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3 years ago

8

Claudia uses 100 N of force on a rope attached to a pulley to lift an anvil that weighs 400 N. What is the Mechanical Advantage

of the pulley?

1 answer:

Mashcka [7]3 years ago

7 0

Answer:

100 N/ 400 N= .25

Explanation:

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How did foucaults pendulum and coriolis effect are used to demonstrate earths tilt?

HACTEHA [7]

They do not demonstrate Earth's tilt. In fact, they're not "used" to demonstrate anything. It works the other way:. When you observe the Coriolis effect and the behavior of the Foucault pendulum, and you try to explain why the behave the way they do, one possible simple explanation for both of them is the Earth's ROTATION. Then, when you also observe the rising and setting of the sun and moon, and you also notice how the NUMBERS all go together, the case for the rotating, spherical Earth gets stronger and stronger.

7 0

4 years ago

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

lutik1710 [3]

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$

where in this problem:

$Q=15 \mu C = 15\cdot 10^{-6} C$ is the charge on the shell

$r=20 cm = 0.20 m$ is the distance from the centre of the shell

Substituting, we find:

$E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C$

4 0

3 years ago

Liam throws a water balloon horizontally at 8.2 m/s out of a window 18 m from the ground.

Alecsey [184]

Time taken by the water balloon to reach the bottom will be given as

$h = \frac{1}{2} gt^2$

here we know that

$h = 18 m$

$g = 9.8 m/s^2$

now by the above formula

$18 = \frac{1}{2}*9.8* t^2$

$18 = 4.9 t^2$

$t = 1.92 s$

now in the same time interval we can say the distance moved by it will be

$d = v_x * t$

$d = 8.2 * 1.92 = 15.7 m$

so it will fall at a distance 15.7 m from its initial position

5 0

3 years ago

water in a cup and a kettle can have the same temperature even though the quantities are different . give reasons

jekas [21]

Answer:

The reason is because both are exposed to a virtually infinite heat sink, due to the virtually infinite mass and of the surrounding environment, compared to the sizes of either the cup or the kettle such that the equilibrium temperature, $T_{(equilibrium)}$ reached is the same for both the cup and the kettle as given by the relation;

$\infty M_{(environ)} \times c_{(environ)} \times (T_2 - T_1) = m_{1} \times c_{(water)} \times (T_3 - T_2) + m_{2} \times c_{(water)} \times (T_4 - T_2)$

Due to the large heat sink, T₂ - T₁ ≈ 0 such that the temperature of the kettle and that of the cup will both cool to the temperature of the environment

Explanation:

4 0

3 years ago

(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{v^2}{r}$

Put the value into the formula

$a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}$

Put the value into the formula

$a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}$

$a=2.32\times10^{15}\ m/s^2$

We need to calculate the rate at which it emits energy because of its acceleration is

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Put the value into the formula

$\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}$

$\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s$

The energy in ev/s

$\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s$

$\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s$

We need to calculate the fraction of its energy that it radiates every second

$\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}$

$\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}$

Hence, The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

5 0

3 years ago