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3 years ago

Calculate the kinetic energy of a 4kg cat running 5m/s

1 answer:

5 0

50 J
Ke = 0.5 * m * v^2 ( m = mass(Kg), V = Velocity(m/s)
= 0.5 * 4 * 5^2 
= 2 * 25 
= 50 J

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Later research indicated that electric current is actually the flow of ____

iren [92.7K]

Answer:

Electrons.

Explanation:

Electricity was discovered before the discovery of electrons by J.J Thompson in 1896. Before the electron, it was thought that it is the positive ions that move through the wire and carry current—that's why today the conventional current represents the flow of positive charges.

After J.J Thompson's discovery of the electrons, it was realized that it is the electrons that actually carry the current through the conductor. But changing the direction of the conventional current didn't seem appropriate, and that's why the convention continues to be used to this day—reminding us that once it were the positive ions that were thought to carry the current.

6 0

3 years ago

Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

3 years ago

Light travels 186 000 miles per second.how many miles dose light travel in one year

Troyanec [42]

(186,000 mi/sec) x (3,600 sec/hr) x (24 hr/da) x (365 da/yr)

= (186,000 x 3,600 x 24 x 365) mi/yr

= 5,865,696,000,000 miles per year (rounded to the nearest million miles)

8 0

4 years ago

A 50 kg pitcher throws a baseball with a mass of 0. 15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is

dsp73

The velocity of the pitcher at the given mass is 0.1 m/s.

The given parameters:

Mass of the pitcher, m₁ = 50 kg
Mass of the baseball, m₂ = 0.15 kg
Velocity of the ball, u₂ = 35 m/s

Let the velocity of the pitcher = u₁

Apply the principle of conservation of linear momentum to determine the velocity of the pitcher as shown below;

m₁u₁ = m₂u₂

$u_1 = \frac{m_2 u_2}{m_1} \\\\u_1 = \frac{0.15 \times 35}{50} \\\\u_1 = 0.105 \ m/s\\\\u_1 \approx 0.1 \ m/s$

Thus, the velocity of the pitcher at the given mass is 0.1 m/s.

Learn more about conservation of linear momentum here: brainly.com/question/13589460

4 0

2 years ago

Which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the bat

RSB [31]

Answer:

a. Capacitance

b. Charge on the plates

e. Energy stored in the capacitor

Explanation:

Let A be the area of the capacitor plate

The capacitance of a capacitor is given as;

$C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\$

where;

V is the potential difference between the plates

The charge on the plates is given as;

$Q = \frac{V\epsilon _0 A}{d}$

The energy stored in the capacitor is given as;

$E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2$

Thus, the physical variables listed that will change include;

a. Capacitance

b. Charge on the plates

e. Energy stored in the capacitor

3 0

3 years ago