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Minchanka [31]
3 years ago
7

You're trying to stop a car that is rolling down-hill. Gravity is pulling it forward with a force of 200 N. If the car has a mas

s of 650 kg, and you want to
it accelerate at -0.15 m/s/s, what force do you need to apply to the car? (Your goal is to find the actual number for the force that you are applying.)
Physics
1 answer:
lara31 [8.8K]3 years ago
8 0

Answer:

no idea .

Explanation:

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Newton's first law of motion is sometimes called the law of _________.
Gre4nikov [31]
Newtons first law of motion is also known as the law of inertia
3 0
3 years ago
A 0.05kg dart is thrown at and sticks into a 0.4 kg block hanging on a string. After the collision the block and dart swing in a
zloy xaker [14]

Answer:

v = 1.4  m /s

Explanation:

We shall apply law of conservation of mechanical energy

The kinetic energy of dart and block   is converted into potential energy of both dart and block .

1 /2 (m+M) v² = ( m +M) gH

.5  x v² =  9.8 x .1

=  v² = 1.96

v = 1.4

v = 1.4  m /s

6 0
2 years ago
The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ
prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

          \nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})

                           = (30)^{2} + 2(-0.25(412 - 0))

                           = 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

          y = 200 e^{\frac{x}{1000}}

      \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}

Now, second derivative of the train is calculated as follows.

         \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}      

       \frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}    

Radius of curvature of the train is as follows.  

   \rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}

               = \frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}

              = 3808.96 m

Now, we will calculate the normal component of the train as follows.

            a_{n} = \frac{\nu^{2}_{B}}{\rho}

                        = \frac{(26.34)^{2}}{3808.96}

                        = 0.1822 m/s^{2}

The magnitude of acceleration of train is calculated as follows.

            a = \sqrt{(a_{t})^{2} + (a_{n})^{2}}

               = \sqrt{(-0.25)^{2} + (0.1822)^{2}}

              = 0.309 m/s^{2}

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is 0.309 m/s^{2}.

6 0
3 years ago
Add a picture of leaves.
iren2701 [21]

Answer:

There you go

Explanation:

 

6 0
3 years ago
Identical twins Kate and Karen are rowing their boat on a hot Summer afternoon when they decide to go for
Ugo [173]
I think it will reduce in speed because friction drags it to the opposite direction and it were the girls mass that was overcoming friction but i think it is newtons 2nd law of motion
5 0
2 years ago
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