1 5kg cat is lifted 2 m into the air.how much gpe does it gain?

A toy gun uses a spring to project a 4.5-g soft rubber sphere horizontally. The spring constant is 8.0 N/m, the barrel of the gu

marishachu [46]

Answer:

1.93 m/s

Explanation:

Parameters given:

Mass = 4.5g = 0.0045kg

Spring constant = 8.0 N/m

Length of barrel = 13 cm = 0.013m

Frictional force = 0.035N

Compression = 5.8 cm = 0.058m

First, we find the P. E. stored in the spring:

P. E. = ½*k*x²

P. E. = ½ * 8 * 0.058² = 0.013J

Then, we find the work done by the frictional force while the sphere is leaving the barrel of the gun:

Work = Force * distance

The distance here is the length of the barrel.

Work = 0.035 * 0.13 = 0.0046 J

The kinetic energy of the sphere can now be found:

K. E. = P. E. - Work done

K. E. = 0.013 - 0.0046 = 0.0084J

We can now find the speed using the formula for K. E.:

K. E. = ½*m*v²

0.0084 = ½ * 0.0045 * v²

v² = 0.0084/0.00255 = 3.733

=> v = 1.93 m/s

4 0

3 years ago

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How do you find the oscillation period in seconds for different pendulum lengths?

GalinKa [24]

To find:

The equation to find the period of oscillation.

Explanation:

The period of oscillation of a pendulum is directly proportional to the square root of the length of the pendulum and inversely proportional to the square root of the acceleration due to gravity.

Thus the period of a pendulum is given by the equation,

$T=2\pi\sqrt{\frac{L}{g}}$

Where L is the length of the pendulum and g is the acceleration due to gravity.

On substituting the values of the length of the pendulum and the acceleration due to gravity at the point where the period of the pendulum is being measured, the above equation yields the value of the period of the pendulum.

Final answer:

The period of oscillation of a pendulum can be calculated using the equation,

$T=2\pi\sqrt{\frac{L}{g}}$

3 0

8 months ago

A force vector points due east and has a magnitude of 140 newtons. A second force is added to . The resultant of the two vectors

ale4655 [162]

Answer:

(a) When the resultant force is pointing along east line, the magnitude and direction of the second force is 280 N East

(b) When the resultant force is pointing along west line, the magnitude and direction of the second force is 560 N West

Explanation:

Given;

a force vector points due east, $F_1$ = 140 N

let the second force = $F_2$

let the resultant of the two vectors = F

(a) When the resultant force is pointing along east line

the second force must be pointing due east

$F = F_1 + F_2\\\\F_2 = F - F_1\\\\F_2 = 420 \ N - 140 \ N\\\\F_2 = 280 \ N$

$F_2 = 280 \ East$

(b) When the resultant force is pointing along west line

the second force must be pointing due west and it must have a greater magnitude compared to the first force in order to have a resultant in west line.

$F = F_2 - F_1\\\\F_2 = F + F_1\\\\F_2 = 420 \ N + 140 \ N\\\\F_2 = 560 \ N$

$F_2 = 560 \ West$

8 0

3 years ago

in the hydrologic cycle, water from the ocean _____ (condenses, precipitates, transpires,or evaporates) into the atmosphere wher

lutik1710 [3]

In the hydrologic cycle, water from the ocean evaporates into the atmosphere where it can condense then <span />

7 0

3 years ago

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what is the pressure exerted, what is the pressure exerted by 50kg girl as she places her weight on one shoe if the heels area i

lana [24]

Answer:

The pressure exerted by the girl is 245,000 N/m²

Explanation:

Given;

mass of the girl, m = 50 kg

area of the girl's shoe, A = 0.002 m²

The pressure exerted by the girl is calculated as follows;

$P = \frac{F}{A} \\\\Where;\\F \ is \ the \ force \ exerted \ by \ girl's \ weight\\\\P = \frac{F}{A} = \frac{mg}{A} = \frac{50 \times 9.8}{0.002} = 245,000 \ N/m^2$

Therefore, the pressure exerted by the girl is 245,000 N/m²

6 0

2 years ago