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3 years ago

a rectangular block weighs 240N. The area of the block in contact with the floor is 20cm2 calculate the pressure on the floor

Physics

1 answer:

Romashka [77]3 years ago

8 0

Answer:

The pressure on the floor is 12 Pascal(Pa).

Explanation:

Given,

Force (F) = 240N

Area (A) = 20 cm^2

Pressure (P) = ?

we know that,

P = F/A

= 240/ 20

= 12 Pa

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3 years ago

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

At x = 1.00 m

$u(1) = \frac{-2}{1} + \frac{4}{1^2}$

= 4J

Kinetic energy = (1/2)mv²

$= \frac{1}{2} * 4(3)^2$

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

$u(5) = \frac{-2}{5} + \frac{4}{5^2}$

$= \frac{4-10}{25} = \frac{-6}{25} J$

= -0.24J

Kinetic energy =

$\frac{1}{2} * 4Vf^2$

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

$Vf = \sqrt{frac{22.4}{2}$

= 3.33 m/s

b) magnitude of force when x = 5.0m

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

$\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}$

$= \frac{2}{x^2} - \frac{8}{x^3}$

At x = 5.0 m

$\frac{2}{5^2} - \frac{8}{5^3}$

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3 years ago

A diver can change his rotational inertia by drawing his arms andlegs close to his body in the tuck position. After he leaves th

NeX [460]

Answer:

3.14946 rad/s

Explanation:

$I_i$ = Intial moment of inertia

$I_f$ = Final moment of inertia

$\omega_i$ = Initial angular velocity

$\omega_f$ = Final angular velocity = $\dfrac{2}{1.33}\times 2\pi\ rad/s$

$\dfrac{I_f}{I_i}=\dfrac{1}{3}$

In this system the angular momentum is conserved

$L_i=L_f\\\Rightarrow I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_i=\dfrac{I_f\omega_f}{I_i}\\\Rightarrow \omega_i=\dfrac{1\times \dfrac{2}{1.33}\times 2\pi}{3}\\\Rightarrow \omega_i=3.14946\ rad/s$

The angular velocity when the diver left the board is 3.14946 rad/s

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2 years ago