HELP

How fast would the car need to go to double its kinetic energy?

boyakko [2]

Answer:

$v_{f} = \sqrt{2}\cdot v_{o}$

Explanation:

Let consider a car travelling at a speed $v_{o}$ . The ratio of final kinetic energy to initial kinetic energy:

$\frac{\frac{1}{2}\cdot m \cdot v^{2}_{f} }{\frac{1}{2}\cdot m \cdot v^{2}_{o}} = 2$

$\frac{v_{f}}{v_{o}} = \sqrt{2}$

The final speed is:

$v_{f} = \sqrt{2}\cdot v_{o}$

3 0

2 years ago

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Two blocks are connected by a light string that passes over two frictionless pulleys. The block of mass m2 is attached to a spri

irina1246 [14]

(BELOW YOU CAN FIND ATTACHED THE IMAGE OF THE SITUATION)

Answer:

$d=\frac{2g(m1-m2)}{k}$

Explanation:

For this we're going to use conservation of mechanical energy because there are nor dissipative forces as friction. So, the change on mechanical energy (E) should be zero, that means:

$E_{i}=E_{f}$

$K_{i}+U_{i}=K_{f}+U_{f}$ (1)

With $K_{i}$ the initial kinetic energy, $U_{i}$ the initial potential energy, $K_{f}$ the final kinetic energy and $U_{f}$ the final potential energy. Note that initialy the masses are at rest so $K_{i} = 0$ , when they are released the block 2 moves downward because m2>m1 and finally when the mass 2 reaches its maximum displacement the blocks will be instantly at rest so $K_{f} =0$ . So, equation (1) becomes:

$U_{i}=U_{f}$ (2)

At initial moment all the potential energy is gravitational because the spring is not stretched so $U_{i}=U_{gi}$ and at final moment we have potential gravitational energy and potential elastic energy so $U_{f}=U_{gf}+U_{ef}$ , using this on (2)

$U_{gi}=U_{gf}+U_{ef}$ (3)

Additional if we define the cero of potential gravitational energy as sketched on the figure below (See image attached), $U_{gi}=0$ and we have by (3) :

$0= U_{gf}+U_{ef}$ (4)

Now when the block 1 moves a distance d upward the block 2 moves downward a distance d too (to maintain a constant length of the rope) and the spring stretches a distance d, so (4) is:

$0=-m1gd+m2gd+\frac{kd^{2}}{2}$

dividing both sides by d

$0=-m1g+m2g+\frac{kd}{2}$

$g(m1-m2)= \frac{kd}{2}$

$d=\frac{2g(m1-m2)}{k}$ , with k the constant of the spring and g the gravitational acceleration.

7 0

2 years ago

Floodplains are most often found for rivers that exist on

Kazeer [188]

I am pretty sure that floodplains are most often found for rivers that exist on 
hilly areas at the base of mountains. In order to give yoy ans example which will make sure that this answer is quite a suitable one, nice example of floodplains
is The Virgin River at the upper end of Zion Canyon. It will definitely help you! Regards.

3 0

2 years ago

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A motor has an internal resistance of 12.1 Ω. The motor is in a circuit with a current of

kotykmax [81]

Answer:

Explanation:

V = I * R

V = 4 * 12.1 = 48.4 v

7 0

2 years ago

Three 20.0 ohm resistors are

V125BC [204]

Answer:

6.67 ohm

Explanation:

From the question given above, the following data were obtained:

Resistor 1 (R₁) =20 ohm

Resistor 2 (R₂) = 20 ohm

Resistor 3 (R₃) = 20 ohm

Equivalent Resistance (R) =?

Since the resistors are arranged in parallel connection, the equivalent resistance can be obtained as follow:

1/R = 1/R₁ + 1/R₂ + 1/R₃

1/R = 1/20 + 1/20 + 1/20

1/R = (1 + 1 + 1) / 20

1/R = 3/20

Invert

R = 20/3

R = 6.67 ohm

Therefore, the equivalent resistance is 6.67 ohm.

5 0

2 years ago