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exis [7]
2 years ago
15

Due Ma duart ded out 25 N 35 N 1-03

Physics
1 answer:
Anuta_ua [19.1K]2 years ago
8 0

Answer:

what's that all about

hehehwhe

Explanation:

dgbjjjedgkigdssfhkkoyddwrhkoyeqaxghjjhasghffhjiopjtewqetujjgda

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What is the mass of an 291pound persond
Luden [163]
Well, it also depends on the height... but say if they were 5'3" and 291 pounds...
their BMI would be

51.55
4 0
2 years ago
1.
MAXImum [283]

Answer:

true

Explanation:

Newton is the measure of the force with turns to be gravity multiplying the mass. Thus, the forces acts on the particles in the direction of the movement of the particles

7 0
3 years ago
Students measured the mass of 25.0 mL of water and found it be 25.4 g. The accepted mass is 25.0 g. What is the percent error of
Andre45 [30]
Well first of all, I think the students may have been correct. 
If they didn't use distilled water, and if it wasn't exactly at 
standard temperature, then the mass of  25.0 mL  could
very well be  25.4 grams.  We don't know that there was
any 'error' in their measurement at all.
But the question says there was, so we'll do the math:

The 'error' was  (25.4 - 25.0) = +0.4 gram

As a fraction of the 'real' value, the error was

                            +0.4 / 25.0  =  +0.016 .

To change a decimal to a percent, move the
decimal point two places that way  ===> .

                           + 0.016  =  +1.6 % .

     
Their measurement was 1.6% too high.

Let's not call it an 'error'.  Let's just call it a 'discrepancy'
between the measured value and the 'accepted' value.  OK ?
4 0
3 years ago
Read 2 more answers
The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given b
Mnenie [13.5K]

Explanation:

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

Q=t^3-2t^2+4t+4

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(t^3-2t^2+4t+4)}{dt}\\\\I=3t^2-4t+4

At t = 1 s,

Current,

I=3(1)^2-4(1)+4\\\\I=3\ A

So, the current at t = 1 s is 3 A.

For lowest current,

\dfrac{dI}{dt}=0\\\\\dfrac{d(3t^2-4t+4)}{dt}=0\\\\6t-4=0\\\\t=0.67\ s

Hence, this is the required solution.

7 0
3 years ago
A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f
Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

T=\dfrac{2\pi}{\omega}

{\omega}=\dfrac{2\pi }{0.2}\ rad/s

{\omega}=31.41\ rad/s

We know that

{\omega}^2=m\ K

K=Spring constant

K=\dfrac{\omega^2}{m}

K=\dfrac{31.41^2}{0.12}\ N/m

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

3 0
3 years ago
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