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kumpel [21]
3 years ago
10

Solve the inequality. x/3 is greater than or equal to - 6. a. x ≥ –9 b. x ≥ 9 c. x ≥ –18 d. x ≤ –18

Physics
1 answer:
NISA [10]3 years ago
6 0
To get x on its own, you times the 3 over to the other side so the 3 cancels out on the LHS. 

~ x greater than or equal to -18

(C)
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2 years ago
A sound wave has frequency 620 Hz and wavelength 10.5 m. What is the speed of sound waves?
olasank [31]

Answer:

14

Explanation:

A sound wave in a steel rail has a frequency of 620 Hz and a wavelength of 10.5 m. What is the speed of sound in steel? U f = ? S f = 3 x 108 m/s) / 0.06 m 14 so it will be 14

7 0
2 years ago
A loaded barge has a mass of 1 500 000 kg and is traveling at 3 m/s. If a tugboat applies an opposing force of 12 000 N for 10 s
yan [13]

Answer:

Explanation:

Initial momentum is 1.5e6(3) = 4.5e6 kg•m/s

An impulse results in a change of momentum

The tug applied impulse is 12000(10) = 120000 N•s or 0.12e6 kg•m/s

The remaining momentum is 4.5e6 - 0.12e6 =  4.38e6 kg•m/s

The barge velocity is now 4.38e6 / 1.5e6 = 2.92 m/s

The tug applies 0.012e6 N•s of impulse each second.

The initial barge momentum will be zero in

t = 4.5e6 / 0.012e6 = 375 s or 6 minutes and 15 seconds

To stop the barge in one minute(60 s), the tug would have to apply

4.5e6 / 60 = 75000 N•s /s or 75 000 N

5 0
2 years ago
A 800 kg safe is 2.1 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses
stellarik [79]

Answer:

k = 17043.5 N/m = 17.04 KN/m

Explanation:

First we need to find the force applied by safe pn the spring:

F = Weight of Safe

F = mg

where,

F = Force Applied by the safe on the spring = ?

m = mass of safe = 800 kg

g = 9.8 m/s²

Therefore,

F = (800 kg)(9.8 m/s²)

F = 7840 N

Now, using Hooke's Law:

F = kΔx

where,

K = Spring Constant = ?

Δx = compression = 46 cm = 0.46 m

Therefore,

7840 N = k (0.46 m)

k = 7840 N/0.46 m

<u>k = 17043.5 N/m = 17.04 KN/m</u>

5 0
3 years ago
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