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Bond [772]
3 years ago
15

Which has the larger kinetic energy, a 10 g bullet fired at 400 m/s or a 80 kg bowling ball rolled at 6.5 m/s ?

Physics
1 answer:
mixer [17]3 years ago
7 0
Formulae for Kinetic energy is:
Kinetic Energy= 1/2xmassx(velocity)^2

For comparison we need to have same units,thus we convert 10g into Kg.
10g/1000=0.01Kg

Input the value of bullet in the formulae;
Kinetic Energy= 1/2x0.01kgx(400)^2
K.E=800J

Input value of the ball:
Kinetic Energy=1/2x80kgx(6.5)^2
K.E=1690J

Which means that th Energy of the ball is more than the bullet.
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Answer:

330 km

55*6= 330 km. An easy formula for this is multiplying time with speed.

Explanation:

330 kilometros

55 * 6 = 330 km. Una fórmula fácil para esto es multiplicar el tiempo por la velocidad.

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Frequency of the light will be equal to 6.97\times 10^{1}Hz

Explanation:

We have given wavelength of the light \lambda =430nm=430\times 10^{-9}m

Velocity of light is equal to v=3\times 10^8m/sec

We have to find the frequency of light

We know that velocity is equal to v=\lambda f, here \lambda is wavelength and f is frequency of light

So frequency of light will be equal to f=\frac{v}{\lambda }=\frac{3\times 10^8}{430\times 10^{-9}}=6.97\times 10^{1}Hz

So frequency of the light will be equal to 6.97\times 10^{1}Hz

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2 years ago
What is the weight of an object with a mass of 19 kg?
Natalija [7]

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Every 2.2 kg is 1 pound. So mulitply 19 * 2.2. It's gonna be equal to 41.8

Explanation:

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3 years ago
A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h
Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

v_{1}=\dfrac{3.0\times10^{3}}{3600}

v_{1}=0.833\ m/s

The force F₁is constant acceleration is also a constant.

F_{1}=ma_{1}

We need to calculate the acceleration

Using formula of acceleration

a_{1}=\dfrac{v}{t}

a_{1}=\dfrac{0.833}{10}

a_{1}=0.083\ m/s^2

Similarly,

F_{2}=ma_{2}

For total force,

F_{3}=F_{2}+F_{1}

ma_{3}=ma_{2}+ma_{1}

The speed of second tugboat is

v=\dfrac{11\times10^{3}}{3600}

v=3.05\ m/s

We need to calculate total acceleration

a_{3}=\dfrac{v}{t}

a_{3}=\dfrac{3.05}{10}

a_{3}=0.305\ m/s^2

We need to calculate the acceleration a₂

0.305=a_{2}+0.083

a_{2}=0.305-0.083

a_{2}=0.222\ m/s^2

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}

\dfrac{F_{1}}{F_{2}}=3.7

F_{1}=3.7F_{2}

Hence, The magnitude of F₁ is 3.7 times of F₂

3 0
2 years ago
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