A 265 g mass attached to a horizontal spring oscillates at a frequency of 3.40 Hz . At t =0s, the mass is at x= 6.20 cm and has
1 answer:
Answer:
The phase constant is 7.25 degree
Explanation:
given data
mass = 265 g
frequency = 3.40 Hz
time t = 0 s
x = 6.20 cm
vx = - 35.0 cm/s
solution
as phase constant is express as
y = A cosФ ..............1
here A is amplitude that is = =
= 6.25 cm
put value in equation 1
6.20 = 6.25 cosФ
cosФ = 0.992
Ф = 7.25 degree
so the phase constant is 7.25 degree
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Answer:
Force that acted on the body was F = 13 N
Explanation:
If once accelerated, the body covers 60 meters in 6 seconds, then its velocity is 60/6 m/s = 10 m/s
When the force was acting (for 10 seconds) the object accelerated from rest (initial velocity vi = 0) to 10 m/s (its final velocity). therefore we can use the kinematic equation for the velocity in an accelerated motion given by:
which in our case becomes;
and we can solve for the acceleration as:
a = 10/10 m/s^2 = 1 m/s^2
Therefore the force acting on the body, based on Newton's 2nd Law expression: F = m * a is:
F = 13 kg * 1 m/s^2 = 13 N
Answer:
T'=92.70°C
Explanation:
To find the temperature of the gas you use the equation for ideal gases:
V: volume = 3000cm^3 = 3L
P: pressure = 1250mmHg; 1 mmHg = 0.001315 atm
n: number of moles
R: ideal gas constant = 0.082 atm.L/mol.K
T: temperature = 27°C = 300.15K
For the given values you firs calculate the number n of moles:
this values of moles must conserve when the other parameter change. Hence, you have V'=2L and P'=3atm. The new temperature is given by:
hence, T'=92.70°C
Answer:
4 Ohms
Explanation
(This is seriously not as hard as it looks :)
You only need two types of calculations:
- replace two resistances, say, R1 and R2, connected in a series by a single one R. In this case the new R is a sum of the two:
- replace two resistances that are connected in parallel. In that case:
I am attaching a drawing showing the process of stepwise replacement of two resistances at a time (am using rectangles to represent a resistance). The left-most image shows the starting point, just a little bit "warped" to see it better. The two resistances (6 Ohm next to each other) are in parallel and are replaced by a single resistance (3 Ohm, see formula above) in the top middle image. Next, the two resistances (9 and 3 Ohm) are nicely in series, so they can be replaced by their sum, which is what happened going to the top right image. Finally we have two resistances in parallel and they can be replaced by a single, final, resistance as shown in the bottom right image. That (4 Ohms) is the <em>equivalent resistance</em> of the original circuit.
Using these two transformations you will be able to solve step by step any problem like this, no matter how complex.
