Can someone help me with this please

Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m. What is the force (in N) of

Lostsunrise [7]

Answer:

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Explanation:

F₂₁ = $\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}$

Where;

F₂₁ is the vector force on q₁ due to q₂

K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²

r₂₁ is the unit vector

|r₂₁| is the magnitude of the unit vector

|q₁| is the absolute charge on point charge one

|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| = $\sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}$

(|r₂₁|)² = 148.25

$F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }$

= 0.050938(0.19107i + 0.54933j) N

= (0.00973i + 0.02798j) N

Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N

7 0

3 years ago

A 6000 kg roller coaster goes around a loop of radius 30 m at 6 m/s. What is the centripetal acceleration?

Margarita [4]

Answer:

The answer to the question is 7200

7 0

3 years ago

A box weighing 200N is pushed on a horizontal floor. What acceleration will result if a horizontal force of 100N is applied on t

Bogdan [553]

Answer:

the friction force in the reverse direction is 200 *0.4=80 N.

the net forward force acting on the box is therefore

Fnet= 100 - 80 N

= 20 N

acceleration = Fnet / mass

=Fnet *g/(weight)

=20 *9.8/200 = 0.98 m/s^2

Explanation:

4 0

3 years ago

Read 2 more answers

A paleontologist estimates that when a particular rock formed, it contained 12 mg of the radioactive isotope potassium-40, which

leva [86]

Answer:

$t = 2.52 billion \:years$

Explanation:

As we know by radioactivity law

$N = N_o e^{-\lambda t}$

so here we will have

$N = 3 mg$

$N_o = 12 mg$

now we will have

$3 = 12 e^{-\lambda t}$

$\lambda t = ln 4$

now we also know that

$\lambda = \frac{ln2}{1.26 \times 10^6 yrs}$

$t = 1.26\times 10^6\times \frac{ln4}{ln2}$

$t = 2.52 billion \:years$

7 0

3 years ago

Alcohol of mass 33.2g and density 0.79kg/m³ or 790kg/m³ is mixed with water of 9g. What is the density of the resulting mixture?

KATRIN_1 [288]

Answer:

0.83 g/cm³

Explanation:

The volume of the alcohol is ...

(33.2 g)/(0.79 g/cm³) ≈ 42.0253 cm³

The density of water is about 1 g/cm³, so the volume of 9 g of it is ...

(9 g)/(1 g/cm³) = 9 cm³

The total volume is ...

42.0253 cm³ +9 cm³ = 51.0253 cm³

The total mass is ...

33.2 g + 9 g = 42.4 g

So, the resulting density is ...

(42.4 g)/(51.0253 cm³) ≈ 0.83 g/cm³

The resulting mixture has a density of about 0.83 g/cm³.

_____

<em>Additional comment</em>

Alcohol dissolves in water, so the total volume will be slightly less than 51.0253 cm³. The attached curve shows the result of mixing ethanol and water.

The weight of a mole of ethanol is about 46 g, of water, about 18.02 g. Then the mole fraction of alcohol is ...

(33.2/46)/(33.2/46 +9/18.02) ≈ 0.59

The volume of the mix is then estimated to be (-1.05 cc/mol)(1.221 mol), or about 1.28 cm³ less than the volume indicated above. That brings the density up to about 0.85 g/cm³.

We're not completely sure of the relevance of this calculation, since many of the applicable parameters are not specified. The point is that <em>the density of the mix will probably be slightly more than the value calculated above</em>. YMMV

3 0

3 years ago