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3 years ago

Using a simple pendulum, we can measure. (a) Acceleration along a circle (b) Acceleration due to gravity (c) Frequency (d) None

Physics

1 answer:

zalisa [80]3 years ago

3 0

Answer:

Option B is the correct answer.

Explanation:

Period of simple pendulum is given by the expression $T=2\pi \sqrt{\frac{l}{g}}$ , where l is the length of pendulum and g is the acceleration due to gravity value.

So if we have a simple pendulum with length l we can find its period. Using the above formula we can calculate acceleration due to gravity value of that place.

So using simple pendulum we can measure acceleration due to gravity value.

Option B is the correct answer.

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You observe two stars over the course of a year (or more) and find that both stars have measurable parallax. (1 arc second is 1/

Ierofanga [76]

Answer:

Star X is much closer since it is at a distance 1 parsec from the Earth.

Explanation:

The angle due to the change in position of a nearby object against the background stars it is known as parallax.

The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth. This angle is gotten when the position of the object is measured in January and then in July according to the configuration of the Earth with respect to the Sun in those months.

The distance between the Earth and the Sun is 150000000 Km. That distance is also known as an astronomical unit (1AU).

The parallax angle can be defined in the following way:

$\tan{p} = \frac{1AU}{d}$

Where d is the distance to the star.

$p('') = \frac{1}{d}$ (1)

Equation (1) can be rewritten in terms of d:

$d(pc) = \frac{1}{p('')}$ (2)

Equation (2) represents the distance in a unit known as parsec (pc).

Case of Star X (p('') = 1):

Using equation 2 the distance of star X can be known:

$d(pc) = \frac{1}{1}$

$d(pc) = 1 pc$

So, star X is at 1 parsec from Earth.

Case of Star Y ( $p('') = \frac{1}{2}$ ):

$d(pc) = \frac{1}{(\frac{1}{2})}$

$d(pc) = 2 pc$

So, star Y is at 2 parsecs from the Earth.

Hence, star X is much closer.

Reminder:

Notice that in equation 2 the distance is inversely proportional to the parallax angle, so if the parallax angle decreases, the distance increases.

5 0

3 years ago

A 4.00 µf capacitor is connected to a 12.0 v battery.

atroni [7]

(a) The capacitance of the capacitor is:
$C=4 \mu F=4 \cdot 10^{-6}F$
and the voltage applied across its plates is
$V=12.0 V$

The relationship between the charge Q on each plate of the capacitor, the capacitance and the voltage is:
$C= \frac{Q}{V}$
and re-arranging it we find the charge stored in the capacitor:
$Q=CV=(4 \cdot 10^{-6} F)(12.0 V)=4.8 \cdot 10^{-5} C$

(b) The electrical potential energy stored in a capacitor is given by
$U= \frac{1}{2}CV^2$
where C is the capacitance and V is the voltage. The new voltage is
$V=1.50 V$
so the energy stored in the capacitor is
$U= \frac{1}{2}(4 \cdot 10^{-6} F)(1.50 V)^2=4.5 \cdot 10^{-6} J$

3 0

3 years ago

Read 2 more answers

What are the element names for the following symbols? Ni,He, Pb

vladimir1956 [14]

Ni is Nickel.
He is Helium.
Pb is Lead.

7 0

3 years ago

Read 2 more answers

A simple hydraulic lift is made by fitting a piston attached to a handle into a 3.0-cm diameter cylinder. The cylinder is connec

stiv31 [10]

Answer:

Approximately $3.1 \times 10^4 \; \rm N$ (assuming that the acceleration due to gravity is $g = 9.81\; \rm kg \cdot N^{-1}$ .)

Explanation:

Let $A_1$ denote the first piston's contact area with the fluid. Let $A_2$ denote the second piston's contact area with the fluid.

Similarly, let $F_1$ and $F_2$ denote the size of the force on the two pistons. Since the person is placing all her weight on the first piston:

$F_1 = W = m \cdot g = 50\; \rm kg \times 9.81 \; \rm kg \cdot N^{-1} =495\; \rm N$ .

Since both pistons fit into cylinders, the two contact surfaces must be circles. Keep in mind that the area of a square is equal to $\pi$ times its radius, squared:

$\displaystyle A_1 = \pi \times \left(\frac{1}{2} \times 3.0\right)^2 = 2.25\, \pi\;\rm cm^{2}$ .
$\displaystyle A_2 = \pi \times \left(\frac{1}{2} \times 24\right)^2 = 144\, \pi\;\rm cm^{2}$ .

By Pascal's Law, the pressure on the two pistons should be the same. Pressure is the size of normal force per unit area:

$\displaystyle P = \frac{F}{A}$ .

For the pressures on the two pistons to match:

$\displaystyle \frac{F_1}{A_1} = \frac{F_2}{A_2}$ .

$F_1$ , $A_1$ , and $A_2$ have all been found. The question is asking for $F_2$ . Rearrange this equation to obtain:

$\displaystyle F_2 = \frac{F_1}{A_1} \cdot A_2 = F_1 \cdot \frac{A_2}{A_1}$ .

Evaluate this expression to obtain the value of $F_2$ , which represents the force on the piston with the larger diameter:

$\begin{aligned}F_2 &= F_1 \cdot \frac{A_2}{A_1} \\ &= 495\; \rm N \times \frac{2.25\, \pi\; \rm cm^2}{144\, \pi \; \rm cm^2} \approx 3.1 \times 10^4\; \rm N\end{aligned}$ .

6 0

3 years ago

1.Seema is the wife of and Rahul is the brother of Robin. Robin is the only uncle of Ram .What is Ram relation with Seema ?

Phantasy [73]

Answer:

Ram is the son of Seema

3 0

3 years ago

Read 2 more answers