Falling rock is accelerating due to gravity so it is again non inertial frame
A hot air balloon moving at 30 degrees east of north with no net force.
since hot air balloon is having no force on it so its acceleration is zero and its an inertial frame
A space shuttle whose boosters just ignited for takeoff.
Since space shuttle is accelerating due to boosters so it is non inertial frame
#3
Here net force on it is due to two given forces
towards right
towards left
so here net force is given as
since force towards left is more so the direction will be towards left
so correct answer is 60 N towards left
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Answer:
T₂ = 123.9 N, θ = 66.2º
Explanation:
To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.
The tension T1 = 100 N, we create a reference frame centered on the pole
X axis
T₁ₓ - = 0
T_{2x}= T₁ₓ
Y axis y
T_{1y} + T_{2y} - 200N = 0
T_{2y} = 200 -T_{1y}
let's use trigonometry to find the component of the stresses
sin 60 = T_{1y} / T₁
cos 60 = t₁ₓ / T₁
T_{1y} = T₁ sin 60
T1x = T₁ cos 60
T_{1y}y = 100 sin 60 = 86.6 N
T₁ₓ = 100 cos 60 = 50 N
for voltage 2 it is done in the same way
T_{2y} = T₂ sin θ
T₂ₓ = T₂ cos θ
we substitute
T₂ sin θ= 200 - 86.6 = 113.4
T₂ cos θ = 50 (1)
to solve the system we divide the two equations
tan θ = 113.4 / 50
θ = tan⁻¹ 2,268
θ = 66.2º
we caption in equation 1
T₂ cos 66.2 = 50
T₂ = 50 / cos 66.2
T₂ = 123.9 N
