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3 years ago

A conservative force does the same work regardless of the path taken.

2 answers:

5 0

A conservative force is a force that when work is done against this force the work done does not depend on the path taken only the initial and final position.

Nimfa-mama [501]3 years ago

5 0

Answer:

True

Explanation:

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Lasers utilize concentrated light waves, focused precisely and used for certain applications. What are some common uses of laser

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3 years ago

Read 2 more answers

A mover loads a crate onto a truck bed 1.6m from the street using a ramp that is 4.6m long. What is a mechanical advantage?

Ne4ueva [31]

Answer:

Mechanical advantage = 2.875

Explanation:

Given:

A diagram is shown below for the above scenario.

Length of ramp (Effort arm) = 4.6 m

Height of truck bed ( Resistance length) = 1.6 m

Mechanical advantage (MA) is the ratio of effort arm and resistance length.

So, mechanical advantage is given as,

$MA=\frac{\textrm{Effort arm}}{\textrm{Resistance length}}= \frac{4.6}{1.6}=2.875$

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3 years ago

asambeis [7]

Answer:

TAJUK

Explanation:

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3 years ago

A 530-g squirrel with a surface area of 935 cm2 falls from a 4.4-m tree to the ground. Estimate its terminal velocity. (Use the

Agata [3.3K]

Answer with Explanation:

We are given that

Mass of squirrel,m=530 g= $\frac{530}{1000}=0.530 kg$

1kg=1000 g

Area=A= $935 cm^2=935\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

Height,h=4.4 m

C=1

$\rho=1.21 kg/m^3$

Width of rectangular prism,b=11.6 cm= $\frac{11.6}{100}=0.116 m$

1 m=100 cm

Length,l=23.2 cm= $0.232 m$

Area= $l\times b=0.116\times 0.232=0.0269 m^2$

Terminal velocity, $v_t=\sqrt{\frac{2mg}{\rho CA}}$

Where $g=9.8 m/s^2$

Using the formula

$v_t=\sqrt{\frac{2\times 0.530\times 9.8}{1.21\times 1\times 0.0269}}$

$v_t=17.86 m/s$

The velocity of person, $v=\sqrt{2gh}$

Using the formula

$v=\sqrt{2\times 9.8\times 4.4}$

$v=9.29 m/s$

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3 years ago

Kinematics practice problems Answers: 4. A race car is traveling at +76 m/s when is slows down at -9 m/s2 for 4

arlik [135]

The new velocity after 4 s is 40 m/s

The height of the spaceship above the ground after 5 seconds is 1,127.5 m

The given parameters for the first question;

initial velocity of the car, u = 76 m/s

acceleration of the car, a = - 9 m/s²

time of motion, t = 4 s

The new velocity after 4 s is calculated as;

v = u + at

v = 76 + (-9)(4)

v = 76 - 36

v = 40 m/s

(5)

The given parameters;

height above the ground, h = 500 m

velocity of spaceship, u = 150 m/s

time of motion, t = 5

The height of the spaceship above the ground after 5 seconds is calculated as;

$h_y = h_0 + ut - \frac{1}{2}gt^2\\\\h_y = 500 + (150\times 5) - (0.5\times 9.8 \times 5^2)\\\\h_y = 1,127.5 \ m$

Learn more here: brainly.com/question/24527971

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2 years ago