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Brut [27]
2 years ago
7

I know the acceleration due to gravity (ie 9.8 m/s2) will have a negative sign when falling down, a positive one when going up.

Physics
1 answer:
rewona [7]2 years ago
6 0

Answer:

Yes

Explanation:

Accerelation is measured by change in velocity. So naturally, if an object is slowing down, its velocity is decreasing so acceleration is negative. If it is speeding up velocity is increasing so positive acceleration.

(Velocity final - Velocity initial)/t

Note that this does not apply only to gravity, but to all linear accelerations

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What does it mean to say that in any system the total energy score stays the<br> same
timurjin [86]

Answer:

energy is conserved

a force sets an object in motion. when the force is multiplied by the time of its application we call the quantity impulse which changes the momentum of that object. what do we call the quantity force x (times) distance, and what quantity can this change?

7 0
3 years ago
a car travels from station A to B at 30kmph during its return trip A it travels 30 km pH for the first half distance and it 70 k
8_murik_8 [283]

Explanation:

If you cannot visualize it, just assume that the distance from station A to B is 420km. Each half is 210km.

When the car travels from A to B, it takes 420/30 = 14 hours.

When the car travels from B to the halfway point, it takes 210/30 = 7 hours.

When the car travels from the halfway point to A, it takes 210/70 = 3 hours.

Total time taken = 14 + 7 + 3 = 24 hours.

Total distance = 420km * 2 = 840km.

Hence, the average speed of the car is 840/24 = 35km/h.

3 0
3 years ago
A ___________ is how strong the push or pull is
KatRina [158]
The answer is force
4 0
3 years ago
_____________ forces are always attractive and the mass of an object determines how strong the ____________ pull is.
dolphi86 [110]
Gravitational, gravitational ! both the option are same
6 0
3 years ago
Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) .
jek_recluse [69]

Explanation:

Given that,

Force, F=((-8i)+6j)\ N

Position of the particle, r=(3i+4j)\ m

(a) The toque on a particle about the origin is given by :

\tau=F\times r

\tau=((-8i)+6j) \times (3i+4j)

Taking the cross product of above two vectors, we get the value of torque as :

\tau=(0+0-50k)\ N-m

(b) Let \theta is the angle between r and F. The angle between two vectors is given by :

cos\theta=\dfrac{r.F}{|r|.|F|}

cos\theta=\dfrac{(3i+4j).((-8i)+6j)}{(\sqrt{3^2+4^2} ).(\sqrt{8^2+6^2}) }

cos\theta=\dfrac{0}{50}

\theta=90^{\circ}

6 0
3 years ago
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