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3 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]3 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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Two manned satellites approaching one another at a relative speed of 0.550 m/s intend to dock. The first has a mass of 2.50 ✕ 10

NNADVOKAT [17]

Answer: Their final relative velocity is -0.412 m/s.

Explanation:

According to the law of conservation,

$m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v$

Putting the given values into the above formula as follows.

$m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v$

$2.50 \times 10^{3} kg \times 0 m/s + 7.50 \times 10^{3} kg \times -0.550 m/s = (2.50 \times 10^{3} kg + 7.50 \times 10^{3} kg)v$

$-4.12 \times 10^{3} kg m/s = (10^{4} kg) v$

v = $\frac{-4.12 \times 10^{3} kg m/s}{10^{4} kg}$

= -0.412 m/s

Thus, we can conclude that their final relative velocity is -0.412 m/s.

8 0

3 years ago

A positive charge is placed in an electric field that points west. What direction is the force on the positive particle? I think

Kisachek [45]

Right. You are true. The direction of the electric field is defined to be
the direction of the force on a small positive charge placed in the field.

3 0

3 years ago

If the Earth got closer to the Sun, how would the gravity between the Sun and Earth change?

Neko [114]

Answer:

I think its 2. It would decrease

Explanation:

Hope this helps

4 0

2 years ago

Read 2 more answers

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

5 0

2 years ago

2. What is stroboscopic motion? -

Oduvanchick [21]

Answer: The illusion of motion that occurs when a stationary object is first seen briefly in one location and, following a short interval, is seen in another location.

Explanation:

6 0

3 years ago