Answer: I'm not sure what it needs to be rounded to, but I got 37.53501401 m/s
Explanation: The formula for speed is speed = distance/time. You plug in the distance (13.40) and the time (0.357), then divide 13.40 by 0.357
I hope this helps! :)
Answer:
r = 0.02 m
Explanation:
from the question we have :
speed = 1 rps = 1x 60 = 60 rpm
coefficient of friction (μ) = 0.1
acceleration due to gravity (g) = 9.8 m/s^{2}
maximum distance without falling off (r) = ?
to get how far from the center of the disk the coin can be placed without having to slip off we equate the formula for the centrifugal force with the frictional force on the turntable force
mv^2 / r = m x g x μ
v^2 / r = g x μ .......equation 1
where
velocity (v) = angular speed (rads/seconds) x radius
angular speed (rads/seconds) = (\frac{2π}{60} ) x rpm
angular speed (rads/seconds) = (\frac{2 x π}{60} ) x 60 = 6.28 rads/ seconds
now
velocity = 6.28 x r = 6.28 r
now substituting the value of velocity into equation 1
v^2 / r = g x μ
(6.28r)^2 / r = 9.8 x 0.1
39.5 x r = 0.98
r = 0.02 m
<span>two objects in contact with each other are the same temperature</span>
Density = (mass) / (volume)
4,000 kg/m³ = (mass) / (0.09 m³)
Multiply each side
by 0.09 m³ : (4,000 kg/m³) x (0.09 m³) = mass
mass = 360 kg .
Force of gravity = (mass) x (acceleration of gravity)
= (360 kg) x (9.8 m/s²)
= (360 x 9.8) kg-m/s²
= 3,528 newtons .
That's the force of gravity on this block, and it doesn't matter
what else is around it. It could be in a box on the shelf or at
the bottom of a swimming pool . . . it's weight is 3,528 newtons
(about 793.7 pounds).
Now, it won't seem that heavy when it's in the water, because
there's another force acting on it in the upward direction, against
gravity. That's the buoyant force due to the displaced water.
The block is displacing 0.09 m³ of water. Water has 1,000 kg of
mass in a m³, so the block displaces 90 kg of water. The weight
of that water is (90) x (9.8) = 882 newtons (about 198.4 pounds),
and that force tries to hold the block up, against gravity.
So while it's in the water, the block seems to weigh
(3,528 - 882) = 2,646 newtons (about 595.2 pounds) .
But again ... it's not correct to call that the "force of gravity acting
on the block in water". The force of gravity doesn't change, but
there's another force, working against gravity, in the water.
The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.
<h3>How to determine the friction factor</h3>
Using the formula
μ = viscosity = 0. 06 Pas
d = diameter = 120mm = 0. 12m
V = velocity = 1m/s and 3m/s
ρ = density = 0.9
a. Velocity = 1m/s
friction factor = 0. 52 × 
friction factor = 0. 52 × 
friction factor = 0. 52 × 0. 55
friction factor 
b. When V = 3mls
Friction factor = 0. 52 × 
Friction factor = 0. 52 × 
Friction factor = 0. 52 × 0. 185
Friction factor 
Loss When V = 1m/s
Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter
Head loss = 0. 289 × 
× 
× 
Head loss = 1. 80 × 10^8
Head loss When V = 3m/s
Head loss = 
× 
× 
×
Head loss = 5. 3× 10^8
Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.
Learn more about friction here:
brainly.com/question/24338873
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