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3 years ago

Where does the kinetic energy people get during the sport come from?

Physics

2 answers:

gizmo_the_mogwai [7]3 years ago

8 0

Answer:

kinetic is the stored energy being released from being dormant

Explanation:

i copied

adelina 88 [10]3 years ago

5 0

Answer:

kinetic is the stored energy being released from being dormant

Explanation:

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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

3 years ago

According to the theory of plate tectonics, the lithosphere is separated into sections that are called tectonic plates. These pl

loris [4]

Answer:

Transform :j

Explanation:

4 0

4 years ago

A 14.0 g wad of sticky clay is hurled horizontally at a 110 g wooden block initially at rest on a horizontal surface. The clay s

ahrayia [7]

Answer:

86.53 m/s

Explanation:

Given:

Mass of clay (m) = 14.0 g = 0.014 kg

Mass of block (M) = 110 g = 0.110 kg

Initial speed of block (U) = 0 m/s

Sliding distance (d) = 7.50 m

Coefficient of friction between block and surface (μ) = 0.650

Let the initial speed of clay be 'u' and speed of clay and block just after collision be 'v'.

Now, momentum is conserved just before and just after collision.

Momentum just before collision = $mu + 0 = mu$

Momentum just after collision = $(m + M)v$

Therefore, $mu=(M+m)v$ --------- (1)

Now, using newton's second law and we find the acceleration of the system.

The frictional force is given as:

$f=\mu mg=-ma\\\\a=-\mu g$

Now, using equation of motion, we can find the velocity just after collision.

$0^2=v^2+2ad\\\\v=\sqrt{-2ad}\\\\v=\sqrt{-2\times (-\mu g)\times d}\\\\v=\sqrt{2\mu gd}$

Plug in the given values and find 'v'. This gives,

$v=\sqrt{2\times 0.650\times 9.8\times 7.50}\\\\v=\sqrt{95.55}=9.77\ m/s$

Now, using equation (1) and substituting the given values, we get:

$0.014u=(0.014+0.110)\times 9.77\\\\u=\frac{0.124\times 9.77}{0.014}\\\\u=86.53\ m/s$

Therefore, the speed of the clay immediately before impact is 86.53 m/s.

7 0

3 years ago

What is the wavelength of an electromagnetic wave

VMariaS [17]

Answer:

Explanation:

Electromagnetic waves are categorized according to their frequency f or, equivalently, according to their wavelength λ = c/f. Visible light has a wavelength range from ~400 nm to ~700 nm.

The electromagnetic spectrum (EMS) is the general name given to the known range of electromagnetic radiation. Wavelengths increase from approximately 10-18 m to 100 km, and this corresponds to frequencies decreasing from 3 × 1026 Hz to 3 ×103 Hz.

3 0

3 years ago

A mail carrier leaves the post ofﬁce and drives 2.00 km to the north. He then drives in a direction 60.0° south of east for 7.00

Oxana [17]

Answer:

12.97 km

Explanation:

In order to find the resultant displacement, we have to resolve each of the 3 displacements along the x and y direction.

Taking north as positive y direction and east as positive x-direction, we have:

- Displacement 1: 2.00 km to the north

$A_x = 0\\A_y = +2.00 km$