What is the percent of change from 74 to 112? rounded to the nearest percent

Mathematics

1 answer:

ruslelena [56]2 years ago

4 0

Answer:

The percent of change from 74 to 112 is 51%.

Step-by-step explanation:

Given:

The percent of change from 74 to 112.

Now, the difference between the numbers $112-74=38$ .

Then, calculating the percentage of the result by $74$ .

$\frac{38}{74}\times100$

$=0.5135\times100$

$=51.35%$ .

Round to the nearest percent of $51.35%$ is $51%$ .

Therefore, the percent of change from 74 to 112 is 51%.

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A particle is projected with a velocity of <img src="https://tex.z-dn.net/?f=40ms%5E-%5E1" id="TexFormula1" title="40ms^-^1" alt

Katena32 [7]

Answer:

$2\sqrt{55}\text{ m/s or }\approx 14.8\text{m/s}$

Step-by-step explanation:

The vertical component of the initial launch can be found using basic trigonometry. In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. Let the vertical component at launch be $y$ . The magnitude of $40\text{ m/s}$ will be the hypotenuse, and the relevant angle is the angle to the horizontal at launch. Since we're given that the angle of elevation is $60^{\circ}$ , we have:

$\sin 60^{\circ}=\frac{y}{40},\\y=40\sin 60^{\circ},\\y=20\sqrt{3}$ (Recall that $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ )

Now that we've found the vertical component of the velocity and launch, we can use kinematics equation $v_f^2=v_i^2+2a\Delta y$ to solve this problem, where $v_f/v_i$ is final and initial velocity, respectively, $a$ is acceleration, and $\Delta y$ is distance travelled. The only acceleration is acceleration due to gravity, which is approximately $9.8\:\mathrm{m/s^2}$ . However, since the projectile is moving up and gravity is pulling down, acceleration should be negative, represent the direction of the acceleration.