When compared to other types of waves, electromagnetic waves differ because they are

Physics

1 answer:

expeople1 [14]3 years ago

7 0

They differ because they are transverse wave. That is their direction of travel is perpendicular to its vibrations.

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Can someone please help me with this question.

Hoochie [10]

Answer:

compressional wave

Explanation:

6 0

3 years ago

Read 2 more answers

The air within a piston equipped with a cylinder absorbs 565 J of heat and expands from an initial volume of 0.12 L to a final v

Alenkinab [10]

Answer:

The change in internal energy of the air within the piston is 490 J.

Explanation:

In thermodynamics the internal energy (ΔU) of a system is the total energy contained in the system and can be considered as the sum of the energy in the form of heat (q, given to or released by the system) and in the form of work (w, make by the system to the environment or from the environment to the system). When the system absorbs heat from the enviroment this value is positive, while when the system realeased heat to the enviroment, the value is negative. In the case of work, when the enviroment make work on the system, the value is positive and, on the contrary, when the system makes work against the environment the value is negative.

ΔU = q + w

w is defined as the negative product between the external pressure (P) and the change in volume (ΔV). [w = - P·ΔV]

⇒ ΔU = q - P·ΔV

In this problem, the system is composed by the cylinder + piston + contained air (Attached)

1) q= 565 J (a positive value because the enviroment delivered heat to the cylinder)

2) P = 1 atm.

3) ΔV = 0.86 atm.L - 0.12 atm.L = 0.74 atm.L.

⇒ ΔU = q + w ⇒ ΔU = 565 J - (1 atm)·(0.74 L) ⇒ ΔU = 565 J - 0.74 atm.L

We have to express the work value in the same units of heat, it means Joules. As one joule is equal to 0.00987 atm.L

⇒ (0.74 atm.L) x ( 1 J/0.00987 atm.L) = 74.97 J.

⇒ ΔU = q + w ⇒ ΔU = 565 J - 74.97 J ⇒ ΔU = 490 J.

Summarizing, the change in internal energy of the air within the piston is 490 J.

3 0

3 years ago

A 2 kg object being pulled across the floor with a speed of 10 m/sec is suddenly

zvonat [6]

Answer:

The frictional force producing this deceleration would have a magnitude of $4\; \rm N$ .

Explanation:

The velocity of this object changed by $\Delta v = (-10\; \rm m\cdot s^{-1})$ in $\Delta t = 5\; \rm s$ . The acceleration of this object would be:

$\begin{aligned}a &= \frac{\Delta v}{\Delta t} \\ &= \frac{-10\; \rm m\cdot s^{-1}}{5\; \rm s} = -2\; \rm m\cdot s^{-2}\end{aligned}$ .

Let $m$ denote the mass of this object. By Newton's Second Law of Motion, the net force on this object would be:

$\begin{aligned}F &= m \, a \\ &= 2\; \rm kg \times (-2\; \rm m\cdot s^{-2}) \\ &= -4\; \rm N\end{aligned}$ .

( $1\; {\rm kg \cdot m \cdot s^{-2} = 1\; {\rm N}$ .)

If the floor is level, friction would be the only unbalanced force on this object. Thus, the magnitude of the frictional force on this object would also be $4\; {\rm N}$ , same as the magnitude of the net force on this object.