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2 years ago

Suppose a force of 1.00 x 104 n is put on a 10.0 cm by 10.0 cm square. What is the pressure?

Physics

1 answer:

ddd [48]2 years ago

8 0

Answer:

$P=1000000Pa=1MPa=1000kPa$

Explanation:

When the force that is applied is normally and uniformly distributed on a surface, the magnitude of the pressure is obtained by dividing the force applied on the corresponding area:

$P=\frac{F}{A}$

Where:

$P=Pressure\hspace{3}in\hspace{3}Pa.\\F=Force\hspace{3}in\hspace{3}N\\A=Area\hspace{3}in\hspace{3}m^2$

Converting cm to m:

$10cm*\frac{1m}{100cm} = 0.1m$

The area of a square is:

$A=l^2=(0.1)^2=0.01m^2$

Therefore:

$P=\frac{1.00\times10^4}{0.01} =1000000Pa=1MPa=1000kPa$

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2 years ago

It is noted that a sample of 134La with a half-life of 6.5 minutes has an activity of 2.6 Ci. What was its activity 72 minutes a

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Answer:

36//

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2 years ago

A penny is placed at the outer edge of a disk (radius = 0.179 m) that rotates about an axis perpendicular to the plane of the di

fgiga [73]

Answer:

$\mu = 0.28$

Explanation:

Time period of the revolution of the disc is given as

$T = 1.60 s$

now the angular speed of the disc is given as

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

so we will have

$\omega = \frac{2\pi}{1.60}$

$\omega = 3.92 rad/s$

now at the rim position of the disc the net force on the penny is due to friction force

So we will have

$\mu mg = m\omega^2 r$

here we will have

$\mu g = \omega^2 r$

$\mu = \frac{\omega^2 r}{g}$

$\mu = \frac{3.92^2 (0.179)}{9.81}$

$\mu = 0.28$

7 0

3 years ago

A hollow plastic sphere is held below the surface of a fresh-water lake by a cord anchored to the bottom of the lake. The sphere

Naya [18.7K]

Answer: a) B = 6811N

b) m = 603.2kg

c) 86.8%

Explanation: Buoyant force is a force a fluid exerts on a submerged object.

It can be calculated as:

$B=\rho_{fluid}.V_{obj}.g$

where:

$\rho_{fluid}$ is density of the fluid the object is in;

$V_{obj}$ is volume of the object;

g is acceleration due to gravity, is constant and equals 9.8m/s²

a) For the hollow plastic sphere, density of water is 1000kg/m³:

$B=10^{3}.0.695.9.8$

B = 6811N

b) Anchored to the bottom, the forces acting on the sphere are Buoyant, Tension and Force due to gravity:

B = T + $F_{g}$

B = T + mg

mg = B - T

$m=\frac{B - T}{g}$

Calculating:

$m=\frac{6811 - 900}{9.8}$

m = 603.2kg

c) When the shpere comes to rest on the surface of the water, there are only buoyant and gravity acting on it:

B = m.g

$\rho_{w}.V_{sub}.g=m.g$

$V_{sub}=\frac{m}{\rho_{w}}$

$V_{sub}=\frac{603.2}{1000}$

$V_{sub}$ = 0.6032m³

Fraction of the submerged volume is:

$\frac{V_{sub}}{V_{obj}}$ = $\frac{0.6932}{0.695}$ = 0.868

3 0

2 years ago

Part D

Montano1993 [528]

Answer:

Explanation:

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4 0

2 years ago