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Mars2501 [29]
3 years ago
13

Point P is located 58 cm to the right of a fixed point charge of 6.7*10^ -16 C. What is the direction and magnitude of the elect

ric field at point P due to this charge? Let the electrostatic constant k = 8.99 * 10 ^ 9 * N * m ^ 2 / (C ^ 2)
A. 8.8 * 10 ^ (- 5) * N/C, to the right
B. 2.4 * 10 ^ (- 6) * N/C , to the left
C. 1.2*10^ (-5) * N/C, to the left
D. 1.8*10^ (-5) N/C, to the right
Physics
1 answer:
professor190 [17]3 years ago
5 0

Do you have the answers?

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mote1985 [20]

Answer:

A) Golgi apparatus

Explanation:

B) v a c u o l e (C) m i t o c h o n d r i a ( E) cell wall

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3 years ago
What would the products of a double-replacement reaction between Na2S
MAVERICK [17]

Answer:

A. NaF and MgS

Explanation:

The reaction equation is given as:

             Na₂S  +  MgF₂  →   2NaF  + MgS

The product of this reaction is  NaF and MgS.

In a double displacement reaction there is an actual exchange of partners to form new compounds. This reaction takes place between ionic compounds.

The driving force for the reaction is:

  • formation of an insoluble precipitate
  • formation of water or any other non-ionizing compound
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8 0
3 years ago
A defibrillator is a device used to shock the heart back to normal beat patterns. To do this, it discharges a 15 μF capacitor th
Mama L [17]

Answer:

Part A: Q = 0.075C

Part B: E = 187.5J

Part C: I = 50A

Part D: ΔQ 1.53C

Explanation:

Part A

Q = C×V

Given

C = 15μF = 15×10‐⁶ F, V = 5kV = 5000V

Q = 15×10‐⁶× 5000 = 0.075C

Part B

Energy = E = 1/2 ×CV² = 1/2 × 15×10‐⁶ × 5000² = 187.5J

Part C

Given R = 100Ω

V = IR, I = V/R = 5000/100 = 50A

Part D

I = ΔQ/Δt

Given Δt= 90ms = 90×10-³s, I = 17A

ΔQ = I × Δt = 17×90×10-³ = 1.53 C

5 0
3 years ago
If we decrease the time it takes for a car to travel over the same distance, this will
Ilya [14]
C) increase the speed
7 0
3 years ago
Read 2 more answers
An alpha particle can be produced in certain radioactive decays of nuclei and consists of two protons and two neutrons. The part
Varvara68 [4.7K]

Answer:

Explanation:

charge, q = 2e = 2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C

mass, m = 4 u = 4 x 1.661 x 10^-27 kg = 6.644 x 10^-27 kg

Radius, r = 4.5 cm = 0.045 m

Magnetic field, B = 1.20 T

(a) Let the speed is v.

v=\frac{Bqr}{m}

v=\frac{1.20\times 3.2\times 10^{-19}\times 0.045}{6.644\times 10^{-27}}

v = 2.6 x 10^6 m/s

(b) Let T be the period of revolution

T=\frac{2\pi r}{v}

T=\frac{2\times 3.14\times 0.045}{2.6\times 10^{6}}

T = 1.09 x 10^-7 s

(c) The formula for the kinetic energy is

K=\frac{B^{2}\times q^{2}\times r^{2}}{2m}

K=\frac{\left ( 1.20\times 3.2 \times 10^{-19}\times 0.045 \right )^{2}}{2\times 6.644\times 10^{-27}}

K = 2.25 x 10^-14 J

(d) Let the potential difference is V.

K = qV

V = \frac{K}{q}

V= \frac{2.25\times 10^-14}{3.2\times 10^{-19}}

V = 70312.5 V

5 0
3 years ago
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