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sergey [27]
3 years ago
8

What factors would change the force of friction?​

Physics
1 answer:
kkurt [141]3 years ago
6 0
The gravity ( im not sure)
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13. Austin rode his bike 10 m/s for two minutes. How far did he travel? A. 200 meters B. 1200 meters C. 1000 meters D. 20 meters
Semmy [17]

Answer:

B. 1200

Explanation:

60 sec in one min in 2 min there will be 120 sec. 10x120=1200

5 0
3 years ago
A teacher wants to demonstrate that the radioactive source emits alpha beta and gamma radiation. Describe a method the teacher c
Mila [183]

By using an electric field, it is feasible to differentiate between these different forms of radiation.

<h3>What is a radioactive source?</h3>

A source that emits radiation like gamma, beta, and alpha rays is said to be radioactive. Using an electric field, we can discriminate between these different forms of radiation.

The field does not deflate the gamma rays, but it does deflate the alpha and beta rays, with the alpha being deflated to the field's negative portion and the beta to its positive part.

Hence, by using an electric field, it is feasible to differentiate between these different forms of radiation.

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brainly.com/question/12741761

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8 0
1 year ago
By counting the number of crests that pass in a given amount of time, a person can calculate the
ladessa [460]
By calculating the crests, you can find the waves' frequency.
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3 0
3 years ago
Is it possible to apply the same amount of force and do different amounts of work?
Tamiku [17]
Yes, with simple machines
5 0
3 years ago
A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t
Jlenok [28]

Answer:

14.36 N

Explanation:

T_{1} = Tension in string 1

T_{2} = Tension in string 2

m_b = mass of the bar = 2.7 kg

W_b = weight of the bar

weight of the bar is given as

W_b = m_{b} g = (2.7) (9.8) = 26.46N

m_m = mass of the bar = 1.35 kg

W_m = weight of the monkey

weight of the monkey is given as

W_m = m_{m} g = (1.35) (9.8) = 13.23N

Using equilibrium of torque about left end

W_{m} (AB) + W_{b} (AB) = T_{2} (AC)\\W_{m} (AB) + W_{b} (AB) = T_{2} (AD - CD)\\(13.23) (1.5) + (26.46)(1.5) = T_{2} (3 - 0.65)\\\\T_{2} = 25.33 N

Using equilibrium of force in vertical direction

T_{1} + T_{2} = W_{b} + W_{m}\\T_{1} + 25.33 = 26.46 + 13.23\\T_{1} = 14.36 N

7 0
3 years ago
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