If a steel spatula experiences a force of 0.24 N when scraping against a teflon frying pan...?
回答 (3)
2018-02-07 4:16 pm
N μ = 0.24 N
N × 0.04 = 0.24 N
Normal force, N = 0.24/0.04 N = 6.0 N
N × 0.04 = 0.24 N
Normal force, N = 0.24/0.04 N = 6.0 N
2018-02-08 12:31 am
...
use this
Ff = mu Fn
0.24 N = 0.04 Fn
Fn = 6 N
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use this
Ff = mu Fn
0.24 N = 0.04 Fn
Fn = 6 N
When you get a good response,
please consider giving a best answer.
This is the only reward we get.
You may have to wait an hour to award BA.
2018-02-07 10:01 pm
Force experienced by the spatula is F =0.24 N
teflan frying pan has friction coefficient mue = 0.04
we know that the force F = N*mue
0.24 = 0.04*N
solving for N
N= 6 N
the normal force N = 6 N
teflan frying pan has friction coefficient mue = 0.04
we know that the force F = N*mue
0.24 = 0.04*N
solving for N
N= 6 N
the normal force N = 6 N
收錄日期: 2021-04-24 00:56:59
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