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3 years ago

Should you be worried if a n eighteen wheeler truck is riding close behind your car? Explain why or why not. Please help me D:

1 answer:

3 0

For some reasons, no. If the driver looks focused and has experience, then it would be okay. Again, it could be dangerous if you bump into the truck, it would cause damage to you and your passengers.

Mostly, I would agree with 'No'. :)

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A window air conditioner that consumes 1.4 kW of electricity when running and has a coefficient of performance of 3 is placed in

Shtirlitz [24]

Answer:

1.4 kJ/s

Explanation:

See attached photo

7 0

3 years ago

How does momentum conservation work for a rocket firing its engines to speed up in deep space?

Mashutka [201]

Idkhhhhhhhhhhubvgbvcccc xzzz. Bcc bbb

3 0

3 years ago

A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic

miss Akunina [59]

Answer:

0.07756 m

Explanation:

Given mass of object =0.20 kg

spring constant = 120 n/m

maximum speed = 1.9 m/sec

We have to find the amplitude of the motion

We know that maximum speed of the object when it is in harmonic motion is given by $v_{max}=A\omega$ where A is amplitude and $\omega$ is angular velocity

Angular velocity is given by $\omega=\sqrt{\frac{k}{m}}$ where k is spring constant and m is mass

So $v_{max}=A\sqrt{\frac{k}{m}}$

$A=V_{max}\sqrt{\frac{m}{k}}=1.9\times \sqrt{\frac{0.2}{120}}=0.07756 \ m$

3 0

3 years ago

Read 2 more answers

A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. The wire ful

balu736 [363]

Answer:

FB = 0.187 N

Explanation:

To find the magnetic force FB in the wire you use the following formula:

$|\vec{F_B}|=ILBsin\theta\\\\L=0.25m\\\\|\vec{B}|=\sqrt{(0.3y)^2+(0.4y)^2}=0.5y \ T$

the angle between B and L is given by:

$\vec{L}\cdot\vec{B}=LBcos\theta\\\\\theta=cos^{-1}(\frac{\vec{L}\cdot\vec{B}}{LB})=cos^{-1}(\frac{0*0.3y+0.25*0.4y}{0.25*0.5y})=36.86\°$

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:

$|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N$

hence, the magnitude of the magnetic force is 0.187N

4 0

3 years ago

A ray is incident on a film of thickness t and the index of refraction n=1.6 at an angle θ = 35 degrees. Find the angle of refra

lidiya [134]

Answer

Given,

refractive index of film, n = 1.6

refractive index of air, n' = 1

angle of incidence, i = 35°

angle of refraction, r = ?

Using Snell's law

n' sin i = n sin r

1 x sin 35° = 1.6 x sin r

r = 21°

Angle of refraction is equal to 21°.

Now,

distance at which refractive angle comes out

d = 2.5 mm

α be the angle with horizontal surface and incident ray.

α = 90°-21° = 69°

t be the thickness of the film.

So,

$tan \alpha = \dfrac{t}{d/2}$

$tan 61^0 = \dfrac{t}{2.5/2}$

t = 2.26 mm

Hence, the thickness of the film is equal to 2.26 mm.

4 0

3 years ago