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katrin [286]
3 years ago
7

José has a monthly budget of $80 to put toward his hobby of building model

Physics
1 answer:
kiruha [24]3 years ago
3 0

Answer:

He should only uses credit cards for important purchases.

Explanation:

If José has a monthly budget for something in particular, he should use it to pay cash. The credit card should be used for important purchases.

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In noisy factory environments, it's possible to use a loudspeaker to cancel persistent low-frequency machine noise at the positi
Vaselesa [24]

Answer:

7.39 m or 3.61 m

Explanation:

\lambda = Wavelength

f = Frequency = 90 Hz

v = Speed of sound = 340 m/s

Path difference of the two waves is given by

s_1-s_2=\frac{\lambda}{2}

Velocity of wave

v=f\lambda\\\Rightarrow \lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{340}{90}\\\Rightarrow \lambda=3.78\ m

s_1=s_2\pm\frac{\lambda}{2}\\\Rightarrow s_1=5.5\pm \frac{3.78}{2}\\\Rightarrow s_1=7.39\ m, 3.61\ m

So, the location from the worker is 7.39 m or 3.61 m

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Lactic acid is produced during low-moderate exercise <br> True <br> False
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The answer to your question is true.
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What is meant by the statement '' density of water is 1000 kilograms per cubic metre ''?
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Convert 141.2 kg to lbs and show all units and work​
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A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t
Jobisdone [24]

Answer:

(a) E=0  :   0 cm from the center of the sphere

(b) E= 227.8*10³ N/C   :    10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C    :    40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C  :    59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

E=\frac{K*Q}{r^{2} } : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

E=k*\frac{Q}{a^{3} } *r :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q:  Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at  0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere  

r>a , We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }

E= 411.84 * 10³ N/C

4 0
3 years ago
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