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elena-14-01-66 [18.8K]
3 years ago
8

When placed near another charge, a 20 microcoulomb charge experiences an attractive force of 0.080 N. What is the electric field

strength at that location?
Physics
1 answer:
prohojiy [21]3 years ago
8 0

Answer:

E = 4000 N/C

Explanation:

Given the following data;

Force = 0.080 N.

Charge, q = 20 microcoulomb = 20 * 10^-6 = 2 * 10^-5 Coulombs

To find the electric field strength;

Mathematically, the electric field strength is given by the formula;

Electric field strength = force/charge

Substituting into the formula, we have;

E = 0.080/0.00002

E = 4000 N/C

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I got the 1.0g of coal producves 35,000 joules which is also 8.36 kcals
8 0
3 years ago
 I will mark you as brainliest if you answer correctly
aleksklad [387]
(A) We can solve the problem by using Ohm's law, which states:
V=IR
where
V is the potential difference across the electrical device
I is the current through the device
R is its resistance
For the heater coil in the problem, we know V=220 V and R=220 \Omega, therefore we can rearrange Ohm's law to find the current through the device:
I= \frac{V}{R}= \frac{220 V}{220 \Omega}=1 A

(B) The resistance of a conductive wire depends on three factors. In fact, it is given by:
R= \rho \frac{L}{A}
where
\rho is the resistivity of the material of the wire
L is the length of the wire
A is the cross-sectional area of the wire
Basically, we see that the longer the wire, the larger its resistance; and the larger the section of the wire, the smaller its resistance.
6 0
3 years ago
A bullet starting from rest accelerates uniformly at a rate of 1,250 meters per square second. What is the bullet's speed after
Tatiana [17]

Answer:

125,000

Explanation:

8 0
3 years ago
Using Excel, or some other graphing software, plot the values of y as a function of x. (You will not submit this spreadsheet. Ho
Evgesh-ka [11]

Answer:

a) > x<-c(1,2,3,4,5)

> y<-c(1.9,3.5,3.7,5.1,6)

> linearmodel<-lm(y~x)

And the output is given by:

> linearmodel

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept)            x  

      1.10         0.98  

b) y = 0.98 x +1.10

And if we compare this with the general model y = mx +b

We see that the slope is m= 0.98 and the intercept b = 1.10

Explanation:

Part a

For this case we have the following data:

x: 1,2,3,4,5

y: 1.9,3.5,3.7,5.1, 6

For this case we can use the following R code:

> x<-c(1,2,3,4,5)

> y<-c(1.9,3.5,3.7,5.1,6)

> linearmodel<-lm(y~x)

And the output is given by:

> linearmodel

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept)            x  

      1.10         0.98  

Part b

For this case we have the following trend equation given:

y = 0.98 x +1.10

And if we compare this with the general model y = mx +b

We see that the slope is m= 0.98 and the intercept b = 1.10

7 0
3 years ago
A 2.1 kg block is dropped from rest from a height of 5.5 m above the top of the spring. When the block is momentarily at rest, t
ella [17]

Answer:

The speed of the block is 8.2 m/s

Explanation:

Given;

mass of block, m = 2.1 kg

height above the top of the spring, h = 5.5 m

First, we determine the spring constant based on the principle of conservation of potential energy

¹/₂Kx² = mg(h +x)

¹/₂K(0.25)² = 2.1 x 9.8(5.5 +0.25)

0.03125K = 118.335

K = 118.335 / 0.03125

K = 3786.72 N/m

Total energy stored in the block at rest is only potential energy given as:

E = U = mgh

U = 2.1 x 9.8 x 5.5 = 113.19 J

Work done in compressing the spring to 15.0 cm:

W = ¹/₂Kx² = ¹/₂ (3786.72)(0.15)² = 42.6 J

This is equal to elastic potential energy stored in the spring,

Then, kinetic energy of the spring is given as:

K.E = E - W

K.E = 113.19 J - 42.6 J

K.E = 70.59 J

To determine the speed of the block due to this energy:

KE =  ¹/₂mv²

70.59 =  ¹/₂ x 2.1 x v²

70.59 = 1.05v²

v² = 70.59 / 1.05

v² = 67.229

v = √67.229

v = 8.2 m/s

8 0
3 years ago
Read 2 more answers
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