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3 years ago

A rod is made up of copper and wood joined together.

Physics

1 answer:

IgorLugansk [536]3 years ago

4 0

my views

Explanation:

Figure (8) shows a rod made up of copper and wood joined together. The rod is heated at the joint in the centre for about a minute. At which point – A, B, C or D, would it show the lowest temperature? PLEASE GIVE REASONS ALSO(MINIMUM 2)

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1. A 700 kg satellite is in orbit 2.4 x 106 meters from the center of the Earth. What is the force of

nevsk [136]

Answer:

$4.8 \cdot 10^4 N$

Explanation:

The force of gravity acting on the satellite is given by:

$F=\frac{GMm}{r^2}$

where

G is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

m is the mass of the satellite

r is the distance of the satellite from the Earth's centre

Here we have

m = 700 kg

$r=2.4\cdot 10^6 m$

Substituting into the equation, we find:

$F=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(700)}{(2.4\cdot 10^6)^2}=4.8 \cdot 10^4 N$

Note that the distance mentioned in the problem (2.4 x 10^6 meters) is not realistic, since it is less than the radius of the Earth (6.37 x 10^6 meters).

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3 years ago

A bus travels east for 3 km the north for 4 km what is its final displacement

lesya [120]

Im about to the math for this right now.

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2 years ago

What must always be included on the graph

Nookie1986 [14]

Clearly visible data points and appropriate labels on each access that include units

4 0

3 years ago

A 1-kilogram mass is attached to a spring whose constant is 14 N/m, and the entire system is then submerged in a liquid that imp

ch4aika [34]

Answer:

Part(a): The equation of motion is $\bf{x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}}$ .

Part(b): The equation of motion is $\bf{x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}}$ .

Explanation:

If ' $m$ ' be the mass of the object, ' $k$ ' be the force constant and ' $\beta$ ' be the damping constant, then the equation of motion of the particle can be written as

$\dfrac{d^{2}x}{dt^{2}} + \dfrac{\beta}{m} \dfrac{dx}{dt} + \dfrac{k}{m}x= 0.........................................(I)$

Given $m$ = 1 Kg, $k$ = 14 $N~m^{-1}$ , $\beta$ = 9. Substituting these values in equation (I),

$\dfrac{d^{2}x}{dt^{2}} + 9~\dfrac{dx}{dt} + 14~x= 0$

Taking a trial solution $x(t) = e^{mt}$ , the auxiliary equation can be written as

$m^{2} + 9m + 14 = 0............................................................(II)$

and its solutions are $m_{1} = -2~and~m_{2} = -7$ , resulting the general solution

$x(t) = C_{1}~e^{-2t} + C_{2}~e^{-7t}....................................................................(III)$

The velocity at any instant of time of the mass is

$v(t) = -2C_{1}~e^{-2t} _7~C_{2}~e^{-7t}..............................................................(IV)$

Part(a):

Given $x(t=0) = 1 m,~and~v(t=0) = 0~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(V)\\&and,& 0 = -2C_{1} - 7C_{2}.......................(VI)$

Solving equations (V) and (VI), we have

$C_{1} = \dfrac{7}{5}~and~C_{2} = \dfrac{-2}{5}$

So the equation of motion is

$x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}$

Part(b):

Given $x(t=0) = 1 m,~and~v(t=0) = - 12~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(VII)\\&and,& -12 = -2C_{1} -7C_{2}.......................(VIII)$

Solving equations (V) and (VI), we have

$C_{1} = -1~and~C_{2} = \dfrac{11}{5}$

So the equation of motion is

$x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}$

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3 years ago

A student throws a ball upward with a velocity of 35 m/s. What is the acceleration of the ball as it rises to the top of its arc

lora16 [44]

Answer:

The acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.

Explanation:

Let suppose that maximum height of the arc is so small in comparison with the radius of the Earth.

Since the ball is launched upwards, then the ball experiments a free-fall motion, that is, an uniform accelerated motion in which the element is accelerated by gravity. Then, the acceleration experimented by the motion remains constant at every instant and position.

Besides, the gravitational acceleration in the Earth and, in consequence, the acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.

7 0

2 years ago