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2 years ago

01.04 Law of Conservation of Energy science question

Physics

1 answer:

mojhsa [17]2 years ago

5 0

Answer:

law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.

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If the velocity of a long jumper during take-offat an angle of 30 degrees is 42 f/s, how fast is he moving forward (Vx) and how

Arturiano [62]

Answer:

Vx= 11.0865(m/s)

Vy= 6.4008(m/s)

Explanation:

Taking into account that 1m is equal to 0.3048 ft, the takeoff speed in m / s will be:

V= 42(ft/s) × 0.3048(m/ft) = 12.8016(m/s)

The take-off angle is equal to 30 °, taking into account the Pythagorean theorem the velocity on the X axis will be:

Vx= 12.8016 (m/s) × cos(30°)= 11.0865(m/s)

And for the same theorem the speed on the Y axis will be:

Vy= 12.8016 (m/s) × sen(30°)= 6.4008(m/s)

5 0

3 years ago

If the magnet is now pushed into the coil and then held stationary inside the coil describe fully what would you observe on the

alexira [117]

Answer:

E = q V B describes the electric field induced

E Proportional to V B

while the magnet is pushed into the coil the induced field (B) will increase (consider 1 turn of the coil)

If V is constant the E-field will increase due to increasing B and the galvanometer will deflect accordingly

When V drops to zero the deflection must again be zero

So one would see a blip due to the deflection of the galvanometer

Note that as V increases the galvanometer will deflect one way and then as V drops to zero the deflection will be opposite (drop to zero when V is zero)

B always increases to a constant value because of the properties of the magnet.

8 0

2 years ago

Kinetic energy varies jointly as the mass and the square of the velocity. A mass of 1515 grams and velocity of 77 centimeters pe

Alexus [3.1K]

Answer:

The second kinetic energy is 162 J.

Explanation:

Given that,

Mass, $m_1=15\ g$

Velocity, $v_1=7\ cm/s$

Kinetic energy, $K_1=147\ ergs$

Mass, $m_2=10\ g$

Velocity, $v_2=9\ cm/s$

We need to find kinetic energy $K_2$ . Kinetic energy is given by :

$K=\dfrac{1}{2}mv^2$

So,

$\dfrac{K_1}{K_2}=\dfrac{m_1}{m_2}\times \dfrac{v_1^2}{v_2^2}\\\\K_2=\dfrac{K_1}{\dfrac{m_1}{m_2}\times \dfrac{v_1^2}{v_2^2}}\\\\K_2=\dfrac{147}{\dfrac{15}{10}\times \dfrac{7^2}{9^2}}\\\\K_2=162\ J$