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Oduvanchick [21]
3 years ago
10

How many ounces are in one ton? 16 ounces 125 ounces 2,000 ounces 32,000 ounces

Physics
2 answers:
anyanavicka [17]3 years ago
7 0

Answer:

32,000 ounces

Explanation:

ale4655 [162]3 years ago
5 0

32,000 ounces. hope it helps

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2.<br> 1,200 kg<br> 8.33 m/s<br> 3.3
Sliva [168]

Answer:

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Explanation:

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3 years ago
Sodium and potassium are positively charged ions called cations, and chloride and phosphate form negatively charged ions called
Alecsey [184]
 Sodium and potassium are positively charged ions called cations, and chloride and phosphate form negatively charged ions called anions. is true!

7 0
3 years ago
HELP ASAP!!! Some bands hate playing in school gyms because sound waves easily reflect off the walls and floor. What could you d
jonny [76]

Answer:

1.) Covering the ceiling and walls with soft perforated boards

2.) Hanging curtains round the hall

3.) Having more opening in the wall.

Explanation:

This is due to what we called Reverberation due to poor acoustic properties.

Reverberation can be reduced by;

1.) Covering the ceiling and walls with soft perforated boards

2.) Hanging curtains round the hall

3.) Having more opening in the wall.

4 0
3 years ago
Who has the most power? athlete A (bench presses 100 kg over 0.6 m in 0.5 s) athlete B (bench presses 150 kg over 0.6 m in 1.0 s
MrRissso [65]

Answer:

Athlete A

Explanation:

Power is the rate of doing work and it is calculated as follows:

Power = work done/time taken = mgh/t

(for work being done against gravity)

So for athlete A

P = (100 kg * 9.8 N/kg* 0.6m)/0.5 s = 1176 W

For athlete B

P = (150 kg * 9.8 N/kg* 0.6m)/1 s = 882 W

For athlete C

P = (200 kg * 9.8 N/kg* 0.6m)/2 s = 588 W

For athlete D

P = (250 kg * 9.8 N/kg* 0.6m)/2.5 s = 588

4 0
2 years ago
You are trying to decide between two new stereo amplifiers. One is rated at 130 W per channel and the other is rated at 200 W pe
NARA [144]

Answer:

The sound level will be 1.870 dB louder.

Explanation:

Given that,

Power = 130 W

Power = 200 W

We need to calculate the sound level

Using formula of sound level

I_{dB}=10\log(\dfrac{I}{I_{0}})

For one amplifier,

I_{1}=10\log(\dfrac{130}{I_{0}})...(I)

For other amplifier,

I_{2}=10\log(\dfrac{200}{I_{0}})...(II)

For difference in dB levels

I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}})-10\log(\dfrac{130}{I_{0}})

I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}}\times\dfrac{I_{0}}{130})

I_{2}-I_{1}=10\log(\dfrac{200}{130})

I_{2}-I_{1}=1.870\ dB

Hence, The sound level will be 1.870 dB louder.

7 0
3 years ago
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