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liraira [26]
3 years ago
7

Lisa has three classes that each last 50 minutes qhat ia the total nuber of minutes of minutes of the three classe?

Physics
2 answers:
mr_godi [17]3 years ago
7 0
That would be a total of 150 minutes.
Bess [88]3 years ago
5 0

the total is 150 minutes of all the classes


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A launched hopper reach to 1.20 m maximum height. How much is it’s launch velocity?
garri49 [273]

The launch velocity is 4.8 m/s

Explanation:

We can solve this problem by applying the law of conservation of energy. In fact, the mechanical energy of the hopper (equal to the sum of the potential energy + the kinetic energy) is conserved. So we can write:

U_i +K_i = U_f + K_f

where:

U_i is the initial potential energy, at the bottom

K_i is the initial kinetic energy, at the bottom

U_f is the final potential energy, at the top

K_f is the final kinetic energy, at the top

We can rewrite the equation as:

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m is the mass of the hopper

g=9.8 m/s^2 is the acceleration of gravity

h_i = 0 is the initial height

u is the launch speed of the hopper

h_f = 1.20 m is the maximum altitude reached by the hopper

v = 0 is the final speed (which is zero when the hopper reaches the maximum height)

Solving the equation for u, we find the launch speed of the hopper:

u=\sqrt{2gh_g}=\sqrt{2(9.8)(1.20)}=4.8 m/s

Learn more about kinetic energy and potential energy:

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Answer:

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A wave x meters long has a speed of y meters per second. The frequency of the wave is
Sergio [31]
The correct answer is (b.) y/x hertz. That is because the formula to get the frequency is f =  v / w. The following values (v=y meters / second; wavelength = x meters) must be substituted to the equation, which leaves you y/x hertz.
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First one is D and Second one is B
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3 years ago
What is the speed of an object after falling (from rest) a distance of 9.80 meters near the surface of the Earth? Assume air
madam [21]

Hi there!

We can use the kinematic equation:

v_f^2 = v_i^2 + 2ad

vf = Final velocity (? m/s)

vi = initial velocity (0 m/s, dropped from rest)

a = acceleration (due to gravity, 9.8 m/s²)

d = distance (9.8 m)

Simplify the equation to solve for vf:

v_f^2 = 0 + 2ad\\\\v_f = \sqrt{2ad}

Substitute in the given values:

v_f = \sqrt{2(9.8)(9.8)} = \boxed{13.86 m/s}

8 0
2 years ago
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