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Thepotemich [5.8K]
2 years ago
9

If the brakes are applied and the speed of the car is reduced to 13 m/s in 17 s , determine the constant deceleration of the car

.
Physics
1 answer:
Svetlanka [38]2 years ago
8 0

Question:  Initially, the car travels along a straight road with a speed of 35 m/s. If the brakes are applied and the speed of the car is reduced to 13 m/s in 17 s, determine the constant deceleration of the car.

Answer:

1.29 m/s²

Explanation:

From the question,

a = (v-u)/t............................ Equation 1

Where a = deceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.

Given: v = 13 m/s, u = 35 m/s, t = 17 s.

a = (13-35)/17

a = -22/17

a = -1.29 m/s²

Hence the deceleration of the car is 1.29 m/s²

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Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit
Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

  $= 822486 \times 10^{-8}$

  $=0.822 \times 10^{-2} \ km/s$

 = 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

  $= 673552 \times 10^{-8}$

  $=0.673 \times 10^{-2} \ km/s$

 = 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

  $= 3371 \times 153.86 \times 10^{-8}$

  = 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

  $= 153.84 \times 2371 \times 10^{-8}$

  $=0.364 \times 10^{-2} \ km/s$

 = 3.64 m/s

       

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3 years ago
When the sound from two sound waves goes up and down (loud, soft, loud, soft) it makes a distinctive sound pattern called beats.
djverab [1.8K]

The beats are actually two new sounds.

Their frequencies are (the sum of the original two frequencies) and (the difference of the original two frequencies).

The existence of the beats is the result of the difference in the frequencies of the original two sounds. <em> (b)</em>

8 0
3 years ago
Read 2 more answers
PLS ANSWER FAST WILL GIVE BRAINLY TIMED TEST
solong [7]
A=F/m
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2 years ago
Jeff puts on a leather jacket over his sweater. The sweater becomes negatively charged. Which statements about Jeff’s situation
KiRa [710]

The sweater has a tendency to attract electrons.

The leather jacket has a lower tendency to attract electrons than the sweater.

Explanation:

The sweater and the leather jackets are made up of distinct fabrics that based on their minutest particles called an atom.

An atom is made up of sub-atomic particles of protons, neutrons and electrons.

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  • Protons are positively charged and are very difficult to lose. They occupy the tiny nucleus with neutrons.
  • A body that becomes negatively charged will be said to have a hihg tendency to attract electrons. Normally atoms are electrically neutral. When additional electrons are added to them, they become negatively charged.
  • In this case, the sweater has a high affinity for electrons and it will attract the ones on the leather jacket.
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Learn more:

Protons, neutrons and electrons brainly.com/question/2757829

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Through what vertical height is a 50 N object moved if 250 J of work is done lifting it against the gravitational field of Earth
zimovet [89]

5m

Explanation:

Given parameters:

Weight of object = 50N

Work done in lifting object = 250J

Unknown:

Vertical height = ?

Solution:

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   Work done = force x distance

  Weight is a force in the presence of gravity;

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Height of lifting = \frac{work done }{weight}

 Height of lifting = \frac{250}{50} = 5m

The vertical height through which the object was lifted is 5m

learn more:

Work done brainly.com/question/9100769

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