Einstein's energy mass equivalence relation say that if the whole given mass is converted to energy then it would be
![E = mc^2](https://tex.z-dn.net/?f=E%20%3D%20mc%5E2)
where
m = mass in kg
c = speed of light in m/s
this is the origination of quantum physics and by this formula we can relate the dual nature of light and particle
So correct relation above will be
![E = mc^2](https://tex.z-dn.net/?f=E%20%3D%20mc%5E2)
The skateboard that has the greatest impulse is Skateboard C (Impulse = 25.74 Ns)
From the question,
We are to determine which skateboard has the greatest impulse. To do this, we will calculate the impulse of each skate board.
Impulse can be calculated by using the formula
I = Ft
Where I is the impulse
F is the force
and t is the time
Force = 33 N
Time = 0. 12 s
∴ Impulse = 33 × 0.12
Impulse = 3.96 Ns
Force = 33 N
Time = 0. 57 s
∴ Impulse = 33 × 0.57
Impulse = 18.81 Ns
Force = 33 N
Time = 0. 78 s
∴ Impulse = 33 × 0.78
Impulse = 25.74 Ns
Hence, the skateboard that has the greatest impulse is Skateboard C (Impulse = 25.74 Ns)
Learn more here: brainly.com/question/21840495
The question above is simply a conversion problem. We are asked to convert from Celsius to Fahrenheit and to Kelvin. To convert from Celsius to Fahrenheit, we do as follows:
°F = (°C<span> × </span>9/5) + 32
°F = (30 × 9/5) + 32
°F = 86
From Celsius to Kelvin,
K = °C + 273.15
K = 30 + 273.15 = 303.15
The distance traveled by the ball before hitting the ground is 7.2 m.
The given parameters:
- Angle of projection, θ = 37⁰
- Horizontal velocity, V = 20 m/s
- Height above the ground, h = 1 m
<h3>Vertical motion of a projectile;</h3>
- The vertical motion of a projectile is affected by gravity.
The time of motion of the ball is calculated as follows;
![h= v_y_i t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2 \\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 1}{9.8} }\\\\t =0.45 \ s](https://tex.z-dn.net/?f=h%3D%20v_y_i%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%5C%5C%5C%5Ch%20%3D%200%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%5C%5C%5C%5Ch%20%3D%200%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20gt%5E2%20%5C%5C%5C%5Ct%20%3D%20%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%20%7D%20%5C%5C%5C%5Ct%20%3D%20%5Csqrt%7B%5Cfrac%7B2%5Ctimes%201%7D%7B9.8%7D%20%7D%5C%5C%5C%5Ct%20%3D0.45%20%5C%20s)
The horizontal distance traveled by the ball before hitting the ground is calculated as follows;
![X = v_x \times t\\\\X = vcos\theta \times t\\\\X = 20 \times cos(37) \times 0.45\\\\X = 7.2 \ m](https://tex.z-dn.net/?f=X%20%3D%20v_x%20%5Ctimes%20t%5C%5C%5C%5CX%20%3D%20vcos%5Ctheta%20%5Ctimes%20t%5C%5C%5C%5CX%20%3D%2020%20%5Ctimes%20cos%2837%29%20%5Ctimes%200.45%5C%5C%5C%5CX%20%3D%207.2%20%5C%20m)
Thus, the distance traveled by the ball before hitting the ground is 7.2 m.
Learn more about horizontal distance here: brainly.com/question/24784992