All categories

3 years ago

A box exerts 10,000 Pa of pressure on the ground. If the box weighs 1000 N, how much area is in contact with the ground?

Physics

1 answer:

lara [203]3 years ago

4 0

Pressure = (total force) / (Area)

10,000 Pa = (1,000 N) / (Area)

Multiply each side by (Area) :

(10,000 Pa) x (Area) = 1,000 N

Divide each side by (10,000 Pa) :

Area = (1,000 N) / (10,000 Pa)

You might be interested in

What is the distance fallen for a freely falling object 1 s after being dropped from a rest position? What is the distance for a

kondor19780726 [428]

<h2>Distance traveled in 1 second after drop is 4.9 m</h2><h2>Distance traveled in 4 seconds after drop is 78.4 m</h2>

Explanation:

We have s = ut + 0.5at²

For a free falling object initial velocity u = 0 m/s and acceleration due to gravity, g = 9.8 m/s²

Substituting

s = 0 x t + 0.5 x 9.8 x t²

s = 4.9t²

We need to find distance traveled in 1 s and 4 s

Distance traveled in 1 second

s = 4.9 x 1² = 4.9 m

Distance traveled in 4 seconds

s = 4.9 x 4² = 78.4 m

Distance traveled in 1 second after drop = 4.9 m

Distance traveled in 4 seconds after drop = 78.4 m

5 0

3 years ago

A 100-m-long wire carrying a current of 4.0 A will be accompanied by a magnetic field of what strength at a distance of 0.050 m

solong [7]

Answer:

1.6 x 10^-5 T

Explanation:

i = 4 A

r = 0.05 m

The magnetic field due to long wire at a distance r is given by

$B = \frac{\mu _{0}}{4\pi }\times \frac{2i}{r}$

B = 10^-7 x 2 x 4 / 0.05

B = 1.6 x 10^-5 T

3 0

3 years ago

a track coach with a meter stick and a stopwatch is trying to determine if a student is walking with constant speed.he should...

Aleksandr [31]

Give you a device that tracks your speed?

7 0

3 years ago

A batter hits a fly ball which leaves the bat 0.89 m above the ground at an angle of 62 ∘ with an initial speed of 29 m/s headin

KatRina [158]

consider the motion in Y-direction

v₀ = initial velocity = 29 Sin62 = 25.6 m/s

a = acceleration = - 9.8 m/s²

t = time of travel

Y = vertical displacement = - 0.89 m

using the equation

Y = v₀ t + (0.5) a t²

- 0.89 = (25.6) t + (0.5) (- 9.8) t²

t = 5.3 sec

consider the motion along the horizontal direction :

v₀ = initial velocity = 29 Cos62 = 13.6 m/s

a = acceleration = 0 m/s²

t = time of travel = 5.3 sec

X = horizontal displacement =?

using the equation

X = v₀ t + (0.5) a t²

X = (13.6) (5.3) + (0.5) (0) t²

X = 72.1 m

d = distance traveled by the center fielder to catch the ball = 107 - x = 107 - 72.1 = 34.9 m

t = time taken = 5.3 sec

v = speed of center fielder

using the equation

v = d/t

v = 34.9/5.3

v = 6.6 m/s

3 0

3 years ago

A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.

alex41 [277]

The magnitude of the unknown height of the projectile is determined as 16.1 m.

<h3>Magnitude of the height</h3>

The magnitude of the height of the projectile is calculated as follows;

H = u²sin²θ/2g

H = (36.6² x (sin 29)²)/(2 x 9.8)

H = 16.1 m

Thus, the magnitude of the unknown height of the projectile is determined as 16.1 m.

Learn more about height here: brainly.com/question/1739912

#SPJ1

3 0

2 years ago