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4 years ago

Rosa uses the formula (vi¡cosθ)tΔ to do a calculation. Which value is Rosa most likely trying to find?

1 answer:

4 0

To break this problem down, let's start with what we know. The equation given finds one component of the velocity and multiplies it by the change in time. This will not find the acceleration that the first two answers say it will, meaning that the answer isn't A or B.

That leaves us with the final two answers, C and D. If the projectile was launched horizontally and we were trying to find the horizontal displacement, we wouldn't need to use cosθ to find the horizontal velocity, meaning that our answer is most likely C) <span>the horizontal displacement of a projectile launched at an angle!</span><span />

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Five marbles roll down a ramp. Each marble reaches the bottom of the ramp at a speed of 3 meters/second. Which marble has the hi

Mashutka [201]

The smallest marble has the most kinetic energy therefore it would be the correct answer

8 0

3 years ago

Read 2 more answers

A 65.0 kg skier is moving at 6.85 m/s on a frictionless, horizontal snow-covered plateau when she encounters a rough patch 4.00

marusya05 [52]

Answer:

v = 4.58 m/s

Explanation:

In order to calculate the speed of the skier when she gets the bottom of the hill, you have to calculate the speed of the skier when she crosses the rough patch.

To calculate the velocity at the final of the rough patch you take into account that the work done by the friction surface is equal to the change in the kinetic energy of the skier:

$W_f=\Delta K\\\\-N\mu_kd=\frac{1}{2}mv^2-\frac{1}{2}mv_o^2=\frac{1}{2}m(v^2-v_o^2)$ (1)

Where the minus sign means that the work is against the motion of the skier.

Wf: friction force

m: mass of the skier = 65.0kg

N: normal force = mg

g: gravitational acceleration = 9.8m/s^2

d: distance of the rough patch = 4.00m

v: speed at the end of the rough patch = ?

vo: initial speed of the skier = 6.85m/s

μk: coefficient of kinetic friction = 0.330

You replace the expression for the normal force in the equation (1), and solve for v:

$-mg\mu_kd=\frac{1}{2}m(v^2-v_o^2)\\\\-g\mu_kd=\frac{1}{2}(v^2-v_o^2)\\\\v=\sqrt{-2g\mu_kd+v_o^2}\\\\v=\sqrt{-2(9.8m/s^2)(0.330)(4.00m)+(6.85m/s)^2}=8.53\frac{m}{s}=4.58\frac{m}{s}$

Then, the speed fot he skier at the bottom of the hill is 4.58m/s

3 0

4 years ago

Which is larger: a megawatt or a kilowatt? how many times larger is it

Mandarinka [93]

A megawatt is larger than a kilowatt. The scientific notation for kilo is 10 to the 3, meaning 1000. A mega is 10 to the 6, meaning one million. That means a kilo watt is a thousand watts, and mega is a million watts

5 0

3 years ago

Read 2 more answers

Tom has built a large slingshot, but it is not working quite right. He thinks he can model the slingshot like an ideal spring wi

Amiraneli [1.4K]

Answer:

8.9

Explanation:

We can start by calculating the initial elastic potential energy of the spring. This is given by:

$U=\frac{1}{2}kx^2$ (1)

where

k = 35.0 N/m is the initial spring constant

x = 0.375 m is the compression of the spring

Solving the equation,

$U=\frac{1}{2}(35.0)(0.375)^2=2.5 J$

Later, the professor told the student that he needs an elastic potential energy of

U' = 22.0 J

to achieve his goal. Assuming that the compression of the spring will remain the same, this means that we can calculate the new spring constant that is needed to achieve this energy, by solving eq.(1) for k:

$k'=\frac{2U'}{x^2}=\frac{2(22.0)}{0.375^2}=313 N/m$

Therefore, Tom needs to increase the spring constant by a factor:

$\frac{k'}{k}=\frac{313}{35}=8.9$

7 0

4 years ago

An object is moving with an initial velocity of 9m/s. It accelerates at a rate 1.5m/sec2 over a distance of 20m. What is it’s ne

puteri [66]

Answer:

11.87

Explanation:

final velocity^2= initial velocity ^2+ 2* Acceleration* distance

Final Velocity^2= 9*9+2*1.5.20

final velocity^2 = 141

final velocity = 11.87

4 0

4 years ago