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almond37 [142]
2 years ago
14

When vehicle wheels are about to lock, the ABS senses the impending lock up and prevents this from occurring. a) True b) False 2

Question 2 Electronic stability control reduces the risk of rollover crashes, such as are common with SUVs. a) True b) False 3 Question 3 _______________have the ability to control wheel lock on all four tires. a) Four-wheel anti-lock brakes b) Rear-wheel anti-lock brakes c) All anti-lock brakes d) None of the above 4 Question 4 _______________is a technology that works to help maintain vehicle control and steering capability in the event of a skid. a) Parking/emergency brake b) Service brakes c) Anti-lock Braking system d) None of the above 5 Question 5 ___________ reduce the impact forces of a crash exerted upon a vehicle's occupants by allowing parts of the front of the vehicle to crush at varying rates. a) Electronic Stability Control b) Anti-lock Braking Systems c) Crumple Zones d) Traction Control Systems 6 Question 6 The Traction Control System (TCS) is designed to help maintain traction on a vehicle in hazardous conditions or situations. a) True b) False 7 Question 7 A driver operating a vehicle with traction control should brake for the feature to activate. a) True b) False 8 Question 8 Air bags in trucks or other vehicles that do not have a rear seat will have an on/off switch that can be activated for children under age 12. a) True b) False 9 Question 9 All vehicles are equipped with Anti-lock Braking Systems. a) True b) False 10 Question 10 Airbags may deploy in the _______ of the passenger or driver, or from the ____of the vehicle.
Physics
1 answer:
Anna35 [415]2 years ago
4 0

The answers to the questions are:

  1. True
  2. True
  3. Four-wheel anti-lock brakes
  4. Anti-lock braking system
  5. Crumple zones
  6. True
  7. False
  8. True
  9. false
  10. Front, sides

<h3>What is ABS?</h3>

ABS is known as an Anti-lock braking systems (ABS) that act in the capacity of steering one in times in emergencies by bringing back traction to one's tires.

Note that it often helps to hinder  wheels from locking up and helping the driver to be able to steer to safety.

When vehicle wheels are about to lock, the ABS senses the impending lock up and prevents this from occurring is true as that is its function.

Note that Electronic stability helps to control helps to the risk of rollover crashes, such as are common with SUVs.

Learn more about vehicle wheels from

brainly.com/question/24233118

#SPJ1

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1. 1. are the charged parts of an atom.
victus00 [196]

Answer:

» e. Electrons and protons

Explaination :

Electrons are negatively charged and protons are positively charged.

  • The neutrons do not have a charge.
5 0
3 years ago
Read 2 more answers
The magnification of a microscope is increased when_________.
azamat

Answer:

Option B

Explanation:

Magnification of Microscope is  

M = M_o \times M_e  

Mo= Magnification of objective lens and  

Me= magnification of the eyepiece.  

Both magnifications( of objective and eyepiece) are inversely proportional to the focal length.  

Magnification,  

M \propto \dfrac{1}{f}

when the focal length is less magnification will be high and when the magnification is the low focal length of the microscope will be more.

Thus. Magnification will increase by decreasing the focal length.

The correct answer is Option B

6 0
4 years ago
Movies and TV shows sometimes portray a person being thrown backwards a sizable distance as a result of being struck by a bullet
MAXImum [283]

Answer:

 R = 4.24 x 10⁻⁴ m

Explanation:

given,

mass of the person = 60.3-kg

mass of the bullet = 10 gram = 0.01 Kg

velocity of bullet = 389 m/s

angle made with the horizontal = 45°

using conservation of momentum.

M v  + m u  = ( M + m ) V

60.3 x 0 + 0.01 x 389 = (60.3 + 0.01) V

V = \dfrac{3.89}{60.31}

V = \dfrac{3.89}{60.31}

V = 0.0645 m/s

for calculation of range

R = \dfrac{V^2sin 2 \theta}{g}

R = \dfrac{0.0645^2sin 2 (45^0)}{9.8}

     R = 4.24 x 10⁻⁴ m

the distance actor fall is  R = 4.24 x 10⁻⁴ m

6 0
4 years ago
Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
solmaris [256]

Answer:

a

\lambda = 3.68 *10^{-36} \  m

b

\lambda_p = 1.28*10^{-14} \ m

Explanation:

From the question we are told that

   The mass of the person is  m =  180 \  kg

    The speed of the person is  v  =  1 \  m/s

    The energy of the proton is  E_ p =  5 MeV = 5 *10^{6} eV  = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \  J

Generally the de Broglie wavelength is mathematically represented as

      \lambda = \frac{h}{m * v }

Here  h is the Planck constant with the value

      h = 6.62607015 * 10^{-34} J \cdot s

So  

     \lambda = \frac{6.62607015 * 10^{-34}}{ 180  * 1  }

=> \lambda = 3.68 *10^{-36} \  m

Generally the energy of the proton is mathematically represented as

         E_p =  \frac{1}{2}  *   m_p  *  v^2_p

Here m_p  is the mass of proton with value  m_p  =  1.67 *10^{-27} \  kg

=>     8.0*10^{-13} =  \frac{1}{2}  *   1.67 *10^{-27}  *  v^2

=>   v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }

=>   v = 3.09529 *10^{7} \  m/s

So

        \lambda_p = \frac{h}{m_p * v_p }

so    \lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }

=>     \lambda_p = 1.28*10^{-14} \ m

     

5 0
3 years ago
An electric furnace runs 13 hours a day to heat a house during January (31 days). The heating element has a resistance of 7.2 an
liq [111]

Answer:

cost of running the furnace during January is $5619.62

Explanation:

given data

runs a day = 13 hours

January days = 31 days

resistance = 7.2 ohm

current = 16.7 A

cost of electricity = $0.10/kWh

to find out

cost of running the furnace during January

solution

first we get her power consumed by furnace that is

Power consumed = \frac{I^2}{R}  ........1

put here value we get

Power consumed = \frac{16.7^2}{7.2}

Power consumed = 38.7347 W

and

Power consumed by furnace in one hour is

Power consumed by furnace in one hour is = Power consumed × 3600

Power consumed by furnace in one hour is = 38.7347 × 3600  

Power consumed by furnace in one hour is 139.445kWh

and

Power consumed by furnace in the month of January is

Power consumed by furnace in the month of January = 139.445kWh × 13 hours × 31 days

Power consumed by furnace in the month of January = 56196.335 kWh

so

cost of running the furnace during January is = $0.10/kWh × 56196.335 kWh

cost of running the furnace during January is $5619.62

4 0
3 years ago
Read 2 more answers
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