All categories

3 years ago

A box has a mass of 150kg and the surface area of the bottom of the box is :

Physics

1 answer:

charle [14.2K]3 years ago

3 0

Answer:

Pressure = 100 Pascals

Explanation:

$pressure = \frac{force}{area} \\ = \frac{150 \times 10}{15} \\ = \frac{1500}{15} \\ = 100 pa$

You might be interested in

Which is an example of a physical change

mote1985 [20]

Water boiling is an example of a physical change. The rest are chemical changes.
Hope that helps!!

6 0

3 years ago

What type of fat may help lower the risk of heart disease

enyata [817]

Polyunsaturated fatty acids which is Omega-3 fatty acids.

8 0

3 years ago

a mass on a spring vibrates in simple harmonic motion at an amplitude of 8.0 cm. if the mass of the object is 0.20kg and the spr

Reil [10]

Answer:

4.06 Hz

Explanation:

For simple harmonic motion, frequency is given by

$f=\frac {1}{2\pi}\times \sqrt{\frac {k}{m}}$ where k is spring constant and m is the mass of the object.

Substituting 0.2 Kg for mass and 130 N/m for k then

$f=\frac {1}{2\pi}\times \sqrt{\frac {130}{0.2}}=4.057670803\\f\approx 4.06 Hz$

5 0

3 years ago

A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc

Harman [31]

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a (→)

F - Ffriction = m*a ⇒ F - (μ*N) = F - (μ*m*g) = m*a ⇒ a = (F - μ*m*g)/m

⇒ a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x ⇒ vf = √(vi² + 2*a*x)

⇒ vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

7 0

3 years ago

Write an expression for a harmonic wave with an amplitude of 0.19 m, a wavelength of 2.6 m, and a period of 1.2 s. The wave is t

zlopas [31]

Answer:

$y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})$

Explanation:

As we know that the wave equation is given as

$y = A sin(\omega t - k x + \phi_0)$

now we have

$A = 0.19 m$

$\lambda = 2.6 m$

so we have

$k = \frac{2\pi}{\lambda}$

$k = \frac{2\pi}{2.6}$

$k = 2.42 per m$

also we have

$T = 1.2 s$

so we have

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{1.2}$

$\omega = 5.23 rad/s$

now we know that at t = 0 and x = 0 wave is at y = 0.19 m

so we have

$\phi_0 = \frac{\pi}{2}$

so we have

$y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})$

6 0

3 years ago