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seraphim [82]
3 years ago
12

A SI unit used to measure force, equal to less than one-quarter of a pound, is the what

Physics
1 answer:
bulgar [2K]3 years ago
4 0

The SI unit of force is the Newton.

1 newton is the force that accelerates a 1 kilogram mass
at the rate of 1 meter per second².

1 pound of force is equivalent to roughly  4.448 newtons.

(1 newton is equivalent to roughly  0.225  pounds of force.)

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A car goes 4 south and 3 m East. Find the resultant vector using Tip-Tail method.
Tema [17]
I think it’s c but check cause I’m not sure
5 0
3 years ago
How is the capacitance of a capacitor related to the charge stored on the capacitor and the potential difference across the capa
tatuchka [14]

Answer:

Capacitance is the ratio of the charge to the potential difference. How is the charge stored on a capacitor related to the capacitance of the capacitor and the potential difference across the capacitor? The charge equals the product of the capacitance and the potential difference.

Explanation:

Hope this helps

6 0
2 years ago
Two cello strings, with the same tension and length, are played simultaneously. Their fundamental frequencies produce audible be
qwelly [4]

Explanation:

Let f₁ is the fundamental frequency, f_1=8\ Hz

Lower pitch frequency, f_2=220\ Hz

Fundamental frequency is, f_1=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu_1}}.....(1)

Lower frequency is, f_2=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu_2}}..............(2)

Dividing equation (1) and (2) as :

\dfrac{f_1}{f_2}=\sqrt{\dfrac{\mu_2}{\mu_1}}

\dfrac{\mu_2}{\mu_1}=(\dfrac{f_1}{f_2})^2

\dfrac{\mu_2}{\mu_1}=(\dfrac{8}{220})^2

\dfrac{\mu_2}{\mu_1}=0.00132

So, the ratio of  linear mass density μ of the string with the higher pitch to that of the string with the lower pitch is 0.00132. Hence, this is the required solution.

3 0
3 years ago
Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge
Naddik [55]

Answer:

4.30 x 10⁵ N/C

Explanation:

Two positive +3.0-μC point charges at opposite corners of the square creates equal and opposite electric field at the center. hence the electric field by these two positive charges at opposite corners becomes zero.

a = length of the square of side = 0.50 m

r = distance of the center from each corner = \frac{a}{sqrt(2)} = \frac{0.50}{sqrt(2)} = 0.354 m

Magnitude of net electric field at the center is given as

E = \frac{2kq}{r^{2}} \\E = \frac{2(8.99\times10^{9})(3\times10^{-6})}{(0.354)^{2}} \\E = 4.30\times10^{5} NC^{-1}

6 0
3 years ago
A Christmas light is made to flash via the discharge of a capacitor. The effective duration of the flash is 0.25 s (which you ca
Sonbull [250]

Answer:

The correct solution is:

(a) 1.375\times 10^{-2} \ J

(b) 4.43\times 10^{-3} \ C

(c) 1.42\times 10^{-3} \ F

(d) 178.57 \ \Omega

Explanation:

The given values are:

Effective duration of the flash,

ζ = 0.25 s

Average power,

P_{avg}=55 \ mW

       =55\times 10^{-3} \ W

Average voltage,

V_{avg}=3.1 \ V

Now,

(a)

⇒ E=P_{avg}\times \zeta

On substituting the values, we get

⇒     =55\times 10^{-3}\times 0.25

⇒     =1.375\times 10^{-2} \ J

(b)

⇒ E=Q\times V_{avg}

then,

⇒ Q=\frac{E}{V_{avg}}

On substituting the values, we get

⇒     =\frac{1.375\times 10^{-2}}{3.1}

⇒     =4.43\times 10^{-3} \ C

(c)

⇒ C=\frac{Q}{V}

⇒     =\frac{4.43\times 10^{-3}}{3.1}

⇒     =1.42\times 10^{-3} \ F

(d)

As we know,

⇒ R=\frac{1}{4C}

⇒     =\frac{1}{4\times 1.42\times 10^{-3}}

⇒     =\frac{1000}{5.6}

⇒     =178.57 \ \Omega

5 0
2 years ago
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