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2 years ago

Compare the time period of two simple pendulums of length 4m and 16m at a place.

Physics

1 answer:

Vlad1618 [11]2 years ago

5 0

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L) is defined as:

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

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

$T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2$

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

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A sample of a gas occupies a volume of 90 mL at 298 K and a pressure of 702 mm Hg. What is the correct expression for calculatin

aleksandr82 [10.1K]

Answer:

Explanation:

Given

volume of sample $V_1=90\ mL$

Temperature $T_1=298\ K$

Pressure $P_1=702\ mm\ of\ Hg$

for different conditions

Temperature $T_2=273\ K$

Pressure $P_2=760\ mm\ of\ Hg$

suppose $V_2$ is the volume of sample

Using ideal gas equation

$PV=nRT$

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$\frac{702\times 90}{298}=\frac{760\times V_2}{273}$

$V_2=76.15\ mL$

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2 years ago

Look at the two question marks between zinc (Zn) and arsenic (As). At the time, no elements were known

marissa [1.9K]

Answer:

Mendeleev predicted the atomic mass of each element along with compounds they each should form.

Explanation:

Based on other elements in the same group he predicted the existence of eka-aluminum and eka-silicon, later to be named gallium (Ga) and germanium (Ge).

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2 years ago

You are asked to design a spring that will give a 1160-kg satellite a speed of 2.50 m>s relative to an orbiting space shuttle

Tju [1.3M]

Given:
m = 1160 kg
g = 9,81 m/s²
v = 2.5 m/s

Unknown:
k spring constant
x spring compression

1. equation to use is Newton's 2nd law and Hook's law:
$F = ma = m(5g) = |kx|$
2. equation to use, the energy of the spring must equal the energy of the satellite:
$\frac{1}{2} kx^2 = \frac{1}{2} mv^2$
Combinig the two equations:
$\frac{1}{2} mv^2 = \frac{1}{2} m(5g)x \\ v^2 = 5gx \\ x = \frac{v^2}{5g} \\ \\ k = \frac{m(5g)}{x} = \frac{m(5g)^2}{v^2}$

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2 years ago

1. Use Coulomb’s Law (equation below) to calculate the approximate force felt by an electron at point A in the schematic below.

Amiraneli [1.4K]

Answer:

Explanation:

From the data it appears that A is the middle point between two charges.

First of all we shall calculate the field at point A .

Field due to charge -Q ( 6e⁻ ) at A

= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴

= 13.82 x 10⁻⁶ N/C

Its direction will be towards Q⁻

Same field will be produced by Q⁺ charge . The direction will be away

from Q⁺ towards Q⁻ .

We shall add the field to get the resultant field .

= 2 x 13.82 x 10⁻⁶

= 27.64 x 10⁻⁶ N/C

Force on electron put at A

= charge x field

= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶

= 44.22 x 10⁻²⁵ N

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2 years ago

A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen

Svetllana [295]

Answer:

$h_p = 30.46\ m$

Explanation:

<u>Free Fall Motion</u>

A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.

The speed vf of the object when a time t has passed is given by:

$v_f=g\cdot t$