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olganol [36]
2 years ago
5

Which factors affect the resistance of a material? Check all that apply.

Physics
1 answer:
shutvik [7]2 years ago
6 0

Answer: Length, thickness, and temperature

Explanation:

I did it

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Find an expression for the acceleration a of the red block after it is released. use mr for the mass of the red block, mg for th
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<span>Assuming pulley is frictionless. Let the tension be ‘T’. See equation below.</span>

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Particles 1 and 2 of charge q1 = q2 = +3.20 × 10−19 C are on a y axis at distance d = 17.0 cm from the origin. Particle 3 of cha
mel-nik [20]

Answer:

(a) 0.17 m

(b) 5.003 m

(c) 6.38 × 10^{-26} N

(d) 7.37 ×10^{-29} N

Explanation:

(a) The minimum value of x will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

h^{2} = b^{2} + p^{2}

h^{2} = 5^{2} + 0.17^{2}   \\h = \sqrt{} 25.03\\h= 5.002 m

<em>Hence, the maximum distance is 5.002 m</em>

(c) For minimum magnitude we use the minimum distance calculated in (a)

Minimum Distance = 0.17 m

For electrostatic force=     F=\frac{kq1q2}{x^{2} }

F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19}  }{0.17^{2} }

F= 6.38×10^{-26} N

(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F =  \frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19}   }{5.002^{2} } }

F= 7.37×10^{-29 N

4 0
2 years ago
A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr
valina [46]
The  spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
T=2 t = 2 \cdot 0.100 s = 0.200 s
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f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz
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v_{max} = A \omega
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Prove the identity <br>Trigonometry grade 10​
g100num [7]

Answer:

and is in photo given.I didn't get time to type.

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