Answer:
those who are attracted to magnets. ex: iron, cobalt and nickel
Answer:
This is true,the rod with smaller elastic modulus will stretch more than larger elastic modulus.
Explanation:
σ=E*ε
ε=δ/L
σ=E*δ/L
δ=(σ*L)/E
σ=F/A
δ=(F*L)/(A*E)
As Force,Area and Length is same
δ∞1/E
From the expression as E increase δ will be small,so there will be more stretch for smaller elastic modulus.
Answer:
The average angular acceleration is ![\alpha =125.487 rad /s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D125.487%20rad%20%2Fs%5E2)
Explanation:
From the question we are told that
From the question we are told that
The length of the bat is
\
The initial linear velocity is ![u = 0 m/s](https://tex.z-dn.net/?f=u%20%3D%200%20m%2Fs)
The time is ![t = 0.15s](https://tex.z-dn.net/?f=t%20%3D%200.15s)
The velocity at t is ![v = 16 m/s](https://tex.z-dn.net/?f=v%20%3D%2016%20m%2Fs)
Generally average angular acceleration is mathematically represented as
![\alpha = \frac{w_f - w_o}{t}](https://tex.z-dn.net/?f=%5Calpha%20%20%3D%20%5Cfrac%7Bw_f%20-%20w_o%7D%7Bt%7D)
Where
is the finial angular velocity which is mathematically evaluated as
![w_f = \frac{v}{l}](https://tex.z-dn.net/?f=w_f%20%3D%20%5Cfrac%7Bv%7D%7Bl%7D)
![w_f = \frac{16}{0.85}](https://tex.z-dn.net/?f=w_f%20%3D%20%5Cfrac%7B16%7D%7B0.85%7D)
![= 18.823 rad/s](https://tex.z-dn.net/?f=%3D%2018.823%20rad%2Fs)
and
is the initial angular velocity which is zero since initial linear velocity is zero
So
![\alpha = \frac{18.823 - 0}{0.15}](https://tex.z-dn.net/?f=%5Calpha%20%20%3D%20%5Cfrac%7B18.823%20-%200%7D%7B0.15%7D)
![\alpha =125.487 rad /s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D125.487%20rad%20%2Fs%5E2)
Answer:
<u><em>1000 units for breakeven</em></u>
Explanation:
Let x be the number of units sold at breakeven.
The total sales at the point would be $2x.
Variable costs would be $1x and fixed costs are $1000.
Total costs are = $1x + $1000
At breakeven: Sales = Costs
Sales =m Costs
$2x = $1x + $1000
$1x = $1000
x = 1000 units.
At 1000 units the sales are equal to the costs ("breakeven").
Answer:
The tank is losing
![v_g = 19.81 \ m/s](https://tex.z-dn.net/?f=v_g%20%3D%2019.81%20%5C%20m%2Fs)
Explanation:
According to the Bernoulli’s equation:
We are being informed that both the tank and the hole is being exposed to air :
∴ P₁ = P₂
Also as the tank is voluminous ; we take the initial volume
≅ 0 ;
then
can be determined as:![\sqrt{[2g (h_1- h_2)]](https://tex.z-dn.net/?f=%5Csqrt%7B%5B2g%20%28h_1-%20h_2%29%5D)
h₁ = 5 + 15 = 20 m;
h₂ = 15 m
![v_2 = \sqrt{[2*9.81*(20 - 15)]](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%5B2%2A9.81%2A%2820%20-%2015%29%5D)
![v_2 = \sqrt{[2*9.81*(5)]](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%5B2%2A9.81%2A%285%29%5D)
as it leaves the hole at the base.
radius r = d/2 = 4/2 = 2.0 mm
(a) From the law of continuity; its equation can be expressed as:
J = ![A_1v_2](https://tex.z-dn.net/?f=A_1v_2)
J = πr²
J =![\pi *(2*10^{-3})^{2}*9.9](https://tex.z-dn.net/?f=%5Cpi%20%2A%282%2A10%5E%7B-3%7D%29%5E%7B2%7D%2A9.9)
J =![1.244*10^{-4} m^3/s](https://tex.z-dn.net/?f=1.244%2A10%5E%7B-4%7D%20%20m%5E3%2Fs)
b)
How fast is the water from the hole moving just as it reaches the ground?
In order to determine that; we use the relation of the velocity from the equation of motion which says:
v² = u² + 2gh
₂
v² = 9.9² + 2×9.81×15
v² = 392.31
The velocity of how fast the water from the hole is moving just as it reaches the ground is : ![v_g = \sqrt{392.31}](https://tex.z-dn.net/?f=v_g%20%3D%20%5Csqrt%7B392.31%7D)
![v_g = 19.81 \ m/s](https://tex.z-dn.net/?f=v_g%20%3D%2019.81%20%5C%20m%2Fs)