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3 years ago

9

The steering wheel is connected to and controls the wheels by the axle.

1 answer:

ozzi3 years ago

5 0

Ummm false hhahhahahahahhaha

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The angle at which light hits a surface (angle of incidence) and the angle at which it is reflected away (angle of reflection) a

ira [324]

Answer:

are always the same

Explanation:

angle of incident is equal to the angle of reflection i = r

the normal is perpendicular to the reflecting surface

8 0

3 years ago

What is the equation for measuring the change in thermal energy?

Daniel [21]

Answer:

Explanation: I think...

Thermal Energy formula Q = mcΔT

Q = Thermal Energy(J)

m = Mass(kg)

c = Specific Heat(J/kg°C)

ΔT = Change in Temperature(°C)

you have to write the equation based on what you are working on

7 0

3 years ago

A pendulum is observed to complete 23 full cycles in 58 seconds. Use the definition of frequency to find the frequency.

Alex17521 [72]

Answer: f = 0.397 Hz

Explanation:

f = cycles per second

f = 23 / 58 = 0.39655...

4 0

3 years ago

Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.3 cm . Two of

sweet [91]

Answer:

$1.44\times 10^{-3}$ N

Explanation:

We are given that three charged particle are placed at each corner of equilateral triangle.

$q_1=-8.2 nC,q_2=-16.4 nC,q_3=8.0nC$

$q_1=-8.2\times 10^{-9} C$

$q_2=-16.4\times 10^{-9} C$

$q_3=8.0\times 10^{-9} C$

Side of equilateral triangle =3.3 cm= $\frac{3.3}{100}=0.033m$

We know that each angle of equilateral angle= $60^{\circ}$

Net force=F = $\sum\frac{kQq }{d^2}$

Where k= $9\times 10^9 Nm^2/C^2$

If we bisect the angle at $q_3$ then we have 30 degrees from there to either charge.

Direction of vertical force due to charge $q_1$ and $q_2$

Therefore, force will be added

Vertical force= $9\times 10^9\times q_3(q_1+q_2)\frac{cos30}{(0.033)^2})$

Vertical net force= $9\times 10^9\times 8\times 10^{-9}(-8.2-16.4)\times 10^{-9}\times 10^6\times\frac{\sqrt3}{2\times 1089}$

Vertical force = $9\times 8(-24.6)\times 10^{-9}\times 10^6\times 1.732\times \frac{1}{2178}$

Vertical force= $-1.41\times 10^{-3}N$ (towards $q_1}$

Horizontal component are opposite in direction then will b subtracted

Horizontal force= $9\times 10^9\times 8\times 10^{-9}(-8.2+16.4) \times 10^{-9}\times \frac{sin30}{(0.033)^2}$

Horizontal force= $0.27\times 10^-3}$ N(towards $q_2$

Net electric force acting on particle 3 due to particle = $\sqrt{F^2_x+F^2_y}$

Net force= $\sqrt{(-1.41\times 10^{-3})^2+(0.27\times 10^{-3})^2}$

Net force= $1.44\times 10^{-3}$ N

3 0

3 years ago

A wire with a linear mass density of 1.17 g/cm moves at a constant speed on a horizontal surface and the coefficient of kinetic

stira [4]

Answer:

The value is $B = 0.2312 \ T$

The direction is into the surface

Explanation:

From the question we are told that

The mass density is $\mu =\frac{m}{L} = 1.17 \ g/cm =0.117 kg/m$

The coefficient of kinetic friction is $\mu_k = 0.250$

The current the wire carries is $I = 1.24 \ A$

Generally the magnetic force acting on the wire is mathematically represented as

$F_F = F_B$

Here $F_F$ is the frictional force which is mathematically represented as

$F_F = \mu_k * m * g$

While $F_B$ is the magnetic force which is mathematically represented as

$F_B = BILsin(\theta )$

Here $\theta =90^o$ is the angle between the direction of the force and that of the current

So

$F_B = BIL$

So

$BIL = \mu_k * m * g$

=> $B = \mu_k * \frac{m}{L} * [\frac{g}{I} ]$

=> $B = 0.25 * 0.117 * [\frac{9.8}{1.24} ]$

=> $B = 0.2312 \ T$

Apply the right hand curling rule , the thumb pointing towards that direction of the current we see that the direction of the magnetic field is into the surface as shown on the first uploaded image

8 0

3 years ago