The answer is <span>d. the sun</span>
Answer:
The final velocity of the object is, 
= 27 m/s
Explanation:
Given,
The acceleration of the object, a = 1000 m/s²
The initial displacement of the object, 
= 0 m
The final displacement of the object, 
= 0.75 m
The initial velocity of the object will be, 
= o m/s
The final velocity of the object, = ?
The average velocity of the object,
v = ( - )/ t
= 0.75 / t
The acceleration is given by the relation
a = v / t
1000 m/s² = 0.75 / t²
t² = 7.5 x 10⁻⁴
t = 0.027 s
Using the I equation of motion,
= u + at
Substituting the values
= 0 + 1000 x 0.027
= 27 m/s
Hence, the final velocity of the object is, = 27 m/s
The average act on her during the deceleration is 4.47 meters per second.
<u>Explanation</u>:
<u>Given</u>:
youngster mass m = 50.0 kg
She steps off a 1.00 m high platform that is s = 1 meter
She comes to rest in the 10-meter second
<u>To Find</u>:
The average force and momentum
<u>Formulas</u>:
p = m * v
F * Δ t = Δ p
vf^2= vi^2+2as
<u>Solution</u>:
a = 9.8 m/s
vi = 0
vf^2= 0+2(9.8)(1)
vf^2 = 19.6
vf = 4.47 m/s .
Therefore the average force is 4.47 m/s.
Answer:
yes
Explanation:
this is simple
the horizontal line is adjacent
the vertical line is opposite
recall that cos x=adj/hyp
adj=hyp(cos x)
while opp=hyp(sin x)
Answer:
x = 0.6034 m
Explanation:
Given
L = 5 m
Wplank = 225 N
Wman = 522 N
d = 1.1 m
x = ?
We have to take sum of torques about the right support point. If the board is just about to tip, the normal force from the left support will be going to zero. So the only torques come from the weight of the plank and the weight of the man.
∑τ = 0 ⇒ τ₁ + τ₂ = 0
Torque come from the weight of the plank = τ₁
Torque come from the weight of the man = τ₂
⇒ τ₁ = + (5 - 1.1)*(225/5)*((5 - 1.1)/2) - (1.1)*(225/5)*((1.1)/2) = 315 N-m (counterclockwise)
⇒ τ₂ = Wman*x = 522 N*x (clockwise)
then
τ₁ + τ₂ = (315 N-m) + (- 522 N*x) = 0
⇒ x = 0.6034 m
