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1 year ago

Traffic officials indicate, it takes longer to ______ when you drive fast.

Physics

2 answers:

Lyrx [107]1 year ago

8 0

Answer:

Stop

Explanation:

when we drive fast it will be take longer time to stop.

As car is moving fast the acceleration will be to decelerate we have to apply brake.

By applying brake the car decelerates very fast as deceleration is the change in velocity per unit time. change in velocity is very high so deceleration will be high.

when the brake is applied the distance travel by the high speed vehicle will be more than the vehicle moving at slow speed.

nignag [31]1 year ago

7 0

The answer in the blank is that it is difficult to accelerate at decelerate the vehicle when it is on a fast speed because having a fast speed makes it difficult to adjust the meter as well as if you try to decelerate the vehicle, it could burn out the tires and engine as it is in the fast speed, in accelerating it, it could also be complicated because it would only make the car faster enough that you may no longer control of how to stop it.

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A metal bar moves through a magnetic field. the induced charges on the bar are

Dmitry [639]

I would say its a positive cgarge

7 0

2 years ago

A bookshelf is 2m long, with supports at its end (p and q). A book weighing 10N is placed in the middle of the shelf. what are t

mel-nik [20]

If the book is placed in the middle, the forces acting on p and q is 5N. If the book is moved 50 cm from q, the forces at p and q can be solved by doing a moment balance

With p as the pivot;

Fq (2 m) = 10 N (0.5 m)

Fq = 2.5 N

Fp = 10 N - 2.5 N = 7.5 N

5 0

2 years ago

If you were to separate all of the electrons and protons in 1.00 g (0.001 kg) of matter, you’d have about 96,000 C of positive c

natima [27]

Answer:

The attractive force between them is $1.296 \times 10^{18}$ N

Explanation:

Given:

Charge $q = 96000$ C

Distance between two charges $r = 8$ m

According to the coulomb's law,

$F = \frac{kq^{2} }{r^{2} }$

Where $k = 9 \times 10^{9}$ = force constant.

$F = \frac{9 \times 10^{9} \times (96000)^{2} }{8^{2} }$

$F = 1.296 \times 10^{18}$ N

Therefore, the attractive force between them is $1.296 \times 10^{18}$ N

3 0

2 years ago

Assume that our computer stores decimal numbers using 16 bits - 10 bits for a sign/magnitude mantissa and 6 bits for a sign/magn

madreJ [45]

Explanation:

a) 7.5= 111.1×2°= 0.1111×2^3

which can also be written as

(1/2+1/4+1/8+1/16)×8

sign of mantissa:=0

Mantissa(9 bits): 111100000

sign of exponent: 0

Exponent(5 bits): 0011

the final for this is:011110000000011

b) -20.25= -10100.01×2^0= -0.1010001×2^5

sign of mantissa: 1

Mantissa(9 bits): 101000100

sign of exponent: 0

Exponent(5 bits): 00101

the final for this is:1101000100000101

c)-1/64= -.000001×2^0= -0.1×2^{-5}

sign of mantissa: 1

Mantissa(9 bits): 100000000

sign of exponent: 0

Exponent(5 bits): 00101

the final for this is:1100000000100101

5 0

2 years ago

Ten days after it was launched toward Mars in December 1998, the Mars Climate Orbiter spacecraft (mass 629 kg) was 2.87×106km fr

Andreas93 [3]

Answer:

3494444444.44444 J

-87077491.39453 J

Explanation:

M = Mass of Earth = $6.371\times 10^{6}\ kg$

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

R = Radius of Earth = $6.371\times 10^{6}\ m$

h = Altitude = $2.87\times 10^9\ m$

m = Mass of satellite = 629 kg

v = Velocity of spacecraft = $1.2\times 10^4\ km/h$

The kinetic energy is given by

$K=\frac{1}{2}629\times \left(1.2\times 10^4\times \dfrac{1000}{3600}\right)^2\\\Rightarrow K=3494444444.44444\ J$

The spacecraft's kinetic energy relative to the earth is 3494444444.44444 J

Potential energy is given by

$U=-\dfrac{GMm}{R_e+h}\\\Rightarrow U=-\dfrac{6.67\times 10^{-11}\times 5.97\times 10^{24}\times 629}{2.87\times 10^9+6.371\times 10^{6}}\\\Rightarrow U=-87077491.39453\ J$

The potential energy of the earth-spacecraft system is -87077491.39453 J

4 0

1 year ago