Suppose a cat climbs a tree to a height of 2 meters. If the cat doubles its height to 4 meters, its potential energy will

Which symbol represents a type of electromagnetic radiation released during radioactive decay?

Tanya [424]

The answer for <span>electromagnetic radiation released during radioactive decay i</span>s C. He

8 0

4 years ago

A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t

Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6× $10^{24}$ be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67× $10^{-11}$ N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v= $\sqrt{\frac{Gmm}{R+h} }$ [(R+h) is the distance of the satellite from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = $h_{i}$ = 98 km = 98000 m

∴Initial Energy ( $E_{i}$ ) = $\frac{1}{2}$ m $v^{2}$ + $\frac{-GMm}{(R+h_{i} )}$

Substituting v= $\sqrt{\frac{GMm}{R+h_{i} } }$ in the above equation and simplifying we get,

$E_{i}$ = $\frac{-GMm}{2(R+h_{i}) }$

Similarly for final condition,

h= $h_{f}$ = 198km = 198000 m

∴Final Energy( $E_{f}$ ) = $\frac{-GMm}{2(R+h_{f}) }$

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = $E_{f}$ - $E_{i}$

= $\frac{GMm}{2}$ ( $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ )

Substituting ,

M = 6 × $10^{24}$ kg

m = 1036 kg

G = 6.67 × $10^{-11}$

R = 6400000 m

$h_{i}$ = 98000 m

$h_{f}$ = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = $\frac{1}{2}$ m[ $v_{f} ^{2}$ - $v_{i} ^{2}$ ]

= $\frac{GMm}{2}$ [ $\frac{1} {R+h_{f} }$ - $\frac{1} {R+h_{i} }$ ]

= -ΔE

= - 484.438 MJ

c) Change in Potential Energy (ΔPE) = GMm[ $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ ]

= 2ΔE

= 968.907 MJ

3 0

4 years ago

A particular car engine operates between temperatures of 440°C (inside the cylinders of the engine) and 20°C (the temperature of

Step2247 [10]

One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.

From the temperature the value is given as

$\eta = 1-\frac{T_L}{T_H}$

Where,

T_L = Cold focus temperature

T_H = Hot spot temperature

Our values are given as,

T_L = 20\° C = (20+273) K = 293 K

T_H = 440\° C = (440+273) K = 713 K

Replacing we have,

$\eta = 1-\frac{T_L}{T_H}$

$\eta = 1-\frac{293}{713}$

$\eta = 0.589$

Therefore the maximum possible efficiency the car can have is 58.9%

4 0

3 years ago

A box is initially sliding across a frictionless floor toward a spring which is attached to a wall. the box hits the end of the

Serjik [45]

The elastic potential energy of a spring is given by
$U= \frac{1}{2}kx^2$
where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.

The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:
$W=-\Delta U = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2$

In our problem, initially the spring is uncompressed, so $x_i=0$ . Therefore, the work done by the spring when it is compressed until $x_f$ is
$W=- \frac{1}{2}kx_f^2$
And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.

8 0

4 years ago

Two models of the same compound are shown. In what way is Model A better than Model B?

astra-53 [7]

The correct answer is Model A shows the three-dimensional shape of the molecule, but Model B does not.

Explanation:

Model A and B show the structure of a molecule. In the case of model A, the structure is represented through the use of three-dimensional shapes, while in model B the structure is represented using the letters of each element and showing how each element is connected to others.

In this context, one feature that makes model A better is that this represents the molecule using a 3D model, which is better to understand how the molecule looks like and what is its structure. Moreover, both models are alike because they show the number of atoms of each element, although model A does not show the types of elements.

7 0

3 years ago

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