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scZoUnD [109]
3 years ago
13

If you have a block with a mass of 2 kilograms and a rope with a tension of 10 Newton’s, what will the acceleration of the block

be?
Physics
1 answer:
omeli [17]3 years ago
7 0

Answer:

5 m/s^2

Explanation:

Assuming the tension in the rope is the only force acting on the block, then we can use Newton's second law:

F=ma

where

F is the net force on the block

m is the mass of the block

a is its acceleration

The tension in the rope is the only force acting on the block, so

F = T = 10 N

The mass of the block is

m = 2 kg

So, solving for a, we find the acceleration:

a=\frac{F}{m}=\frac{10}{2}=5 m/s^2

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Theweight ofof body is 420N .Calculate its mass​
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Answer:

42.87

Explanation:

Google

<u>OR</u>

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420/9.8 which equals 42.87.

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The above graph shows the speed of a car over time. During which time period was the car stopped?
rjkz [21]
The answer is C, as there is not increase or decrease in speed during that time frame.
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According to collision theory, which condition(s) must be met in order for molecules to react?
kondaur [170]

Answer:

B because molecules occurs by the reaction

5 0
3 years ago
Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times g
ollegr [7]

Answer:

The  beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

<u>velocity of wave on the string with greater tension;</u>

v_1 = \sqrt{\frac{T_1}{\mu }

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} =  \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

v_2^2 = \frac{T_2v_1^2}{T_1}  \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s

<u>Fundamental frequency of wave on the string with greater tension;</u>

<u />f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1}  =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz<u />

Beat frequency = F₂ - F₁

                          = 270.6 - 258

                          = 12.6 Hz

Therefore, the  beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

6 0
4 years ago
Convert 3.5 km/hr to m/s
Nastasia [14]

Answer:

0.972222........

Explanation:

1. Convert km/h to m/h by multiplying by 1000

2. Divide by 60 so that you have 58.3333... meters per minute

3. Divide by 60 again so now you have 0.972222... m/s

8 0
3 years ago
Read 2 more answers
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