Answer:
21.21 m/s
Explanation:
Let KE₁ represent the initial kinetic energy.
Let v₁ represent the initial velocity.
Let KE₂ represent the final kinetic energy.
Let v₂ represent the final velocity.
Next, the data obtained from the question:
Initial velocity (v₁) = 15 m/s
Initial kinetic Energy (KE₁) = E
Final final energy (KE₂) = double the initial kinetic energy = 2E
Final velocity (v₂) =?
Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:
KE = ½mv²
NOTE: Mass (m) = constant (since we are considering the same car)
KE₁/v₁² = KE₂/v₂²
E /15² = 2E/v₂²
E/225 = 2E/v₂²
Cross multiply
E × v₂² = 225 × 2E
E × v₂² = 450E
Divide both side by E
v₂² = 450E /E
v₂² = 450
Take the square root of both side.
v₂ = √450
v₂ = 21.21 m/s
Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.
Answer:
ΔL = 3.82 10⁻⁴ m
Explanation:
This is a thermal expansion exercise
ΔL = α L₀ ΔT
ΔT = T_f - T₀
where ΔL is the change in length and ΔT is the change in temperature
Let's reduce the length to SI units
L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m
let's calculate
ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))
ΔL = 3.8236 10⁻⁴ m
using the criterion of three significant figures
ΔL = 3.82 10⁻⁴ m
For A 53 g ice cube at −30◦C is dropped into a container of water at 0◦C, the amount of water that freezes onto the ice? is mathematically given as
x = 9.93 g
<h3>What is the amount of water that freezes onto the ice?</h3>
Where
Energy received = energy given out
Generally, the amount of water is mathematically given as
(53)(0.5)(30) = (80)(x)
Therefore
x = (49)(0.5)(16)/(80)
x = 9.93 g
In conclusion, the mass of water
x = 9.93 g
Read more about mass
brainly.com/question/15959704
Answer:
B
Explanation:
While answer C may sound correct, Answer B is makes more sense. We know you cant use High-beam lights when u cant see ongoing traffic because it could affect the other driver coming across from you. Its good to use it when legal and safe, but in that term I still don't believe there's no reason for HIGH-beamed. That's this leaves B, when you are on u lighted streets.
<h3><u>Answer;</u></h3>
Radius = 0.0818 m
Angular velocity = 2.775 × 10^7 rad/sec
<h3><u>Explanation;</u></h3>
The mass of proton m=1.6748 × 10^-27 kg;
Charge of electron e= 1.602 × 10^-19 C;
kinetic energy E= 2.7 MeV
= 2.7 × 10^6 × 1.602 × 10^-19 J;
= 4.32 × 10^-13 Joules
But; K.E =0.5m*v^2,
Hence v=√(2K.E/m)
Velocity = 2.27 × 10^7 m/s
Angular velocity, ω = v/r
Therefore; V = ωr
Hence; V = √(2K.E/m) = ωr
r= √(2E/m)/w = √E*√(2*m)/(eB)
= √E * √(2×1.6748×10^-27)/(1.602×10^-19 ×2.9)
but E = 4.32 × 10^-13 Joules
r = 0.0818 m
Angular speed
Angular velocity, ω = v/r , where r is the radius and v is the velocity
Therefore;
Angular velocity = 2.27 × 10^7 / 0.0818 m
= 2.775 × 10^7 rad /sec
