Applications of pressure

A playground slide is inclined 40°. If a boy with a mass of 32 kg slides down for -3meters. How much work is done by gravity on

telo118 [61]

Answer:

the work done by gravity on the boy is 604.62 J

Explanation:

Given;

distance the boy slides, d = 3 m

angle of inclination of the playground, θ = 40⁰

mass of the boy, m = 32 kg

The vertical height, h, above the ground through which the boy falls represents the height of the triangle which is the opposite side.

The distance through which the boy slides, d, represents the hypotenuse side of the right triangle.

$sin \theta = \frac{opposite }{hypotenuse} \\\\sin \theta = \frac{h}{d} \\\\h = dsin\theta\\\\h = 3 \times sin(40^0)\\\\h = 1.928 \ m$

The work done by gravity on the boy is calculated as;

W = P.E = mgh

= 32kg x 9.8m/s² x 1.928m

= 604.62 J

Therefore, the work done by gravity on the boy is 604.62 J

8 0

3 years ago

What length of a certain metal wire of diameter 0.15 mm is needed for the wire to have a resistance of 15 ω? the resistivity of

Murljashka [212]

The resistance of the cylindrical wire is $R=\frac{\rho l}{A}$ .

Here $R$ is the resistance, $l$ is the length of the wire and $A$ is the area of cross section. Since the wire is cylindrical $A=\frac{\pi d^2}{4}$ . Rearranging the above equation,

$A=\frac{\rho l}{R}\\ \frac{\pi d^2}{4}=\frac{\rho l}{R}\\ l=\frac{\pi d^2 R}{4 \rho}$

Here $d=0.15, R=15, \rho=1.68(10^{-8})$ .

Substituting numerical values,

$l=\frac{\pi 0.15^2(10^{-6}) (15)}{4 (1.68)(10^{-8})} \\ l=15.78$

The length of the wire is $15.78 \;\;m$

5 0

3 years ago

Why do electric fish need to force electric charges to move?

Anastasy [175]

The electric eel generates large electric currents by way of a highly specialized nervous system that has the capacity to synchronize the activity of disc-shaped, electricity-producing cells packed into a specialized electric organ. Hope this helps!! Good luck

3 0

2 years ago

A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod

ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = $8.0\times 10^{- 5}\ T$