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Vlada [557]
2 years ago
9

A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration?

Physics
1 answer:
Scrat [10]2 years ago
4 0

Answer:

{ \boxed{ \bold{ \sf{Acceleration \: ( \: a) = 8 \: m/ {s \: }^{2} }}}}

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♨ Question :

  • A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration ?

♨ \underbrace{ \sf{Required \: Answer : }}

☄ Given :

  • Initial velocity ( u ) = 0
  • Final velocity ( v ) = 60.0 m/s
  • Time ( t ) = 7.50 s

☄ To find :

  • Acceleration ( a )

✒ We know ,

\boxed{ \underline{ \bold{ \sf{Acceleration \: ( \: a) =  \frac{Final velocity ( v )  - Initial velocity ( u)}{t} }}}}

Substitute the values and solve for a.

➛ \sf{a =  \frac{60.0 - 0}{7.50}}

➛ \sf{a =  \frac{60.0}{7.50}}

➛ \boxed{ \boxed{ \sf{a = 8 \: m/ {s \: }^{2} }}}

---------------------------------------------------------------

✑ Additional Info :

  • When a certain object comes in motion from rest , in the case , initial velocity ( u ) = 0
  • When a moving object comes in rest , in the case , final velocity ( v ) = 0
  • If the object is moving with uniform velocity , in the case , u = v.
  • If any object is thrown vertically upwards in the case , a = -g
  • When an object is falling from certain height , in the case , final velocity at maximum height ( v ) = 0.

Hope I helped!

Have a wonderful time ツ

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Answer:

Paying for employees seminars and workshops related to their careers

Explanation:

To motivate personal development among employees, several things can be done. Among them, giving employees chance to present their own solutions to problems, exposing the employees to several global challenges and how to handle them, paying for employees seminars and workshops related to their own careers for professional development among other things.

8 0
3 years ago
57. Example of the law of force and acceleration
7nadin3 [17]

Answer:

Newton's Second Law of Motion says that acceleration (gaining speed) happens when a force acts on a mass (object). Riding your bicycle is a good example of this law of motion at work. Your bicycle is the mass. Your leg muscles pushing pushing on the pedals of your bicycle is the force.

Explanation:

6 0
2 years ago
Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25
mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

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3 years ago
Which type of water body does water move the slowest?
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7 0
3 years ago
Would a measured force of (46.5 0.8 N  ) be in agreement with a theoretically calculated force of (48.4 0.6 N  ) ? Show your w
OverLord2011 [107]

Answer:

A measured force of (46.5 0.8 N  ) would not be in agreement with a theoretically calculated force of (48.4 0.6 N  )

Explanation:

From the question we are told that

  Measured force is  F_M  =  [46.5 \pm 0.8 \  N ]

   Calculated force is  F_c =  [48.4 \pm 0.6 \  N ]

Generally the measured force in interval form is

     46.5 - 0.8  < F_M  <  46.5 + 0.8

=>  45.7   < F_M  < 47.3

Generally the calculated  force in interval form is

     48.4 - 0.6  < F_c  <  48.4 + 0.6

=>  47.8   < F_M

Generally looking both interval we see that they do not intersect at any point Hence  

A measured force of (46.5 0.8 N  ) would not be in agreement with a theoretically calculated force of (48.4 0.6 N  )      

8 0
2 years ago
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