I desperately need help with my Physics Exam I am failing this class will mark Brainliest to whoever helps the most: PART 1 (I d
on't want to overwhelm anyone so I'm posting my questions in parts) Also the questions marked are incorrect, so just work with the 3 remaining answers.
10000000 thank you's
10000000 thank you's
1 answer:
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Answer:
T = 153.72 N
Explanation:
For this exercise we must use the conditions of translational and rotational equilibrium.
Let's set a frame of reference on the hinge, start by writing the rotational equilibrium relationship, suppose counterclockwise rotation is positive
We look for the components of the cable tension with trigonometry
cos 37 = Tₓ / T
sin 37 = / T
Tₓ = T cos 37
T_{y} = T sin 37
the expression for rotational equilibrium is
T_{y} L + Tₓ 0 - W L / 2 - W_light 0.55 = 0
where L is the length L= 1.8 m,
T_{y} = (W L/2 + W_lght 0.55) / L
T sin 37 = Mg /2 + m_light g 0.55 / L
T = (M / 2 + m_light 0.55 / L) g / sin 35
let's calculate
T = (15/2 + 4.9 0.55 / 1.8) 9.8 / sin 35
T = 153.72 N
Answer:
Work done = 13605.44
Explanation:
Data provided in the question:
For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J
The relation between Energy (U) and the elongation (s) is given as:
U = ................(1)
where,
k is the spring constant
on substituting the valeus in the above equation, we get
3.0 =
or
k = 13605.44 N/m
now
for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m
using the equation 1, we have
U =
or
U = 26.149 J
Also,
Work done = change in energy
or
W = 26.149 - 3.0 = 23.149 J
Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants ( = 0) zero, so let's use the equation
y = t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
=
- gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch