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3 years ago

Please help!!

Physics

2 answers:

Sveta_85 [38]3 years ago

6 0

Answer:

If you push hard, the block's bottom will move out from under its center of gravity and the block will tip over. The taller the block, the easier it is to push the bottom out from under the center of gravity. Keeping a car's center of gravity low is important to avoiding disaster during turns. But if you don't keep the car's center gravity low it will be inversely proportional and the acceleration can be damaged with force, and its mass could get disproportionate.

Explanation:

I'm sorry i tried my best. :/

baherus [9]3 years ago

4 0

Answer:

Mass is the correct answer.

Explanation:

Many drivers report a more positive handling response and a definite improvement when reducing unsprung mass. You want to keep unsprung weight to as little as possible. This minimizes the momentum and energies that your suspension has to counter. In effect, it can make your shocks more sensitive.

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Answer:

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An object with a mass of 10.5kg is traveling at a velocity of 9m/s. What is the kinetic energy of

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KE = 1/2mv^2
1/2(10.5)(9)^2
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2 years ago

A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra

Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

$sin(\alpha) = \frac{Vyi}{Vi}$

$Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}$

Vy in this problem will follow this equation =

$Vy(t) = Vyi -g.t$

where g is the gravity acceleration

$Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t$

This is equation (1)

For Y(t) :

$Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}$

We suppose yi = 0

$Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}$

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

$Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s$

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

$Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} } .(0.675s)^{2} \\Y(0.675s) =2.236 m$

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

$Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m$

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

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3 years ago

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a 0.145 kg baseball pitched at 39.0 m/s is hit on a horizontal line drive straight back toward the pitcher at 52.0 m/s. if the c

velikii [3]

consider the velocity towards the pitcher as positive

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F = average force between bat and ball

Using impulse-change in momentum equation

F t = m (v - v₀ )

F (3 x 10⁻³) = (0.145) (52 - (- 39))

F = 4398.33 N

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3 years ago