Answer:
v₂ = 5.7 m/s
Explanation:
We will apply the law of conservation of momentum here:
where,
Total Initial Momentum = 340 kg.m/s
m₁ = mass of bike
v₁ = final speed of bike = 0 m/s
m₂ = mass of Sheila = 60 kg
v₂ = final speed of Sheila = ?
Therefore,
<u>v₂ = 5.7 m/s </u>
Answer: 10 m/s^2
Explanation:
1) The second law of Newton gives the definition and formula to calculate the net force:
Net force acting on an object = mass * acceleration.
2) From that, when you know the net force acting of the object and its mass, you can solve for the acceleration:
acceleration = Net force / mass
acceleration = 50 N / 5 kg = 10 m/s^2, which is the answer.
Answer:
v = 344.1 m / s
d = 1720.5 m
Explanation:
For this problem we must calculate the speed of sound in air at 22ºC
v = 331 RA (1+ T / 273)
we calculate
v = 331 RA (1 + 22/273)
v = 344.1 m / s
the speed of the wave is constant,
v = d / t
d = v t
we calculate
d = 344.1 5
d = 1720.5 m
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]
"1 watt" means 1 joule of energy per second.
75 W means 75 joules/sec .
Energy = (75 Joule/sec) x (12 min) x (60 sec/min)
Energy = (75 x 12 x 60) (Joule-<em>min-sec</em> / <em>sec-min</em>)
<em>Energy = 54,000 Joules</em>
