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Tresset [83]
4 years ago
15

A stack of bricks weighs 170 KN (Kilo-newtons). The stack exerts 180 KPa (Kilo-pascals) of pressure on the ground. What is the a

rea upon which this pressure is exerted (in square ft)?
Physics
1 answer:
Alex17521 [72]4 years ago
5 0

Answer:

10.12square feet

Explanation:

Pressure exerted on the object is defined as the ratio of the force exerted on it to its unit area. Mathematically, Pressure = Force/Area

Given the force = 170kN

Pressure = 180KPa

Area = Force /Pressure

Area = 170kN/180KPa

Area = 0.94N/Pa

Note that 1Newton/Pascal = 10.764square feet

Therefore 0.94N/Pa = x

x = 0.94× 10.764

x = 10.12square feet

Therefore the area upon which this pressure is exerted is 10.12sqft.

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Draw two unique free body diagrams that each show a net force of 30 N to the left.​
Reil [10]

Answer:

just trace a picture of it.

7 0
3 years ago
A major-league pitcher can throw a ball in excess of 40.1 m/s. If a ball is thrown horizontally at this speed, how much will it
mote1985 [20]

Answer:

The ball will drop 0.881 m by the time it reaches the catcher.

Explanation:

The position of the ball at time "t" is described by the position vector "r":

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

When the ball reaches the catcher, the position vector will be "r final" (see attached figure).

The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.

Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):

x = x0 + v0x · t

17.0 m = 0 m + 40.1 m/s · t

t = 17.0 m/ 40. 1 m/s = 0.424 s

With this time, we can calculate the y-component of the vector "r final", the drop of the ball:

y = y0 + v0y · t + 1/2 · g · t²

Initially, there is no vertical velocity, then, v0y = 0.

y = 1/2 · g · t²

y = -1/2 · 9.8 m/s² · (0.424 s)²

y = -0.881 m

The ball will drop 0.881 m by the time it reaches the catcher.

8 0
3 years ago
Kon'nichiwa~<br>please help me with this question!!​
andrezito [222]

Let's check the relationship

\\ \rm\hookrightarrow g=\dfrac{GM}{r^2}

\\ \rm\hookrightarrow g\propto G

So

  • Raindrops will fall faster . .
  • Also walking on ground would become more difficult as g increases.

Option C is wrong by now .Let's check D once

\\ \rm\hookrightarrow T\propto \dfrac{1}{\sqrt{g}}

  • So time period of simple pendulum would decrease.

4 0
2 years ago
As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p
coldgirl [10]

Answer:

Explanation:

charge, q = 1.6 x 10^-19 C

distance, r = 911 nm = 911 x 10^-9 m

The Coulomb's force is given by

F=\frac{Kq_{1}q_{2}}{r^{2}}

F=\frac{9\times 10^{9}\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{\left (911\times 10^{-9}  \right )^2}

F = 2.78 x 10^-16 N

The force between the electron and the proton is 2.78 x 10^-16 N.

4 0
3 years ago
The gold has a density of 19300 kg/m3 calculate the mass of one gold bar 1= 2.54cm
icang [17]
Fair enough, but you'll have to tell us the volume of the bar first.
5 0
3 years ago
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