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Tanya [424]
3 years ago
13

Why is it better to drive a large SUV in icy conditions with chains on the tires, instead of a small car with old tires? Explain

your answer using the terms: Force of Friction, Normal Force, Coefficient of friction, mass, rough, and smooth.
Plisss I need help
Physics
1 answer:
Step2247 [10]3 years ago
4 0

Answer:

It is better to drive a large SUV in stead of a smaller car in icy conditions, because the large SUV contains enough friction of force to drive in the snow on the ground. The mass of the car also plays a huge effect on this because then it does not slide off the road. The large tires on the big SUV make you ride smoothly through the snow and ice. With driving a smaller car though, the old tires have already been through a lot of driving, so this makes driving the car rough on the road. This also makes it so there is not a normal force in the wheels, so this means the car does not go at its normal speed, and this could cause an accident, or the rough tires could cause a fire when driving.

Trust me on this. I am 17, and I drive a Nissan Armada as my first car. Life changing. Happy Holidays!

Explanation:

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The efficiency of a light source is the percentage of its energy input that gets radiated as visible light. if some of the blue
Anestetic [448]

The efficiency of a light source is the percentage of its energy input that gets radiated as visible light if some of the blue light in an led is used to cause a fluorescent material to glow the overall efficiency of the LED decreases.

How efficient is LED?

Different wavelengths that correlate to different visible colours are used in LED light therapy. Various shades pierce the skin at different rates.

  1. Your skin's outermost layer is impacted by blue light.
  2. Yellow light is more enveloping.
  3. Red light penetrates your skin more deeply.
  4. The deepest penetrating light is near-infrared.

Different LED hues have various effects. For instance, according to experts red LED light therapy has the potential to reduce inflammation and boost collagen formation, which declines with age and is crucial for maintaining youthful-looking skin.

Acne-causing bacteria may be destroyed by a blue LED light therapy (P. acnes).

Hence the answer is the overall efficiency of the LED decreases.

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brainly.com/question/1263539

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7 0
2 years ago
the resistance of a wire of length 80cm and of uniform area of cross-section 0.025cmsq., is found to be 1.50 ohm. Calculate spec
vivado [14]
Specific\ resistance\ =resistivity\\
From\ formula\ on \ resistance:\ R= \frac{pL}{A}\ p-resistivity,\ L-length,\\ A-area\ of\ cross\ section\\
p= \frac{R*A}{L}= \frac{1,5Ohm*0,025*10^{-4}m^2 }{80* 10^{-2}m }=0,0003515625 Ohm*m
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3 years ago
An amusement park ride consists of a rotating circular platform 10.9 m in diameter from which 10 kg seats are suspended at the e
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4 0
3 years ago
If a 0.08 kg cell phone falls off a table at 15 m/s, then what is its kinetic energy right before it hits the ground?
Mariana [72]

The kinetic energy of the phone right before it hits the ground is 9J.

<h3>Kinetic energy of the phone</h3>

The kinetic energy of the phone right before it hits the ground is calculated as follows;

K.E = ¹/₂mv²

where;

  • m is mass of the phone
  • v is velocity of the phone

K.E = ¹/₂(0.08)(15)²

K.E = 9 J

Thus, the kinetic energy of the phone right before it hits the ground is 9J.

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7 0
2 years ago
A circular bird feeder 19.0 cm in radius has rotational inertia 0.130 kg·m2. It's suspended by a thin wire and is spinning slowl
soldi70 [24.7K]

Answer:

N₂=20.05 rpm

Explanation:

Given that

R= 19 cm

I=0.13 kg.m²

N₁ = 24.2 rpm

\omega_1=\dfrac{2\pi \times 24.2}{60}\ rda/s

ω₁= 2.5 rad/s

m= 173 g = 0.173 kg

v=1.2 m

Initial angular momentum L₁

L₁ =  Iω₁  - m v r       ( negative sign because bird coming opposite to motion of the wire motion)

Final linear momentum L₂

L₂=  I₂ ω₂

 I₂ = I + m r²

The is no any external torque that is why angular momentum will be conserve

L₁ = L₂

Iω₁  - m v r =  I₂ ω₂

Iω₁  - m v r =  ( I + m r²) ω₂

Now by putting the all values

Iω₁  - m v r =  ( I + m r²) ω₂

0.13 x 2.5 - 0.173 x 1.2 x 0.19 =  ( 0.13 + 0.173 x  0.19²) ω₂

0.325  - 0.0394 = 0.136 ω₂

ω₂ = 2.1 rad/s

\omega_2=\dfrac{2\pi \times N_2}{60}

N₂=20.05 rpm

3 0
3 years ago
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