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o-na [289]
3 years ago
9

A car slows down at -5.00 m/s^2 until it comes to a stop after travelling 15.0 m. How much time did it take to stop?

Physics
1 answer:
Xelga [282]3 years ago
5 0
In physics, there are already derived equation that are based on Newton's Law of Motions. The rectilinear motions at constant acceleration have the following equations:

x = v₁t + 1/2 at²
a = (v₂-v₁)/t

where
x is the distance travelled
v₁ is the initial velocity
v₂ is the final velocity
a is the acceleration
t is the time

Now, we solve first the second equation. Since it mentions that the car comes eventually to a stop, v₂ = 0. Then,

-5 = (0-v₁)/t
-5t = -v₁
v₁ = 5t

We use this new equation to substitute to the first one:
x = v₁t + 1/2 at²
15 = 5t(t) + 1/2(-5)t²
15 = 5t² - 5/2 t²
15 = 5/2 t²
5t² = 30
t² = 30/5 = 6
t = √6 = 2.45

Therefore, the time it took to travel 15 m at a deceleration of -5 m/s² is 2.45 seconds.

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Explanation:

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Consequently, covering the bulb with a towel is a serious mistake.

7 0
3 years ago
The velocity of a 48.0 g shell leaving a 2.95 kg rifle is 391. m/s. What is the recoil velocity of the rifle?
Monica [59]

Hi there!

\large\boxed{  -6.36m/s}

Use the equation:

v_{2} = -\frac{m_{1}}{m_{2}} v_{1}

Where m2 and v2 deal with the larger object, and m1 and v1 with the smaller object. Plug in the given values:

v2 = ?

m1 = 0.048 kg (converted)

m2 = 2.95

v1 = 391

v_{2} = -\frac{0.048}{2.95} *391

v_{2} = -6.36m/s

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The pulling force between two objects like earth and you is _____. (1 point
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A hydraulic system is being used to lift a 1500-kg car. If the large piston under the car has a diameter of 50 cm, the small pis
Ivanshal [37]

Answer:

W = 122.3 J

Explanation:

First, we need to find out the force applied to the smaller piston. We know that the pressure applied to smaller piston must be equally transmitted to the larger piston. Therefore,

P₁ = P₂

F₁/A₁ = F₂/A₂

F₂ = F₁(A₂/A₁)

where,

F₁ = Force of Larger Piston = Weight of car = mg = (1500 kg)(9.8 m/s²)

F₁ = 14700 N

F₂ = Force applied to smaller piston = ?

A₁ = Area of larger piston = πd₁²/4

A₂ = Area of smaller piston = πd₂²/4

Therefore,

F₂ = (14700 N)[(πd₂²/4)/(πd₁²/4)]

F₂ = (14700 N)(d₂²/d₁²)

where,

d₁ = diameter of large piston = 50 cm

d₂ = diameter of small piston = 4 cm

Therefore,

F₂ = (14700 N)[(4 cm)²/(50 cm)²]

F₂ = 94.08 N

Now, for the work done on the car:

Work Done = W = F₂ d

where,

d = displacement of car = 1.3 m

Therefore,

W = (94.08 N)(1.3 m)

<u>W = 122.3 J</u>

7 0
3 years ago
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