Answer:
The torque on the wrench is 4.188 Nm
Explanation:
Let r = xi + yj where is the distance of the applied force to the origin.
Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,
r = 0.18i + 0.055j
The applied force f = 88i - 23j
The torque τ = r × F
So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j
= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j
= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0 since i × i = 0, j × j = 0, i × j = k and j × i = -k
= 0 - 4.14k + 0.0484(-k) + 0
= -4.14k - 0.0484k
= -4.1884k Nm
≅ -4.188k Nm
So, the torque on the wrench is 4.188 Nm
<span>It's pretty easy problem once you set it up.
Earth------------P--------------Moon
"P" is where the gravitational forces from both bodies are acting equally on a mass m
Let's define a few distances.
Rep = distance from center of earth to P
Rpm = distance from P to center of moon
Rem = distance from center of earth to center of moon
You are correct to use that equation. If the gravitational forces are equal then
GMearth*m/Rep² = Gm*Mmoon/Rpm²
Mearth/Mmoon = Rep² / Rpm²
Since Rep is what you're looking for we can't touch that. We can however rewrite Rpm to be
Rpm = Rem - Rep
Mearth / Mmoon = Rep² / (Rem - Rep)²
Since Mmoon = 1/81 * Mearth
81 = Rep² / (Rem - Rep)²
Everything is done now. The most complicated part now is the algebra,
so bear with me as we solve for Rep. I may skip some obvious or
too-long-to-type steps.
81*(Rem - Rep)² = Rep²
81*Rep² - 162*Rem*Rep + 81*Rem² = Rep²
80*Rep² - 162*Rem*Rep + 81*Rem² = 0
We use the quadratic formula to solve for Rep:
Rep = (81/80)*Rem ± (9/80)*Rem
Rep = (9/8)*Rem and (9/10)*Rem
Obviously, point P cannot be 9/8 of the way to the moon because it'll be
beyond the moon. Therefore, the logical answer would be 9/10 the way
to the moon or B.
Edit: The great thing about this idealized 2-body problem, James, is
that it is disguised as a problem where you need to know a lot of values
but in reality, a lot of them cancel out once you do the math. Funny
thing is, I never saw this problem in physics during Freshman year. I
saw it orbital mechanics in my junior year in Aerospace Engineering. </span>
sylent_reality
· 8 years ago
Answer: Change in ball's momentum is 1.5 kg-m/s.
Explanation: It is given that,
Mass of the ball, m = 0.15 kg
Speed before the impact, u = 6.5 m/s
Speed after the impact, v = -3.5 m/s (as it will rebound)
We need to find the change in the magnitude of the ball's momentum. It is given by :
So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.
Read more on Brainly.com - brainly.com/question/12946012#readmore
Complete question is;
A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 24 m/s. The landing incline below her falls off with a slope of θ = 59◦ . The acceleration of gravity is 9.8 m/s².
What is the magnitude of the relative angle φ with which the ski jumper hits the slope? Answer in units of ◦
Answer:
14.08°
Explanation:
The time covered will be given by the formula;
t = (2V_x•tan θ)/g
t = (2 × 24 × tan 59)/9.8
t = 8.152 s
Now, the slope of the flight path at the point of impact will be given by the formula;
tan α = V_y/V_x
We are given V_x = 24 m/s
V_y will be gotten from the formula;
v = gt
Thus;
V_y = gt
V_y = 9.8 × (8.152) = 78.89 m/s
Thus;
tan α = 78.89/24
tan α = 3.2871
α = tan^(-1) 3.2871
α = 73.08°
Thus ;
Relative angle φ = α - θ = 73.08 - 59 = 14.08°
Answer:
Explanation:
Given that,
Spring constant = 16N/m
Extension of spring
x = 8cm = 0.08m
Mass
m = 5g =5/1000 = 0.005 kg
The ball will leave with a speed that makes its kinetic energy equal to the potential energy of the compressed spring.
So, Using conservation of energy
Energy in spring is converted to kinectic energy
So, Ux = K.E
Ux = ½ kx²
Then,
Ux = ½ × 16 × 0.08m²
Ux = 0.64 J
Since, K.E = Ux
K.E = 0.64 J
