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Natalka [10]
3 years ago
6

What is the force of friction between the dragsters tires and the track if the dragster mass is 300kg and the coefficient of the

friction force is 0.8
Physics
1 answer:
Nuetrik [128]3 years ago
7 0

Ff=μN=μmg=0.8*300*10=2400 N

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Would you rather have no air conditioning or no heating? Explain your answer
TEA [102]
No air conditioning, I already don’t have it, I only have heating :(
4 0
3 years ago
Read 2 more answers
What does the membrane in the cochlea do<br> what is the membranes function in the ear
elena-14-01-66 [18.8K]

The cochlea detects sound and is tonotopically organized

The cochlea is the auditory sense organ within the inner ear that is responsible for hearing. The cochlea is a coiled, fluid-filled tube that is split into three chambers by two membranes, Reissner's membrane and the basilar membrane

6 0
2 years ago
An autotransformer is used to reduce the voltage of a 100-kilovolt amp, 480-volt secondary of an isolated type transformer, to s
tigry1 [53]

Answer:

42KVA

Explanation:

Given data

High Voltage (HV)= 480V

Low Voltage (LV)= 277V

Fo find

Size of transformer=?

Solution

To find the size of transformer here we use the co-ratio.The Co-ratio is given as:

Co-Ratio= (HV - LV)/HV

where

HV is High Voltage

LV is Low Voltage

Now put the values we get

Co- Ratio=(480-277)/480=.42

So the size of transformer is 42KVA

5 0
3 years ago
A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc
adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

f_{c_1 = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

f_{c_1 = 3 × 10¹⁰ / 2( 0.0711  )

f_{c_1 = 3 × 10¹⁰ cm/s / 0.1422  cm

f_{c_1 =  21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

so

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( (f_{c_1 / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀e^{-\alpha _1l_{min = -100dB

we substitute

20log₁₀e^{-(310.7np/m)l_{min = -100dB

l_{min = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0
3 years ago
The uniform rods AB and BC weigh 24 ky and kg, respectively,and the small wheel at C is of negligible weight. If the wheel ismov
victus00 [196]

The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

<h3>What is Potential and Kinetic energy?</h3>

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

T_{1}  + V_{1} = T_{2}  + V_{2}

h_{ab}  = h_{bc}  = 0.18m

potential energy at position 1,

V1 = m_{ab} gh_{ab}  + m_{bc} gh_{bc}

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18

V1 = 11.30112

kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} +  \frac{1}{2} m_{bc} v^{2}_{G} +  \frac{1}{2} lw^{2}_{bc}

l_{ab} =\frac{m_{ab} l^{2}_{ab}  }{3}

= 1/3 *4 * (0.36)²

=0.10368kg m²

l =\frac{m_{bc} l^{2}_{bc}  }{12}

= 1/12 *4 * (0.6)²

=0.12kg m²

on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

vb = 3.2549 m/s

to learn more about Kinetic and potential energy go to - brainly.com/question/18963960

#SPJ4

5 0
1 year ago
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