Answer:
The temperature change in Celsius is 46°C.
The temperature change in Fahrenheit is 82.8°F.
Explanation:
A degree of Celsius scale is equal to that of kelvin scale; therefore,
A degree in Fahrenheit is 1.8 times the Celsius degree; therefore
Hence, the temperature change in Celsius is 46°C, and the temperature change in Fahrenheit is 82.8°F.
Answer:
The energy of the system is 15 J.
Explanation:
Given that,
Energy E = 2.5 J
Amplitude = 10 cm
We need to calculate the spring constant
Using formula of mechanical energy of the system
Put the value into the formula
If the block is replaced by a block with twice the mass of the original block
Amplitude = 6 cm
We need to calculate the energy
Using formula of mechanical energy
Put the value into the formula
Hence, The energy of the system is 15 J.
Answer:
Ft
Explanation:
We are given that
Initial velocity=u=0
We have to find the magnitude of p of the momentum of the particle at time t.
Let mass of particle=m
Applied force=F
Acceleration,
Final velocity ,
Substitute the values
We know that
Momentum, p=mv
Using the formula
Answer: Hello there!
We know this:
The distance between the cars at t= 0 is D.
car 2 has an initial velocity of v0 and no acceleration.
car 1 has no initial velocity and a acceleration of ax that starts at t = 0
then we could obtain the acceleration of the car 1 by integrating the acceleration over the time; this is v(t) = ax*t where there is not a constant of integration because the car 1 has no initial velocity.
Because the cars are moving against each other, we want to se at what time t they meet, this is equivalent to see:
position of car 1 + position of car 2 = D
and in this way we could ignore constants of integration :D
for the position of each car we integrate again:
P1(t) = (1/2)ax*t^2 and P2(t) = v0t
v0t + (1/2)ax*t^2 = D
v0t + (1/2)ax*t^2 - D = 0
now we can solve it for t using the Bhaskara equation.
that we cant solve witout knowing the values for v0, D and ax. But you could replace them in that equation and obtain the time, where you must remember that you need to choose the positive solution (because this quadratic equation has two solutions).
Now we want to know the velocity of car 1 just before the impact, this can be calculated by valuating the time in the as the time that we just found in the velocity equation for the car 1, this is:
where again, you need to replace the values of v0, D and ax.
Explanation:
It is given that,
Mass of bumper car, m₁ = 202 kg
Initial speed of the bumper car, u₁ = 8.5 m/s
Mass of the other car, m₂ = 355 kg
Initial velocity of the other car is 0 as it at rest, u₂ = 0
Final velocity of the other car after collision, v₂ = 5.8 m/s
Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁
Using the conservation of linear momentum as :
p₁ = m₁v₁ = -342 kg-m/s
So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.