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3 years ago

Tom says that insulation keeps out the cold. Explain why this statement is incorrect. What should Tom have said?

Physics

2 answers:

ludmilkaskok [199]3 years ago

7 0

It does in the winter and keeps the heat in. During the summer it keeps in the cold air and keeps out the heat.

Zina [86]3 years ago

4 0

There is no such thing as"cold", in the same way that there is no such thing
as "darkness" or "quietness". "Darkness" is the absence of light, "quietness"
is the absence of sound, and "cold" is the absence of heat.

Tom should have said that insulation <em>keeps the heat in</em> .

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Which is not a device for reproducing sound?

Alona [7]

Phonograph. Hope that helps

7 0

3 years ago

A plastic rod 1.3 m long is rubbed all over with wool, and acquires a charge of -3e-08 coulombs. we choose the center of the rod

Anestetic [448]

(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
$L/8=1.3m/8=0.1625 m$

(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:
$c_6 = 0.0812m+0.1625m=0.2437 m$

(c) The total charge is $Q=-3 \cdot 10^{-8}C$ . To get the charge on each piece, we should divide this value by 8, the number of pieces:
$Q/8=-3\cdot 10^{-8}C/8=-3.75\cdot 10^{-9}C$

(d) We have to calculate the electric field at x=0.7 generated by piece 6. The charge on piece 6 is the value calculated at point (c):
$q= -3.75\cdot 10^{-9}C$
If we approximate piece 6 as a single charge, the electric field is given by
$E=k_e \frac{q}{d^2}$
where $k_e=8.99\cdot 10^9Nm^2C^{-2}$ and d is the distance between the charge (center of piece 6, located at 0.2437m) and point a (located at x=0.7m). Therefore we have
$E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m$
poiting towards the center of piece 6, since the charge is negative.

(e) missing details on this question.

5 0

3 years ago

An air-track glider with a mass of 239 g is moving at 0.81 m/s on a 2.4 m long air track. It collides elastically with a 513 g g

HACTEHA [7]

Answer:

Glider it stops just when it reaches the end of the runway

Explanation:

This is a shock between two bodies, so we must use the equations of conservation of the amount of movement, in the instant before the crash and the subsequent instant, with this we calculate the second glider speed, as the shock that elastic is also keep it kinetic energy

Po = pf

Ko = Kf

Before crash

Po = m1 Vo1 + 0

Ko = ½ m1 Vo1²

After the crash

Pf = m1 Vif + Vvf

Kf = ½ m1 V1f² + ½ m2 V2f²

m1 V1o = m1 V1f + m2 V2f (1)

m1 V1o² = m1 V1f² + m2 V2f² (2)

We see that we have two equations with two unknowns, so the system is solvable, we substitute in 1 and 2

m1 (V1o -V1f) = m2 V2f (3)

m1 (V1o² - V1f²) = m2 V2f²

Let's use the relationship (a + b) (a-b) = a² -b²

m1 (V1o + V1f) (V1o -V1f) = m2 V2f²

We divide with 3 and simplify

(V1o + V1f) = V2f (4)

Substitute in 3, group and clear

m1 (V1o - V1f) = m2 (V1o + V1f)

m1 V1o - m2 V1o = m2 V1f + m1 V1f

V1f (m1 -m2) = V1o (m1 + m2)

V1f = V1o (m1-m2 / m1+m2)

We substitute in (4) and group

V2f = V1o + (m1-m2 / m1 + m2) V1o

V2f = V1o [1+ + (m1-m2 / m1 + m2)]

V2f = V1o (2m1 / (m1+m2)

We calculate with the given values

V1f = 0.81 (239-513 / 239 + 513)

V1f = 0.81 (-274/752)

V1f = - 0.295 m/s

The negative sign indicates that the planned one moves in the opposite direction to the initial one

V2f = 0.81 [2 239 / (239 + 513)]

V2f = 0.81 [0.636]

V2f = 0.515 m / s

Now we analyze in the second glider movement only, we calculate the energy and since there is no friction,

Eo = Ef

Where Eo is the mechanical energy at the lowest point and Ef is the mechanical energy at the highest point

Eo = K = ½ m2 vf2²

Ef = U = m2 g Y

½ m2 v2f² = m2 g Y

Y = V2f² / 2g

Y = 0.515²/2 9.8

Y = 0.0147 m

At this height the planned stops, let's use trigonometry to find the height at the end of the track of the track

tan θ = Y / x

Y = x tan θ

The crash occurs in the middle of the track whereby x = 1.2 m

Y = 1.2 tan 0.7

Y = 0.147 m

As the two quantities are equal in glider it stops just when it reaches the end of the runway

7 0

3 years ago

A truck driver sees a dog running into the road. He immediately brakes. Describe the energy transfers in the truck as it comes t

Eduardwww [97]

Answer: find the answer in the explanation as kinetic energy converts to potential energy.

Explanation:

Before the truck driver sees a dog running into the road, The mechanical energy state of the truck will be kinetic energy at maximum.

Immediately he applied the brakes, the mechanical energy of the truck will be combination of kinetic energy and potential energy.

The kinetic energy will gradually decrease as potential energy continue to increase till it reaches maximum potential energy.

The truck will come to a stop at maximum potential energy

3 0

3 years ago

A car is being driven down a level track with no net force on it. This motion is best described as

Alex

A)Linear motion

If there is not net force on the car, then by the Newton Second Law, the acceleration is zero, and the only valid option for zero acceleration is A).

3 0

2 years ago