Answer:
Explanation:
Let h be the height .
initial velocity in first case u = 0
final velocity v = 6 m /s
acceleration due to gravity g = 9.8 m /s²
v² = u² + 2 g h
6² = 0 + 2 x 9.8 x h
h = 1.837 m .
For second case u = 3 m /s
v² = u² + 2 gh
= 3² + 2 x 1.837 x 9.8
= 9 + 36
= 45 m
v = 6.7 m /s
Explanation:
First we will convert the given mass from lb to kg as follows.
157 lb = 
= 71.215 kg
Now, mass of caffeine required for a person of that mass at the LD50 is as follows.
= 12818.7 mg
Convert the % of (w/w) into % (w/v) as follows.
0.65% (w/w) = 
= 
= 
Therefore, calculate the volume which contains the amount of caffeine as follows.
12818.7 mg = 12.8187 g = 
= 1972 ml
Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.
Gradpoint ? And the answer is water.
Answer:
F = 147,78*10⁻⁹ [N]
Explanation:
By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .
The angle β ( angle between the line running through one of the charges in y axis and the charge in x axis) is
tan β = 0,5/0,7
tan β = 0,7142 then β = arctan 0,7142 ⇒ β = 35 ⁰
cos β = 0,81
d = √ (0,5)² + (0,7)² d1stance between charges
d = √0,25 + 0,49
d = √0,74 m
d = 0,86 m
Now Foce between two charges is:
F = K* q₁*q₂/ d² (1)
Where K = 9*10⁹ N*m²/C²
q₁ = 2,5* 10⁻⁹C
q₂ = 3,0*10⁻⁹C
d² = 0,74 m²
Plugging these values in (1)
F = 9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74 [N*m²/C²]*C*C/m²
F = 91,21 * 10⁻⁹ [N]
And Fx = F*cos β
Fx = 91,21 * 10⁻⁹ *0,81
Fx =73,89*10⁻⁹ [N]
Then total force acting on charge located at x = 0,7 m is:
F = 2* Fx
F = 2*73,89*10⁻⁹ [N]
F = 147,78*10⁻⁹ [N]
Answer:
<em>The magnitude of the magnetic field will act in a direction towards me.</em>
<em></em>
Explanation:
When a charged particle enters a magnetic field, it is deflected. The direction of travel of the particle is deflected, but the kinetic energy of the particle is not affected. <em>The force experienced by a charged particle as it enters a magnetic field that acts perpendicular to the path of the velocity of the particle, will produce a force that is perpendicular to both the direction of travel of the particle and the direction of the magnetic field.</em> In this case, the proton moves in the y-direction, the magnetic field is in the x-direction, therefore the force experienced by the particle will be towards me.
