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Nutka1998 [239]

3 years ago

8

Two square air-filled parallel plates that are initially uncharged are separated by 1.2 mm, and each of them has an area of LaTe

X: 190~mm^2190 m m 2. How much charge must be transferred from one plate to the other if 1.1 nJ of energy are to be stored in the plates

1 answer:

RideAnS [48]3 years ago

7 0

Answer:

5.5x 10^-11 C

Explanation:

Pls see attached file

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Which type of electromagnetic radiation carries the most energy and has the highest frequency?

jek_recluse [69]

Answer:

Gamma rays

Explanation:

Gamma rays is at the end of the electromagnetic spectrum, and has the highest energy. It propagates through space at 3x10^8 m/s and has the smallest wavelength and the highest frequency. It is given off by atoms of element as they undergo nuclear disintegration.

8 0

3 years ago

Read 2 more answers

A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. A

Julli [10]

Answer:

1) p₀ = 0.219 kg m / s, p = 0, 2) Δp = -0.219 kg m / s, 3) 100%

Explanation:

For the first part, which is speed just before the crash, we can use energy conservation

Initial. Highest point

Em₀ = U = mg y

Final. Low point just before the crash

Emf = K = ½ m v²

Em₀ = Emf

m g y = ½ m v²

v = √ 2 g y

Let's calculate

v = √ (2 9.8 0.05)

v = 0.99 m / s

1) the moment before the crash is

p₀ = m v

p₀ = 0.221 0.99

p₀ = 0.219 kg m / s

After the collision, the car's speed is zero, so its moment is zero.

p = 0

2) change of momentum

Δp = p - p₀

Δp = 0- 0.219

Δp = -0.219 kg m / s

3) the reason is

Δp / p = 1

In percentage form it is 100%

3 0

3 years ago

If the energy bar was his only fuel, how far could a 68 kg person walk at 5 km/h?

goldfiish [28.3K]

An energy bar contains of 20grams of carbohydrates and 1gram is equal to 17 KJ. If the energy bar was his only fuel the total energy available is equal to (17,000 x 20) = 340,000j. Estimate KJ per hour is equal to 1539KJ, it has Time equal to (340,000/1539,000) is equal to 0.2209 hour and his Distance is equal to (5,000 x .2209) = 1.1045km.

3 0

3 years ago

A fourth-grade teacher wants to ensure students understand the life cycle process of plants. The teacher decides to incorporate

Annette [7]

Answer:

There are several options that the teacher can use to incorporate the concept into students' understanding.

Explanation:

1. The students can draw all the plants that they know.

2. Children can be asked to bring the flowers to school so that they can identify the plants themselves.

3. The children can plat the flowers in makeshift pots and then take the best plants and transplant them in the garden or elsewhere.

4. The children can take occasional trips and observe and record any changes to the plants.

4. The teacher can ask the students to draw the flowers and emphasize on the productive parts like the stamens, leaves, pistils, stems.

4 0

3 years ago

A horizontal spring-mass system has low friction, spring stiffness 165 N/m, and mass 0.6 kg. The system is released with an init

AURORKA [14]

a) 19.4 cm

b) 3.2 m/s

Explanation:

a)

A horizontal spring-mass system has a motion called simple harmonic motion, in which the mass oscillates following a periodic function (sine or cosine) around an equilibrium position.

As the system oscillates back and forth, its total mechanical energy (sum of elastic potential energy and kinetic energy) will remain conserved (since we consider friction negligible). The elastic potential energy at any point is given by:

$U=\frac{1}{2}kx^2$

where

k is the spring constant

x is the displacement of the system

While the kinetic energy at any point is

$K=\frac{1}{2}mv^2$

where

m is the mass

v is the speed

So the total mechanical energy of the system is

$E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2$

For this system, when it is initially released,

m = 0.6 kg

k = 165 N/m

x = 7 cm = 0.07 m

v = 3 m/s

So the total energy is

$E=\frac{1}{2}(0.6)(3)^2+\frac{1}{2}(165)(0.07)^2=3.1 J$

Since friction is negligible, this total energy remains constant. Therefore, when the system reaches its maximum stretch during the motion, the kinetic energy will be zero and all the mechanical energy will be elastic potential energy; so we will have:

$E=U=\frac{1}{2}kx_{max}^2$

where $x_{max}$ is the maximum stretch. Solving for $x_{max}$ ,

$x_{max}=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(3.1)}{165}}=0.194 m$

So, 19.4 cm.

b)

The maximum speed in a spring-mass oscillating system is reached when the kinetic energy is maximum, and therefore, since the total energy is conserved, when the elastic potential energy is zero:

$U=0$

which means when the displacement is zero:

x = 0

So, when the system is transiting through the equilibrium position.

Therefore, the total mechanical energy is equal to the maximum kinetic energy:

$E=K=\frac{1}{2}mv_{max}^2$

where

m is the mass

$v_{max}$ is the maximum speed

Here we have:

E = 3.1 J

m = 0.6 kg

Therefore, solving for the maximum speed,

$v_{max}=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(3.1)}{0.6}}=3.2 m/s$

6 0

3 years ago