0° C = 273.15 K
Notice that the K temp is 273.15 more than the C temp.
So, to get any Celsius temp, subtract 273.15 from the Kelvin temp.
Answer:
530.43m/s
Explanation:
Given data
Mass m = 0.23kg
PE= 61J
From the expression for potential energy
PE= mgh
after it was shot
PE=KE
KE= 1/2mv^2
So
PE= 1/2mv^2
substitute
61= 1/2*0.23*v^2
61=0.5*0.23*v^2
61=0.115v^2
divide both sides by 0.115
v= 61/0.115
v=530.43 m/s
Hence the velocity is 530.43m/s
Answer:
14 m/s
18 m/s
22 m/s
Explanation:
The car is moving at constant acceleration, so we can find its velocity at time t using the equation:
where
u = 10 m/s is the initial velocity
is the acceleration
t is the time
Substituting the different times, we find:
At t = 2 s:
At t = 4 s:
At t = 6 s:
First, use a high-quality measurement tool. Next, measure carefully. Finally, repeat the measurement a few times. Hope it helps!
Answer:
The mass of the block, M =T/(3a +g) Kg
Explanation:
Given,
The upward acceleration of the block a = 3a
The constant force acting on the block, F₀ = Ma = 3Ma
The mass of the block, M = ?
In an Atwood's machine, the upward force of the block is given by the relation
Ma = T - Mg
M x 3a = T - Ma
3Ma + Mg = T
M = T/(3a +g) Kg
Where 'T' is the tension of the string.
Hence, the mass of the block in Atwood's machine is, M = T/(3a +g) Kg