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3 years ago

Please help!!!! Write 5 harmful effects of friction.

1 answer:

8 0

_____ _____hello mate_______

here are five harmful effects of friction

1) It makes us use more fuel

2) The gears in your care will wear down if not lubricated with oil

3) The friction up the device so, due to heating machine can damaged

4) Causes the production of heat between surfaces when the object move
eg : car brakes and tyres heat up due friction

5) Even our shoes wear out due to friction with the ground ; Friction with air causes air craft , meteors to heat

HOPE THE ANSWER WILL HELP U!!!! :-) :-) :-)

:-) :-)

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A diffraction grating with 161 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. At what angles i

Tresset [83]

Answer:

$\theta_1 = 0.400^o$

$\theta_2 =0.378^o$

Explanation:

From the question we are told that

The number of slits per cm is k = $161\ slits\ per\ cm = 161 \ slits\ per\ 0.01 m$

The order of the maxima is n = 1

The wavelength are $\lambda_1 = 434 nm = 434 *10^{-9} \ m \ \ \ , \lambda_2 = 410nm = 410 *10^{-9} \ m$

The spacing between the slit is mathematically represented as

$d = \frac{ 0.01}{k}$

=> $d = \frac{ 0.01}{161}$

=> $d = 6.211 *10^{-5} \ m$

Generally the condition for constructive interference is

$n\lambda = d \ sin \theta$

At $\lambda_1$

$\theta _1 = sin^{-1} [ \frac{1 * 434 *10^{-9}}{6.211 *10^{-5}} ]$

$\theta_1 = 0.400^o$

At $\lambda_2$

$\theta _2 = sin^{-1} [ \frac{1 * 410 *10^{-9}}{6.211 *10^{-5}} ]$

$\theta_2 =0.378^o$

4 0

3 years ago

PLEASE HELP!! WILL GIVE POINTS

AURORKA [14]

Answer:

49 kg is the mass of the couch.

Explanation:

GPE = mgh

9800 = m * 10 * 20

9800 = 200m

m = (9800/20) = 49 m

Thenks and mark me brainliest :))

3 0

3 years ago

Use the information below to answer questions

Ulleksa [173]

Answer:

The charges are q₁ = 2 × 10⁻⁸ C and q₂ = 3 × 10⁻⁸ C

Explanation:

Here is the complete question

Two identical tiny balls have charge q1 and q2. The repulsive force one exerts on the other when they are 20cm apart is 1.35 X 10-4 N. after the balls are touched together and then represented once again to 20cm, now the repulsive force is found to be 1.40 X 10-4 N. find the charges q1 and q2.

Solution

The force F = 1.35 × 10⁻⁴ N when the charges are separated a distance of r = 20 cm = 0.2 m is given by

F = kq₁q₂/r₁²

q₁q₂ = Fr₁²/k

q₁q₂ = 1.35 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.054/9 × 10⁻¹³ C² = 0.006 × 10⁻¹³ C² = 6 × 10⁻¹⁶ C²

q₁q₂ = 6 × 10⁻¹⁶ C² (1)

When the charges are brought together, the charge is now q = (q₁ + q₂)/2

The new repulsive force F = 1.406 × 10⁻⁴ N at a distance of r₂ = 20 cm = 0.2 m is then

F₂ = kq²/r₂²

q² = F₂r₂²/k = 1.406 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.00625 × 10⁻¹³ C² = 6.25 × 10⁻¹⁶ C²

q² = 6.25 × 10⁻¹⁶ C²

q = √(6.25 × 10⁻¹⁶) C

q = 2.5 × 10⁻⁸ C

(q₁ + q₂)/2 = 2.5 × 10⁻⁸ C

(q₁ + q₂) = 2 × 2.5 × 10⁻⁸ C

q₁ + q₂ = 5 × 10⁻⁸ C (2)

q₁ = 5 × 10⁻⁸ C - q₂ (3)

Substituting equation (3) into (1), we have

(5 × 10⁻⁸ C - q₂)q₂ = 6 × 10⁻¹⁶ C²

Expanding the bracket, we have

(5 × 10⁻⁸ C)q₂ - q₂² = 6 × 10⁻¹⁶ C²

So, q₂² - (5 × 10⁻⁸ C)q₂ + 6 × 10⁻¹⁶ C² = 0

Using the quadratic formula to find q₂

$q_{2} = \frac{-(-5 X 10^{-8} )+/- \sqrt{(-5 X 10^{-8} )^{2} - 4X1X6 X 10^{-16} } }{2X1}\\ = \frac{5 X 10^{-8} )+/- \sqrt{25 X 10^{-16} - 24 X 10^{-16} } }{2}\\= \frac{5 X 10^{-8} )+/- \sqrt{1 X 10^{-16} } }{2}\\= \frac{5 X 10^{-8} )+/- 1 X 10^{-8} }{2}\\= \frac{5 X 10^{-8} + 1 X 10^{-8} }{2} or \frac{5 X 10^{-8} - 1 X 10^{-8} }{2}\\= \frac{6 X 10^{-8} }{2} or \frac{4 X 10^{-8}}{2}\\= 3 X 10^{-8} C or 2 X 10^{-8} C$

q₁ = 5 × 10⁻⁸ C - q₂

q₁ = 5 × 10⁻⁸ C - 3 × 10⁻⁸ C or 5 × 10⁻⁸ C - 2 × 10⁻⁸ C

q₁ = 2 × 10⁻⁸ C or 3 × 10⁻⁸ C

So the charges are q₁ = 2 × 10⁻⁸ C and q₂ = 3 × 10⁻⁸ C

5 0

3 years ago

A gyre is a set of currents that form:

tangare [24]

A gyre is a set of currents that form b. a loop. The circulation of gyres are affected by global wind patterns, landmasses, and the planet's rotation. The circulation is also affected by temperature, as warm water goes up and cold water sinks. There are five major gyres in the world: <span>North Atlantic, South Atlantic, Indian, North Pacific, and South Pacific.</span>

7 0

3 years ago

Read 2 more answers

A +1.0 nC charge is at x = 0 cm, a -1.0 nC charge is at x = 1.0 cm and a 4.0 nC at x= 2 cm. What is the electric potential energ

lesantik [10]

Answer:

- 2.7 x 10^-6 J

Explanation:

q1 = 1 nC at x = 0 cm

q2 = - 1 nC at x = 1 cm

q3 = 4 nC at x = 2 cm

The formula for the potential energy between the two charges is given by

$U=\frac{Kq_{1}q_{2}}{r}$

where r be the distance between the two charges

By use of superposition principle, the total energy of the system is given by

$U = U_{1,2}+U_{2,3}+U_{3,1}$

$U=\frac{Kq_{1}q_{2}}{0.01}+\frac{Kq_{2}q_{3}}{0.01}+\frac{Kq_{3}q_{1}}{0.02}$

$U=-\frac{9\times10^{9}\times 1\times10^{-9}\times 1\times10^{-9}}}{0.01}-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.01}+-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.02}$

U = - 2.7 x 10^-6 J

3 0

3 years ago