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3 years ago

A 0.22-caliber handgun fires a 27-g bullet at a velocity of 755 m/s. is the wave nature of matter significant for bullets?

1 answer:

6 0

<span>All matter can exhibit wave-like behavior, but the question asks if it is significant. A bullet is a very, very large object compared to the light wave-particles where the wave nature is typically observed. So for an object of this magnitude, it is insignificant.</span>

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Answer:

Vacuum. A sound vacuum was created, i believe.

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Amount of pressure of liquid increases with ?

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2 years ago

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A man pushes an 250-N crate at constant speed a distance of 30.0 m upward along a rough slope that makes an angle of 60° with th

mel-nik [20]

Answer:

6495.19 Joule

Explanation:

F = Weight of the crate = 250 N

d = Distance the cart is pushed = 30 m

θ = Angle of inclination = 60°

The weight of the crate will be resloved into two components

Fdsinθ and Fdcosθ

Work done by the force of gravity is

W = Fdsinθ

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3 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago

A six pack of your favorite beverage (12 fl. oz. cans, assumed to have the same characteristics as water) is placed in a cooler

nasty-shy [4]

Answer:

$Q = 169 BTU$

Explanation:

As we know that volume is given as

$V = 12 Fl oz$

so it is given in liter as

$V = 12 fl oz = 0.355 Ltr$

now we have six pack of such volume

so total volume is given as

$V = 6\times 0.355 Ltr$

$V = 2.13 ltr$

so its mass is given as

$m = 2.13 kg$

now the change in temperature is given as

$\Delta T = 70 - 34 = 36 ^oF$

$\Delta T = 20^oC$

now the heat given to the liquid is given as

$Q = ms\Delta T$

$Q = 2.13(4186)(20)$

$Q = 1.78 \times 10^5 J$

$Q = 169 BTU$

4 0

3 years ago