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3 years ago

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If a 50 kg student is standing on the edge of a cliff. Find the student’s gravitational potential energy if the cliff is 40 m hi

gh.

1 answer:

Oduvanchick [21]3 years ago

3 0

you can find it using the equation: potential energy=mass*gravitational acceleration*height.

energy=50kg*9.8N/kg*40m=19600Nm=19600J or 19.6kJ

Sometimes they use 10 instead of 9.8 for the g constant.

Rember to make me Brainliest!!!

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What is the process that changes smaller atoms into larger ones in a star

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3 years ago

a plane is flying due east in still air at 395 km/h. suddenly, the plane is hit by wind blowing at 55km/h toward the west. what

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3 years ago

Which term describes the increase in a material's volume due to an increase in temperature?

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3 years ago

Read 2 more answers

Calculate the distance traveled by a projectile as a function of launch angle. Compare the distances for two projectiles launche

DaniilM [7]

Answer:

$R = x_{max} = \frac{v^2\sin(2\theta)}{g}\\\frac{R_1}{R_2} = \frac{\sin(2\theta_1}{\sin(2\theta_2}$

Explanation:

Using kinematics equations:

$\Delta x = v_{0x}t\\\Delta y = -\frac{1}{2}gt^2+v_{0y}t$

Use $\Delta y = 0$ due to condition of distance traveled.

Solving second equation for time, there are two solutions. t=0 and

$t=\frac{2v_{0y}}{g}$

Use the expression in the first equation to have

$R = \frac{2v^2 \cos\theta\sin\theta}{g}$

Using trigonometric identities, you have the answer of the distance.

By doing the ratio for two different angles, you have the second answer. Due to sine function properties, the distances can be the same to complementary angles. Example, for 20° and 70°, the distance is the same.

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3 years ago

Salmon often jump waterfalls to reach their

natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

$v_x = u cos \theta$

where u is the initial speed and $\theta=37.7^{\circ}$ . The horizontal distance travelled by the salmon is

$d=v_x t = (ucos \theta)t$

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

$t=\frac{d}{v_x}=\frac{d}{u cos \theta}$ (1)

Along the vertical direction, the equation of motion is

$y=h+u_y t -\frac{1}{2}gt^2$

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

$u_y = u sin \theta$ is the vertical component of the initial velocity of the salmon

$g=9.81 m/s^2$ is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

$y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}$

and we can solve this formula for u, the initial speed of the salmon:

$y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s$

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4 years ago