Solution1:
0.25 min = 15 secs
a = (v - ) / t
v = 0 (stops to rest), t = 15
⇒ a = -/15
x = t + a
⇒ 10.5 = * 15 + * * 225
10.5 = * 15 -
10.5 = * 7.5
= 1.4 (m/s)
⇒ a = - (m/s^2)
Solution2:
Answer:
85.654988MHz
Explanation:
f = C/λ
λ (Lambda) = Wavelength in meters
c = Speed of Light (299,792,458 m/s)
f = Frequency (MHz)
299,792,458 m/s. ÷ 3.5. = 85.654988MHz
Answer:
Explanation:
Total frictional force on boxes
= total weight x coefficient of friction
= ( 11 + 7 ) x 9.8 x .02 = 3.53 N
Net force on boxes
= 8.9 - 3.53
= 5.37 N
acceleration = 5.37 /( 11 + 7 )
= 0.3 m / s ²
b ) Let us consider movement of block A ( 11 kg )
acceleration a = .3 m/s²
friction force on block A
11 X 9.8 X .02
= 2.156 N
Net force on block A
8.9 - friction force - reaction force by block B
= 8.9 - 2.156 - F_ c
force = mass x acceleration
8.9 - 2.156 - F_ c = 11 x .3
F_ c = 3.444 N
c )
If force is applied from the side of box B
We consider all forces on box B
frictional force on it
= 7 x 9.8 x .02
= 1.372 N
Net force on it
8.9 - 1.372 - F_c
force = mass x acceleration
= 8.9 - 1.372 - F_c = 7 x .3 ( Acceleration of box B will be the same that is 0.3 m/s² )
F_c = 5.428 N
The nut distance from the Bullwinkle after uniform motion is 21 m.
We need to know about the uniform motion to solve this problem. The uniform motion is an object's motion under acceleration. It should follow the rule
vt = vo + a . t
vt² = vo² + 2a . s
s = vo . t + 1/2 . a . t²
where vt is final velocity, vo is initial velocity, a is acceleration, t is time and s is displacement.
From the question above, the parameters given are
m = 0.5 kg
s = 15 m
vx = 12 m/s
vo = 0 m/s
a = g = 9.8 m/s²
Find the time taken of the nut for landing
s = vo . t + 1/2 . a . t²
15 = 0 + 1/2 . 9.8 . t²
t² = 3.06
t = 1.75 s
Find the distance of nut in horizontal direction
vx = x / t
12 = x / 1.75
x = 12 . 1.75
x = 21 m
Find more on uniform motion at: brainly.com/question/28040370
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Answer:
a
b
Explanation:
From the question we are told that
The mass of the rock is
The length of the small object from the rock is
The length of the small object from the branch
An image representing this lever set-up is shown on the first uploaded image
Here the small object acts as a fulcrum
The force exerted by the weight of the rock is mathematically evaluated as
substituting values
So at equilibrium the sum of the moment about the fulcrum is mathematically represented as
Here is very small so
and
Hence
=>
substituting values
The mechanical advantage is mathematically evaluated as
substituting values