Circular motion formulas

A pendulum clock depends on the period of a pendulum to keep correct time. Suppose a pendulum clock is keeping correct time and

Mashutka [201]

Answer:

time period is increased so the clock will become SLOW

Explanation:

As we know that the time period of the simple pendulum is given by the formula

$T = 2\pi\sqrt{\frac{L}{g}}$

here we know that

L = distance of the pendulum bob from the hinge

g = acceleration due to gravity

now here the bob slide down so that the length of the pendulum is being increased

so time period T of the pendulum is also increased

so here the pendulum will take more time to oscillate or to complete one oscillation

so clock will become SLOW

8 0

3 years ago

As an airplane is taking off at an airport its position is closely monitored by radar. The following three positions are measure

Sergeeva-Olga [200]

Answer: acceleration a = 25m/s^2

Explanation:

Given that:

The plane travels with constant acceleration

x1 = 241.22 m at t1 = 3.70 s

x2 = 297.60 m at t2 = 4.20 s

x3 = 360.23 m at t3 = 4.70 s.

We need to calculate the velocity in the two time intervals.

Interval 1:

Average Velocity v1 = ∆x/∆t = (x2 - x1)/(t2-t1)

v1 = (297.60-241.22)/(4.20-3.70) = 112.76m/s

Interval 2:

Average Velocity v2 = ∆x/∆t = (x3-x2)/(t3-t2)

v2 = (360.23-297.60)/(4.70-4.20)

v2 = 125.26m/s

Acceleration:

Acceleration a = ∆v/∆t

∆v = v2-v1 = 125.26m/s-112.76m/s = 12.5m/s

∆t = change in average time of the two intervals = (t3-t1)/2 = (4.70-3.70)/2 = 0.5s

a = 12.5/0.5 = 25m/s^2

4 0

3 years ago

Consider an ideal gas of 7 moles that is in contact with a thermal reservoir of temperature 475 K. The gas is enclosed in a cont

kozerog [31]

Answer:

(a). The initial pressure is $5.5\times10^{4}\ Pa$

(b). The final pressure is $1.8\times10^{4}\ Pa$

Explanation:

Given that,

Number of moles = 7

Temperature = 475 K

Initial volume = 0.50 m³

Expanded volume = 1.50 m³

We need to calculate the initial pressure

Using formula of pressure

$P_{i}=\dfrac{nRT_{i}}{V_{i}}$

Put the value into the formula

$P_{i}=\dfrac{7\times8.31\times475}{0.50}$

$P_{i}=55261.5\ Pa$

$P_{i}=5.5\times10^{4}\ Pa$

We need to calculate the final pressure

Using formula of pressure

$P_{f}V_{f}=nRT_{f}$

After expansion,

$\dfrac{P_{f}V_{f}}{P_{i}V_{i}}=\dfrac{nRT_{f}}{nRT_{i}}$

$P_{f}=\dfrac{T_{f}}{T_{i}}\times\dfrac{P_{i}V_{i}}{V_{f}}$

Put the value into the formula

For thermal process,

$T_{i}=T_{f}$

$P_{f}=\dfrac{5.5\times10^{4}\times0.50}{1.50}$

$P_{f}=18333.33\ Pa$

$P_{f}=1.8\times10^{4}\ Pa$

Hence, (a). The initial pressure is $5.5\times10^{4}\ Pa$

(b). The final pressure is $1.8\times10^{4}\ Pa$

3 0

4 years ago

For the different values given for the radius of curvature RRR and speed vvv, rank the magnitude of the force of the roller-coas

nirvana33 [79]

Explanation:

The force of the roller-coaster track on the cart at the bottom is given by :

$F=\dfrac{mv^2}{R}$ , m is mass of roller coaster

Case 1.

R = 60 m v = 16 m/s

$F=\dfrac{(16)^2m}{60}=4.26m\ N$

Case 2.

R = 15 m v = 8 m/s

$F=\dfrac{(8)^2m}{15}=4.26m\ N$

Case 3.

R = 30 m v = 4 m/s

$F=\dfrac{(4)^2m}{30}=0.54m\ N$

Case 4.

R = 45 m v = 4 m/s

$F=\dfrac{(4)^2m}{45}=0.36m\ N$

Case 5.

R = 30 m v = 16 m/s

$F=\dfrac{(16)^2m}{30}=8.54m\ N$

Case 6.

R = 15 m v =12 m/s

$F=\dfrac{(12)^2m}{15}=9.6m\ N$

Ranking from largest to smallest is given by :

F>E>A=B>C>D

5 0

3 years ago

According to Newton's second law of motion what is force equal to

lubasha [3.4K]

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the netforce, and inversely proportional to the mass of the object

8 0

3 years ago

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