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Lera25 [3.4K]
1 year ago
5

Consider a force of 750 n (roughly the weight of an adult human). over what area (in cm2) would this force need to be applied in

order achieve a solid pressure (stress) of 25 mpa (the pressure required to rupture average human skin)? [recall : 1 mpa = 106 pa , and 1 pa = 1 n/m2]
Physics
1 answer:
polet [3.4K]1 year ago
7 0
p = \frac{f}{a}
P=25x10^6 andF=750.So plug in everything to solve for A. which is 3x10^-5m^2 OR 0.3mm^2
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A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn
EastWind [94]

(a) 12.0 V

In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, C=5.00 \mu F, the charged stored on the capacitor at the end of the process is

Q=CV=(5.00 \mu F)(12.0 V)=60 \mu C

When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

V=\frac{Q}{C}

and since neither Q nor C change, the voltage V remains the same, 12.0 V.

(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

C=\frac{\epsilon_0 \epsilon_r A}{d}

where:

\epsilon_0 is the permittivity of free space

\epsilon_r is the relative permittivity of the material inside the capacitor

A is the area of the plates

d is the separation between the plates

As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: C'=\frac{C}{2}. The new voltage across the plate is given by

V'=\frac{Q}{C'}

and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is

V'=\frac{Q}{C'}=\frac{Q}{C/2}=2 \frac{Q}{C}=2V

So, the new voltage is

V'=2 (12.0 V)=24.0 V

(c) (ii) 3.0 V

The area of each plate of the capacitor is given by:

A=\pi r^2

where r is the radius of the plate. In this case, the radius is doubled: r'=2r. Therefore, the new area will be

A'=\pi (2r)^2 = 4 \pi r^2 = 4A

While the separation between the plate was unchanged (d); so, the new capacitance will be

C'=\frac{\epsilon_0 \epsilon_r A'}{d}=4\frac{\epsilon_0 \epsilon_r A}{d}=4C

So, the capacitance has increased by a factor 4; therefore, the new voltage is

V'=\frac{Q}{C'}=\frac{Q}{4C}=\frac{1}{4} \frac{Q}{C}=\frac{V}{4}

which means

V'=\frac{12.0 V}{4}=3.0 V

3 0
1 year ago
There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the differe
Usimov [2.4K]

An example that illustrates the difference is the circular motion

Explanation:

Let's start by reminding the definition of the two quantities:

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Speed can be  calculate as:

speed = \frac{d}{t}

where:

d is the distance travelled

t is the time taken

- Velocity is instead a vector quantity, given by:

velocity = \frac{d}{t}

where;

d is the displacement of the object (displacement is a a vector connecting the initial position to the final position of motion)

t is the time taken

Since it is a vector, velocity has both a magnitude and a direction, therefore it also takes into account the direction of motion of the object.

For an object in motion in a straight line, speed and velocity are the same. However, this is not always the case.

In fact, an example of motion in which the two quantities are different is the circular motion. Consider for example the object making one complete revolution along the circle. Therefore, its average speed is the ratio between the length of the perimeter (the distance) divided by the time taken:

speed = \frac{2\pi r}{t}

where r is the radius of the circle.

However, the displacement of the object is zero (because the object returns to the starting point), and so the average velocity is also zero:

velocity = \frac{0}{t}=0

Learn more about speed and velocity:

brainly.com/question/8893949

brainly.com/question/5063905

brainly.com/question/5248528

#LearnwithBrainly

5 0
2 years ago
The δe of a system that releases 12.4 j of heat and does 4.2 j of work on the surroundings is __________ j.
Vedmedyk [2.9K]

The answer for this problem is clarified through this, the system is absorbing (+). And now see that it uses that the SURROUNDINGS are doing 84 KJ of work. Any time a system is overshadowing work done on it by the surroundings the sign will be +. So it's just 12.4 KJ + 4.2 = 16.6 KJ.

5 0
2 years ago
A pickup truck starts from rest and maintains a constant acceleration a0. After a time t0, the truck is moving with speed 25 m/s
Anastaziya [24]

Answer:

the correct answer is c     v₁> 12.5 m / s

Explanation:

This is a one-dimensional kinematics exercise, let's start by finding the link to get up to speed.

            v² = v₀² + 2 a₁ x

as part of rest v₀ = 0

           a₁ = v² / 2x

           a₁ = 25² / (2 120)

           a₁ = 2.6 m / s²

now we can find the velocity for the distance x₂ = 60 m

           v₁² = 0 + 2 a1 x₂

           v₁ = Ra (2 2,6 60)

           v₁ = 17.7 m / s

these the speed at 60 m

we see that the correct answer is c     v₁> 12.5 m / s

5 0
2 years ago
A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler
aleksandrvk [35]

Answer:

Part a)

t_1 = \frac{\omega_1}{\alpha}

Part b)

\theta = \frac{1}{2}\alpha t_1^2

Part c)

t = \frac{t_1}{5}

Explanation:

Part a)

As we know that it is having constant torque so here the time taken by it to accelerate is given as

\omega_f = \omega_i + \alpha t

\omega_1 = 0 + \alpha t_1

t_1 = \frac{\omega_1}{\alpha}

Part b)

angular displacement is given as

\theta = \omega_i t_1 + \frac{1}{2}(\alpha) t_1^2

\theta = 0 + \frac{1}{2}\alpha t_1^2

\theta = \frac{1}{2}\alpha t_1^2

Part c)

As we know that the angular deceleration produced by the brakes is given as

\alpha_d = - 5\alpha

now we have

\omega_f = \omega _i + \alpha t

0 = \omega_1 - 5 \alpha_1 t

t = \frac{\omega_1}{5 \alpha}

As we know that

t_1 = \frac{\omega_1}{\alpha}

so we have

t = \frac{t_1}{5}

6 0
2 years ago
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