How much force is required to accelerate a 5kg mass pushed by a 10n force

1. A pendulum has a period of 3 seconds. What's its frequency? 2. A pendulum has a frequency of 0.25 Hz. What's its period? 3. A

RideAnS [48]

Answer:

(1) 0.333 Hz

(2) 4 sec

(3) 2 sec, 0.5 Hz

Explanation:

(1) We have given time period of pendulum is 3 sec

So T = 3 sec

Frequency will be equal to $f=\frac{1}{T}=\frac{1}{3}=0.333Hz$

(2) Frequency of the pendulum is given f = 0.25 Hz

Time period is equal to $T=\frac{1}{f}=\frac{1}{0.25}=4sec$

(3) It is given that a pendulum makes 10 back and forth swings in 20 seconds

So time taken to complete 1 back and forth swings = $=\frac{20}{10}=2sec$

So time period T = 2 sec

Frequency will be equal to $f=\frac{1}{T}=\frac{1}{2}=0.5Hz$

6 0

3 years ago

Read 2 more answers

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, $r=4\ AU=1.496\times 10^{11}\ m$

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

$T^2=\dfrac{4\pi^2}{GM}r^3$

M is the mass of the sun

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3$

$T^2=\sqrt{9.96\times 10^{14}}\ s$

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0

3 years ago

How many discovered planets are there

Allisa [31]

We have discovered 786 planets. Most of which were only recently discovered.

5 0

2 years ago

Help ASAP

EastWind [94]

Answer:

Elastic potential energy, $E=2.35\times 10^{-8}\ J$

Explanation:

Charge, $q=9.4\times 10^{-10}\ C$

Potential, V = 50 V

It is required to find the electric potential energy in a capacitor stored in it. The formula of the electric potential energy in a capacitor is given by :

$E=\dfrac{1}{2}qV\\\\E=\dfrac{1}{2}\times 9.4\times 10^{-10}\times 50\\\\E=2.35\times 10^{-8}\ J$

So, the electric potential energy stored in the capacitor is $2.35\times 10^{-8}\ J$

8 0

3 years ago

What potential difference is required to cause 4.00 a to flow through a resistance of 330 ω?

Alisiya [41]

We can solve the problem by using Ohm's law, which states that an Ohmic conductor the following relationship holds:
$\Delta V = I R$
where
$\Delta V$ is the potential difference applied to the resistor
I is the current flowing through it
R is the resistance

In our problem, I=4.00 A and $R=330 \omega$ , so the potential difference is
$\Delta V = IR=(4.00 A)(330 \omega)=1320 V$

7 0

3 years ago