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2 years ago

To determine the y-component of a projectile’s velocity, what operation is performed on the angle of the launch?

2 answers:

3 0

To determine the y component of velocity of a projectile sine operation is performed on the angle of launch.

Answer: sine

Explanation:

Thus $a_x=0,a_y=g$

The initial velocity u can be resolved along two directions.

Along the X direction initial velocity = u cos θ

Along y direction initial velocity= u sin θ

From the equation of motion $v= u+at$

Thus velocity along x direction $v_x$ =u cos θ

Velocity along y direction $v_y$ = u sinθ -gt

Sign of g is negative.

Citrus2011 [14]2 years ago

3 0

Answer:

Sine

Explanation:

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In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is

Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

$\theta$ = Angle = 20°

y = -20 cm

Velocity components

$u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s$

$u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s$

Acceleration components

$a_x=0$

$a_y=-9.81\ m/s^2$

$y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0$

$t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629$

Time taken is 0.26088 seconds

$x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m$

The distance the beetle travels on the ground is 0.3677181864 m

6 0

2 years ago

A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second

Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

$v_x = 15 m/s$

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

$d= v_x t =(15 m/s)(5 s)=75 m$

5 0

2 years ago

Two objects, each of weight W, hang vertically by spring scales as shown in the figure. The pulleys and the strings attached to

valentinak56 [21]

The reading on the scale is the tension on the string that connects the two objecst. In order to support the blocks it must pull the weights by a force magnitude of W. So, the tension of the rope is W. Therefore, the reading on the scale is W, D.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

6 0

2 years ago

Read 2 more answers

Which statement is not a good practice when working inside a computer case?

lozanna [386]

Answer:

a. be sure to hold expansion cards by the edge connectors

Explanation:

Removal of loose jewelry is a good safety practice. Also not touching a microchip with a magnetized screwdriver is also a good practice.

But holding expansion cards by the edge connectors is not a good practice, so it is the odd one in the question. Therefore answer option a provides the correct and best answer to the question

3 0

2 years ago

Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the crit

Dennis_Churaev [7]

Complete Question:

A beam of white light is incident on the surface of a diamond at an angle $\theta_{a}$ , since the index of refraction depends on the light's wavelength, the different colors that comprise white light will spread out as they pass through the diamond. For example, the indices of refraction in diamond are $n_{red} = 2.410$ for red light and $n_{blue} = 2.450$ for blue light. Thus, blue light and red light are refracted at different angles inside the diamond. The surrounding air has $n_{air} = 1.000$ .

Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the critical angle θcrit, the light will not be refracted out into the air, but instead it will be totally internally reflected back into the diamond. Find θcrit. Express your answer in degrees to four significant figures.

Answer:

$\theta_{crit} = 24.09^{0}$

Explanation:

Only the blue refracted ray is related to the critical angle in this question

$n_{air} = 1.000$

$n_{blue} = 2.450$

The relationship between the critical angle( $\theta_{crit}$ ), $n_{air}$ and $n_{blue}$ can be given as $sin \theta_{crit} = \frac{n_{air} }{n_{blue} }$

$sin \theta_{crit} = \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} 0.4082^{0}\\ \theta_{crit} = 24.09^{0}$

6 0

2 years ago