Answer:
F * t = m * Δv
F * t = m * (v2 - v1) where v2 is the new velocity and v1 is the current velocity
m * (v2 - v1) = (F * t)
Plugging in the values,
0.3kg * v2 - 0.3kg * 20 m/s = 5 * 0.03 s
0.3 * v2 - 6 = 0.15
0.3 * v2 = 0.15 + 6
v2 = 6.15 / 0.3
v2 = 20.5 m/s
The new velocity of motion is 20.5 m/s.
A piece of glass weights 25 g in air, 16.77g in water at 4C and 16.89g in water at 60C. Find the mean coefficient of cubical expansion of water between 4C anf 60C, taking the coefficient of linear expansion of glass as 8*10^6
sorry i think so i dont exactly know i am sorry
Explanation:
As a certain amount of water is cooled from room temperature until it reaches 4 °C, its volume drops. The density reduces as the volume increases below 4 °C. Therefore, water's greatest density occurs at 4 degrees Celsius.
What coefficient of cubical expansion of water?Water has an unusual trait called anomalous expansion, which causes it to enlarge rather than compress when the temperature drops from 4 °C to 0 °C, and it becomes less dense.
At temperatures not far below zero degrees Celsius, water ice is unique for having low coefficients of static and dynamic friction that range from 0.04-0.02, but as the temperature drops, these numbers rise.
Therefore, When a metal sheet's temperature rises by one degree, its coefficient of area expansion is calculated as the increase in surface area per unit of original surface area.
Learn more about coefficient here:
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Compounds are made from the atoms of two or more______?
Answer:
elements
not really an explanation
HELP ME ASAP PLZZ
Which of the following is not a measurement taken by a radiosonde?
a. atmospheric composition
b. atmospheric pressure
c. wind speed
d. wind direction
Answer:
A atmospheric composition
A wheel of mass 50 kg has a radius of 0.4 m. It is making 480 rpm. What is the
torque necessary to bring it to rest in 40 seconds?
Solution:
Answer:
The torque necessary to bring the wheel to rest in 40 seconds is 10.4 N·m
Explanation:
The question is with regards to rotational motion
The rotary motion parameters are;
The mass of the wheel = 50 kg
The radius of the wheel = 0.4 m
The rate of rotation of the wheel = 480 rpm
The time in which the wheel is to be brought to rest = 40 s
The rotational rate of the wheel in rotation per second is given as follows;
480 r.p.m = 480 r.p.m × 1 minute/(60 seconds) = 8 revolution/second
1 revolution = 2·π radians
Therefore, we have the angular velocity, ω, given as follows;
ω = 2·π × 8 revolutions/second ≈ 50.3 rad/s
The angular acceleration, α, is given as follows;
[tex]\alpha = \dfrac{\Delta \omega}{\Delta t} = \dfrac{\omega _2 - \omega_1}{t_2 - t_1}[/tex]
Whereby the wheel is brought to rest from its initially constant rotational motion in 40 seconds, we have;
ω₁ ≈ 50.3 rad/s, ω₂ = 0 rad/s, and t₂ - t₁ = 40 seconds
Plugging in the values for the variables of the equation for the angular acceleration, "α", we get;
[tex]\alpha = \dfrac{0 - 50.3 \ rad/s}{40 \ s} \approx 1.3 \ rad/s^2[/tex]
The torque on the wheel, τ, is given as follows;
τ = m·r²·α
Where;
m = The mass of the object = 50 kg
r = The radius of the wheel = 0.4 m
α = The acceleration of the wheel ≈ 1.3 rad/s²
Therefore;
τ = 50 kg × (0.4 m)² × 1.3 rad/s² ≈ 10.4 N·m
The torque necessary to bring the wheel to rest in 40 seconds = τ ≈ 10.4 N·m.
Answer:
-10.048 N m
Explanation:
find not true when the water freezes...
Answer: i think its c
Explanation:
Answer:
a
Explanation:
someone please help with this
Answer:
The new force is 2/3 of the original force
Explanation:
Coulomb's Law
The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.
Written as a formula:
[tex]\displaystyle F=k\frac{q_1q_2}{d^2}[/tex]
Where:
[tex]k=9\cdot 10^9\ N.m^2/c^2[/tex]
q1, q2 = the objects' charge
d= The distance between the objects
Suppose the first charge is doubled (2q1) and the second charge is one-third of the original charge (q2/3). Now the force is:
[tex]\displaystyle F'=k\frac{2q_1*q_2/3}{d^2}[/tex]
Factoring out 2/3:
[tex]\displaystyle F'=\frac{2}{3}k\frac{q_1*q_2}{d^2}[/tex]
Substituting the original force:
[tex]F'=\frac{2}{3}F[/tex]
The new force is 2/3 of the original force