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2 years ago

Round the following off to required number of significant figures.

Physics

2 answers:

Ray Of Light [21]2 years ago

6 0

140 but I’m not sure. Have a great day

Stells [14]2 years ago

4 0

Answer:

140

hope this helps

have a good day :)

Explanation:

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While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. how much time does it ta

Yuri [45]

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. The time taken by the car will be 4 seconds. the correct answer is option(c).

When acceleration is constant, the rate of change in velocity is also constant. In the absence of any acceleration, velocity remains constant. When acceleration is positive, velocity becomes more significant.

Let a denote acceleration, u denote initial velocity, v denote final velocity, and t denote time.

The equation of motion is stated as,

v = u + at

v² = u² + 2as

A car travels across a distance of 60.0 m while accelerating constantly from 12.0 m/s to 18.0 m/s.

Then the time taken by the car will be,

u = 12 m/s

v = 18 m/s

s = 60 m

Put these in the equation v² = u² + 2as.

18² = 12² + 2 x a x 60

a = 1.5

Then the time will be

18 = 12 + 1.5t

1.5t = 6

t = 4 seconds

Hence, the time taken is 4 s.

The complete question is:

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. How much time does it take?

1.00 s

2.50 s

4.00 s

4.50 s

To know more about acceleration refer to: brainly.com/question/12550364

#SPJ4

8 0

1 year ago

Answer it asap I promise i will mark them the Brainly

nataly862011 [7]

Answer:

A) The acceleration is zero

B) The total distance is 112 m

Explanation:

Velocity vs Time Graph

It shows the behavior of the velocity as time increases. If the velocity increases, then the acceleration is positive, if the velocity decreases, the acceleration is negative, and if the velocity is constant, then the acceleration is zero.

The graph shows a horizontal line between points A and B. It means the velocity didn't change in that interval. Thus the acceleration in that zone is zero.

A. To calculate the acceleration, we use the formula:

$\displaystyle a=\frac{v_2-v_1}{t_2-t_1}$

Let's pick the extremes of the region AB: (0,8) and (12,8). The acceleration is:

$\displaystyle a=\frac{8-8}{12-0}=0$

This confirms the previous conclusion.

B. The distance covered by the body can be calculated as the area behind the graph. Since the velocity behaves differently after t=12 s, we'll split the total area into a rectangle and a triangle.

Area of rectangle= base*height=12 s * 8 m/s = 96 m

Area of triangle= base*height/2 = 4 s * 8 m/s /2= 16 m

The total distance is: 96 m + 16 m = 112 m

4 0

2 years ago

A rifle fires a bullet at a velocity of 78 m/s, 40 degrees above the horizontal. Determine the height (h) above the starting pos

qwelly [4]

Answer:

H = Vy t - 1/2 g t^2 height of an object with an initial "vertical" velocity

at t sec after firing

Vy = 78 m/s * sin 40 = .643 * 78 m/s = 50.1 m/s

H = 50.1 * 6 - 1/2 * 9.8 * 6^2 = 300 m - 176 m = 124 m

7 0

2 years ago

1. On this graph, what is the change in velocity between 5 seconds and 8 seconds?

alukav5142 [94]

Can I see the graph so I can help you

8 0

2 years ago

Blocks A and B are identical metal blocks. Initially block A is neutral, and block B has a net charge of 8.7 nC. Using insulatin

JulsSmile [24]

Explanation:

(a) Since, it is given that the blocks are identical so distribution of charge will be uniform on both the blocks.

Hence, final charge on block A will be calculated as follows.

Charge on block A = $\frac{(8.7 + 0 nC}{2}$

= 4.35 nC

Therefore, final charge on the block A is 4.35 nC.

(b) As it is given that the positive charge is coming on block A . This means that movement of electrons will be from A to B.

Thus, we can conclude that while the blocks were in contact with each other then electrons will flow from A to B.

6 0

2 years ago

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