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

A person walks the path shown below. The total trip consists of four straight-line paths.

dmitriy555 [2]

At the end of the walk, the person's resultant displacement is 495.1 m at 63⁰ south of west.

<h3>What is resultant displacement?</h3>

The resultant displacement of an object is the change in position of the object. It can be described as the shortest distance connecting the final position of the object to the initial position of the object.

<h3>Net horizontal displacement </h3>

Path 1 = 40 m

Path 2 = 0 m

Path 3 = 110 m x cos(30) = 95.26 m

Path 4 = 180 m x cos(60) = 90 m

Total horizontal displacement, X = 40 m + 0 m + 95.26 m + 90 m = 225.26 m

<h3>Net vertical displacement </h3>

Path 1 = 0 m

Path 2 = 230 m

Path 3 = 110 m x sin(30) = 55 m

Path 4 = 180 m x sin(60) = 155.885 m

Total horizontal displacement, Y = 0 m + 230 m + 55 m + 155.885 m = 440.885 m

<h3>Resultant displacement</h3>

R = √(X² + Y²)

R = √(225.26² + 440.885²)

R = 495.1 m

<h3>Direction of the displacement</h3>

θ = arc tan (Y/X)

θ = arc tan (440.885/225.26)

θ = 63⁰

Thus, at the end of the walk, the person's resultant displacement is 495.1 m at 63⁰ south of west.

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3 0

9 months ago

You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body

stellarik [79]

A) The change in internal chemical energy is $1.15\cdot 10^7 J$

B) The time needed is 1 minute

Explanation:

First of all, we start by calculating the power output of you and the bike, given by:

$P=Fv$

where

F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed

v = 8.0 m/s is the velocity

Substituting,

$P=(80)(8.0)=640 W$

The energy output is related to the power by the equation

$P=\frac{E}{t}$

where:

P = 640 W is the power output

E is the energy output

$t = 30 min \cdot 60 = 1800 s$ is the time elapsed

Solving for E,

$E=Pt=(640)(1800)=1.15\cdot 10^6 J$

Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by

$\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J$

B)

From the previous part, we found that in a time of

t = 30 min

the amount of internal chemical energy converted is

$E=1.15\cdot 10^7 J$

Here we want to find the time t' needed to convert an amount of chemical energy of

$E'=3.8\cdot 10^5 J$

So we can setup the following proportion:

$\frac{t}{E}=\frac{t'}{E'}$

And solving for t',

$t'=\frac{E't}{E}=\frac{(3.8\cdot 10^5)(30)}{1.15\cdot 10^7}=1 min$

Learn more about power and energy:

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3 0

2 years ago

A spider begins to spin a web by first hanging from a ceiling by his fine, silk fiber. He has a mass of 0.025 kg and a charge of

Rasek [7]

Answer:

a) (5.59 × 10³) N/C

b) 0.226 N directed away from the spider.

Explanation:

a) Electric field, E, felt as a result of point charge, Q, at a distance, d away is given by

E = kQ/d²

So, magnitude of the electric field due to the charge on the second spider at the position of the first spider

Q = 4.2 µC = 4.2 × 10⁻⁶ C

k = Coulomb's constant = 8.99 × 10⁹ Nm²/C

d = 2.6 m

E = (8.99 × 10⁹ × 4.2 × 10⁻⁶)/2.6²

E = 5.59 × 10³ N/C

b) Tension in the silk fiber above the spider is the net force due to the weight of spider one and the force of repulsion due the two charges.

Force due to the two charges = Eq

where q now represents the charge of the first spider at the first point, feeling the electric field calculated in (a)

F = 5.59 × 10³ × 3.4 × 10⁻⁶ = 0.01901 N directed upwards. (That is, F = + 0.019 N)

Weight of the spider = mg = 0.025 × 9.8 = 0.245 N directed downwards. (That is, W = -0.245 N)

Net force, T = mg - F = 0.245 - 0.019 = 0.226 N (that is, 0.226 N, directed upwards, away from the spider).

8 0

3 years ago

which layer of the atmosphere is directly above the troposphere A. exosphere B. mesosphere C. troposphere or D. stratosphere

lozanna [386]

The layer above it would be D. the Stratosphere

8 0

2 years ago

Read 2 more answers

An electron moves at 2.80 ×106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.

d1i1m1o1n [39]

Answer:

$a = 3.73*10^{16}m/s^2$

Explanation:

The magnetic force is given by the equation,

$\vec{F}_{magnetic} = q (\vec{v}X\vec{B})$

$\vec{F}_{magnetic} = qvBsin\theta$

Where $\theta$ is the angle between velocity vector and the magnetic field,

That angle is 90°.

We know as well that

F=ma, replacing the mass and the acceleration in our previous equation we have

$ma = qvBsin\theta$

$a= \frac{qvBsin\theta}{m}$

Our values are,

$q= 1.6*10^{-19}c$

$v=2.8*10^6m/s$

$B=7.6*10^{-2}T$

$m=9.11*10^{-31}kg$

Substituting,

$a = \frac{(1.6*10^{-19})(2.8*10^6)(7.6*10^{-2})}{9.11*10^{-31}}$

$a = 3.73*10^{16}m/s^2$

4 0

3 years ago