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3 years ago

Calculate the potential energy of a 4 kg cat crouched 3 meters off the ground

Physics

2 answers:

AnnZ [28]3 years ago

8 0

Answer : Potential energy of the cat is 117.6 J.

Explanation :

It is given that,

Mass of the cat, m = 4 kg

Height from the ground, h = 3 m

We know that the potential energy of the cat is possessed by the virtue of its height under the action of gravity.

Mathematically, it is given as :

$PE=mgh$

$PE=4\ kg\times 9.8\ m/s^2\times 3\ m$

$PE=117.6\ J$

Hence, this is the required solution

Tom [10]3 years ago

7 0

Gravitational potential energy = mass × gravity × height

Ep = (4)(9.81)(3)

Energy = 117.72 Joules

= 1.2x10^2 Joules

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A 12.0 μF capacitor is charged to a potential of 50.0 V and then discharged through a 265 Ω resistor. A)How long does the capaci

larisa [96]

(a) The time for the capacitor to loose half its charge is 2.2 ms.

(b) The time for the capacitor to loose half its energy is 1.59 ms.

<h3>Time taken to loose half of its charge</h3>

q(t) = q₀e-^(t/RC)

q(t)/q₀ = e-^(t/RC)

0.5q₀/q₀ = e-^(t/RC)

0.5 = e-^(t/RC)

1/2 = e-^(t/RC)

t/RC = ln(2)

t = RC x ln(2)

t = (12 x 10⁻⁶ x 265) x ln(2)

t = 2.2 x 10⁻³ s

t = 2.2 ms

<h3>Time taken to loose half of its stored energy</h3>

U(t) = Ue-^(t/RC)

U = ¹/₂Q²/C

(Ue-^(t/RC))²/2C = Q₀²/2Ce

e^(2t/RC) = e

2t/RC = 1

t = RC/2

t = (265 x 12 x 10⁻⁶)/2

t = 1.59 x 10⁻³ s

t = 1.59 ms

Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.

Learn more about energy stored in capacitor here: brainly.com/question/14811408

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6 0

2 years ago

A car moving around a circular track with constant speed (Select all that apply.) has an acceleration with a direction that cann

BARSIC [14]

Answer:

Explanation:

Has an acceleration directed towards center of its path

5 0

3 years ago

A student gives a brief push to a block of dry ice. A moment later, the block moves across a very smooth surface at a constant s

Monica [59]

Hey There,

Question: "<span>A student gives a brief push to a block of dry ice. A moment later, the block moves across a very smooth surface at a constant speed. When drawing the free body diagram for the block of dry ice moving at a constant speed, the forces that should be included are: (select all that apply)"

Answer: C. Force Of Friction
B. Force

If This Helps May I Have Brainliest?</span>

7 0

3 years ago

Read 2 more answers

Green light (λ = 518 nm) strikes a single slit at normal incidence. What width slit will produce a central maximum that is 3.00

MariettaO [177]

Answer:

6.9066 × 10⁻⁵ m

Explanation:

For constructive interference, the expression is:

$d\times sin\theta=m\times \lambda$

Where, m = 1, 2, .....

d is the distance between the slits.

The formula can be written as:

$sin\theta=\frac {\lambda}{d}\times m$ ....1

The location of the bright fringe is determined by :

$y=L\times tan\theta$

Where, L is the distance between the slit and the screen.

For small angle , $sin\theta=tan\theta$

So,

Formula becomes:

$y=L\times sin\theta$

Using 1, we get:

$y=L\times \frac {\lambda}{d}\times m$

Thus, the distance between the central maximum is 3.00 cm

First bright fringe , m = 1 occur at 3.00 / 2 = 1.50 cm

Since,

1 cm = 0.01 m

y = 0.0150 m

Given L = 2.00 m

λ = 518 nm

Since, 1 nm = 10⁻⁹ m

So,

λ = 518 × 10⁻⁹ m

Applying the formula as:

$0.0150\ m=2.00\ m\times \frac {518\times 10^{-9}\ m}{d}\times 1$

<u>⇒ d, distance between the slits = 6.9066 × 10⁻⁵ m</u>

7 0

3 years ago

In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so

Pepsi [2]

Answer:

(a) 181.05 m/s²

(b) 13.2°

Explanation:

Given:

Radius of the circle (R) = 0.610 m

Angular acceleration (α) = 67.6 rad/s²

Angular speed (ω) = 17.0 rad/s

(a)

Radial acceleration of the ball is given as:

$a_r=\omega^2R$

Plug in the given values and solve for $a_r$ . This gives,

$a_r=(17.0\ rad/s)^2\times (0.610\ m)\\\\a_r=289\times 0.610\ m/s^2\\\\a_r=176.29\ m/s^2$

Now, tangential acceleration is given by the formula:

$a_t=R\alpha$

Plug in the given values and solve for $a_t$ . This gives,

$a_t=(0.610\ m)(67.6\ rad/s^2)\\\\a_t=41.236\ m/s^2$

Now, the magnitude of total acceleration is given as the square root of the sum of the squares of tangential and centripetal accelerations. Therefore,

$a_{Total}=\sqrt{(a_r)^2+(a_t)^2}$

Plug in the given values and solve for total acceleration, $a_{Total}$ . This gives,

$a_{Total}=\sqrt{(176.29)^2+(41.236)^2}\\\\a_{Total}=181.05\ m/s^2$

Therefore, the magnitude of total acceleration is 181.05 m/s².

(b)

Angle of total acceleration relative to radial direction is given by the formula:

$\theta=\tan^{-1}(\frac{a_t}{a_r})\\\\\theta=\tan^{-1}(\frac{41.236}{176.29})\\\\\theta=13.2\°$

Therefore, the total acceleration makes an angle of 13.2° relative to radial direction.

4 0

3 years ago