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3 years ago

Calculate the acceleration of a galloping horse going from 2 m/s to 12 m/s in 2 seconds.

Physics

1 answer:

olchik [2.2K]3 years ago

8 0

A)
5m/s^2

(12m/s-2m/s)
__________ = 5m/s^2
2s

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Are mushrooms, producers, consumers, or decomposers?

swat32

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2 years ago

In case 1, a force f is pushing perpendicular on an object a distance l/2 from the rotation axis. in case 2 the same force is pu

Serggg [28]

In case 1, the torque is given by the product between the force and the arm:
$\tau_1 = F \cdot \frac{L}{2}$

In case 2, the torque is given by the product between the component of the force perpendicular to the arm and the arm itself, so we have:
$\tau_2 = F \cos 30^{\circ} L=F \frac{ \sqrt{3} }{2} L = \sqrt{3} F \frac{L}{2}= \sqrt{3} \tau_1$

and since $\sqrt{3}$ is larger than 1, than the torque in case 2 is larger than the torque in case 1.

6 0

3 years ago

Point charge q1 of 30 nC is separated by 50 cm from point charge q2 of -45 nC. As shown in the diagram, point a is located 30 cm

Angelina_Jolie [31]

Answer:

E1 = 2996.667N/C E2 = 11237.5N/C

Explanation:

E1 = kQ1/r^2

=8.99 x 10^9 x 30 x 10^-9/(30x10^-2)^2

= 2996.667N/C

E2 = kQ2/r^2

= 8.99 x 10^9 x 50 x 10^-9/(20x10^-2)^2

= 11237.5N/C

The direction are towards the point a

6 0

2 years ago

A car starting from rest (i.e. initial velocity = 0.0 m/s), moves in the positive X-direction with a constant average accelerati

olga_2 [115]

Answer:

Final speed of the car, v = 24.49 m/s

Explanation:

It is given that,

Initial velocity of the car, u = 0

Acceleration, $a=3.1\ m/s^2$

Time taken, t = 7.9 s

We need to find the final velocity of the car. Let it is given by v. It can be calculated using first equation of motion as :

$v=u+at$

$v=0+3.1\times 7.9$

v = 24.49 m/s

So, the final speed of the car is 24.49 m/s. Hence, this is the required solution.

8 0

2 years ago

You see lightning and 30 seconds later you hear thunder. how far away is the thunderstorm? take the speed of sound to be 339 m/s

Jobisdone [24]

Let the observer be 'd' distance away from the thunderstorm and let light take 't' time to reach the observer
Since the speed of sound and light remains constant in a particular medium, we can use
Speed = Distance/Time

For light,
3 x 10^8 = d/t
t = d/(3 x 10^8) -1

For sound,
339 = d/(t + 30) -2

Putting value from 1 in 2.
d = 10^4 m(approx)

3 0

2 years ago