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bija089 [108]
3 years ago
12

If the voltage output of a digital manifold absolute pressure (MAP)sensor is 2 volts, the approximate engine vacuum is _______ i

nches Hg.
A. 15

B. 10

C. 0.0

D. 5
Physics
1 answer:
LenaWriter [7]3 years ago
3 0
The correct answer to this question is 15
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A 500 N force accelerates an object at 20 m s-2. What is its mass?
avanturin [10]

<u>Answer</u>: The mass of the object is 25kg.

The given question deals with Newton's second law of motion and its applications.

<u>Explanation:</u> Given force, F=500N

                                 acceleration, a=20 m/s^{2}

  From Newton's 2nd law of motion , we have

                             F=ma where m=mass of the object

                         ⇒500=m×20

                         ⇒m=500/20=25

                 ∴ Mass of the object is 25 kg .

<u> </u><u>Reference Link: </u>brainly.com/question/1141170

#SPJ2

8 0
2 years ago
A dog is 60m away while moving at constant velocity of 10m/s towards you. Where is the dog after 4 seconds?
evablogger [386]
20m away

the dog was 60m away from. you subtract 40m since it is 10m/s x 4 seconds
4 0
3 years ago
Read 2 more answers
A 300 gg ball on a 70-cmcm-long string is swung in a vertical circle about a point 200 cmcm above the floor. The string suddenly
Ratling [72]

Answer:

the   tension in the string an instant before it broke = 34 N

Explanation:

Given that :

mass of the ball m = 300 g = 0.300 kg

length of the string r = 70 cm = 0.7 m

At highest point, law of conservation of energy can be expressed as :

\frac{1}{2} mv^2 = mgh\\\\v = \sqrt{2gh}\\\\v = \sqrt{2*(9.8 \  m/s^2)*(6.00 \ m - 2.00 \ m)}\\\\

v = 8.854 \ m/s

The tension in the string is:

T = \frac{mv^2}{r}\\\\T = \frac{(0.300 \ kg)*(8.854 \ m/s^2)}{0.70 \ m}\\\\T = 33.59 N\\\\T = 34 \ N

Thus, the   tension in the string an instant before it broke = 34 N

6 0
3 years ago
A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7.
dalvyx [7]

Answer:

(a) W_c=127.008 J

(b) W_g=148.176 J

(c) K.E. = 21.168 J

(d) v=3.4293m.s^{-1}

Explanation:

Given:

  • mass of a block, M = 3.6 kg
  • initial velocity of the block, u=0 m.s^{-1}
  • constant downward acceleration, a_d= \frac{g}{7}

\Rightarrow That a constant upward acceleration of \frac{6g}{7} is applied in the presence of gravity.

∴a=- \frac{6g}{7}

  • height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

F_c= M\times a

F_c=3.6\times (-6)\times\frac{9.8}{7}

F_c=-30.24 N

∴Work by the cord on the block,

W_c= F_c\times d

W_c= -30.24\times 4.2

We take -ve sign because the direction of force and the displacement are opposite to each other.

W_c=-127.008 J

(b)

Force on the block due to gravity:

F_g= M.g

∵the gravity is naturally a constant and we cannot change it

F_g=3.6\times 9.8

F_g=35.28 N

∴Work by the gravity on the block,

W_g=F_g\times d

W_g=35.28\times 4.2

W_g=148.176 J

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

K.E.= W_g+W_c

K.E.=148.176-127.008

K.E. = 21.168 J

(d)

From the equation of motion:

v^2=u^2+2a_d\times d

putting the respective values:

v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }

v=3.4293m.s^{-1} is the speed when the block has fallen 4.2 meters.

6 0
3 years ago
A car engine applies a force of 65,000 N, how much work is done by the engine as it pushed a car a distance of 75 m?
kupik [55]

Answer:

workdone = force \times distance \\  = 65000 \times 75 \\  = 4,875,000 \: J

3 0
3 years ago
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