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2 years ago

15

an object starts from rest with a constant acceleration of 8.0m/s^2 along a straight line.find the average speed for 5 second in

terval

1 answer:

frozen [14]2 years ago

6 0

Answer:

40 m/s

Explanation:

If you multiply the acceleration with time you get the average speed.

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35 POINTS! Say If You Drop A Ball from 100 Centimeters. When The ball Bounces, The Ball Does Not Bounce Back Up To 100 Centimete

diamong [38]

some ball when you bounce it it comes back up but according to gravity the energy goes away

8 0

3 years ago

One of the checks that you could do for problem 1) would be to check the output resistance of the Wheatstone bridge to make sure

Ray Of Light [21]

Answer:

Explanation:

Let the four resistances of th wheat stone bridge is

P, Q, R and S and the value of each is 350 ohm.

Here, P and Q are in series.

R' = P + Q = 350 + 350 = 700 ohm

Then R and S are in series

R' = R + S = 350 + 350 = 700 ohm

Now R' and R'' are in parallel.

So, the equivalent resistance is

Req = R' x R'' / ( R' + R'')

Req = 700 / 2 = 350 ohm

Thus, the reading of ohmmeter is 350 ohm.

6 0

3 years ago

What are some tasks that organelles perform inside of a cell?

natta225 [31]

Answer:

Different organelles play different roles in the cell — for instance, mitochondria generate energy from food molecules; lysosomes break down and recycle organelles and macromolecules; and the endoplasmic reticulum helps build membranes and transport proteins throughout the cell.

Explanation:

Hopefully this helped!

4 0

3 years ago

The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of

adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = $ma_a (0.35)$

$1.8v^2(1.1)$ - 1200(9.8)(1.25) = 1200a(0.35)

$1.8v^2(1.1)$ - 14700 = 420 a ------- equation (1)

$F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a$ --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

${1.1 * 1200 \ a} - 14700 = 420 a$

1320 a - 14700 = 420 a

1320 a - 420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

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3 years ago

Let x define the position of an object such that x =

timama [110]

See the attached picture for answers

7 0

3 years ago