A freely suspended magnet does not point exactly at the geographical N-S direction

Physics

1 answer:

elena-s [515]4 years ago

6 0

Because their is nothing at the geographical poles that attracts the magnet

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An object is moving initially with a velocity of 4.7 m/s . After 3.9 s the object's velocity is -2.1 m/s . What is the object's

IgorC [24]

Answer: The acceleration of the object is 0.67m/s^2 west.

Explanation: Here we are given the initial velocity and final velocity as well as the time taken. Acceleration is the change in velocity per unit time, thus the equation becomes.

a=dv/t

a=vf-vi/t

a=-2.1-4.7/3.9

a= 0.66m/s^2 west

8 0

4 years ago

Two liquids, A and B, have equal masses and equal initial temperatures. Each is heated for the same length of time over identica

DochEvi [55]

Answer:

So the specific heat of the liquid B is greater than that of A.

Explanation:

Liquid A is hotter than the liquid B after both the liquids are heated identically for the same duration of time from the same initial temperature then according to heat equation,

$Q=m.c.\Delta T$

where:

m = mass of the body

c = specific heat of the body

$\Delta T=$ change in temperature of the body

The identical heat source supplies the heat for the same amount of time then the quantity of heat supplied is also equal.

So for constant heat, constant mass the temperature change is inversely proportional to the specific of heat of the liquid.

$\Delta T=\frac{Q}{m} \times \frac{1}{c}$

$\Delta T\propto\frac{1}{c}$

So the specific heat of the liquid B is greater than that of A.

5 0

3 years ago

A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication

koban [17]

Answer:

r = 4.24x10⁴ km.

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}$

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>

From equation (1), r₁ is:

$r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}$