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3 years ago

The reason that oscillating systems are good to use for clocks is:

Physics

1 answer:

Neporo4naja [7]3 years ago

6 0

Answer:

A. The period of an oscillation does not depend upon amplitude.

Explanation:

The period of a spring-mass system is:

T = 1/f = 2π√(m/k)

where f is the frequency, m is the mass, and k is the spring constant.

The answer isn't B. There are no frictionless systems in the real world.

The answer isn't C or D. As shown, the frequency is a function of both the mass and the spring constant.

The answer isn't E. Turning motion into heat is not an advantage for a clock.

The correct answer is A. The period of the system does not depend on the amplitude.

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PLEASE HELP AS FAT WILL GIVE BRAILIEST TO MOST SPECIFIC AND CORRECT ANSWER

Bas_tet [7]

Answer:

Time

Explanation:

8 0

4 years ago

What is the speed of a cyclist that rides west 88 km in 32 minutes?

suter [353]

Answer:

The speed of the cyclist is 2.75 km/min.

Explanation:

Given

The distance d = 88 km

Time t = 32 minutes

To determine

We need to find the speed of a cyclist.

In order to determine the speed of a cyclist, all we need to do is to divide the distance covered by a cyclist by the time taken to cover the distance.

Using the formula involving speed, time, and distance

$s=\frac{d}{t}$

where

s = speed

d = distance covered

t = time taken

substitute d = 88, and t = 32 in the formula

$s=\frac{d}{t}$

$s=\frac{88}{32}$

Cancel the common factor 8

$s=\frac{11}{4}$

$s=2.75$ km/min

Therefore, the speed of the cyclist is 2.75 km/min.

8 0

3 years ago

The masses of the earth and moon are 5.98 x 1024 and 7.35 x 1022 kg, respectively. Identical amounts of charge are placed on eac

dybincka [34]

Answer:

The magnitude of charge on each is $5.707\times 10^{13} C$

Solution:

As per the question:

Mass of Earth, $M_{E} = 5.98\times 10^{24} kg$

Mass of Moon, $M_{M} = 7.35\times 10^{22} kg$

Now,

The gravitational force of attraction between the earth and the moon, if 'd' be the separation distance between them is:

$F_{G} = \frac{GM_{E}M_{M}}{d^{2}}$ (1)

Now,

If an identical charge 'Q' be placed on each, then the Electro static repulsive force is given by:

$F_{E} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}$ (2)

Now, when the net gravitational force is zero, the both the gravitational force and electro static force mut be equal:

Equating eqn (1) and (2):

$\frac{GM_{E}M_{M}}{d^{2}} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}$

$(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22}) = (9\times 10^{9}){Q^{2}}$

$\sqrt{\farc{(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22})}{9\times 10^{9}}} = Q$

Q = $\pm 5.707\times 10^{13} C$

7 0

3 years ago

A freight train consists of two 8.00 \times 10^5~\text{kg}8.00×10 5 kg engines and 45 cars with average masses of 5.50 \time

erma4kov [3.2K]

Answer:

F₁ = 4.29 x 10⁵ N

Explanation:

The total force required to move the freight train with the given acceleration is given by the following formula:

F = ma + f

where,

F = Total Force Required from both engines = ?

m = equivalent mass of system = 2(8 x 10⁵ kg) + 5.5 x 10⁵ kg = 21.5 x 10⁵ kg

a = required acceleration = 5 x 10⁻² m/s²

f = force of friction = 7.5 x 10⁵ N

Therefore,

F = (21.5 x 10⁵ kg)(0.05 m/s²) + 7.5 x 10⁵ N

F = 8.575 x 10⁵ N

Now, for identical forces in each engine can be given as:

Force exerted by each engine = F₁ = F/2

F₁ = 8.575 x 10⁵ N/2

3 0

3 years ago

How much electromagnetic energy is contained in each cubic meter near the Earth's surface if the intensity of sunlight under cle

WITCHER [35]

Answer:

344.8 x10^-8J/m³

Explanation:

Using=> energy intensity/ speed oflight

= 1000/2.9x10^8

= 344.8 x10^-8J/m³

3 0

3 years ago