Physics: Astartled armadillo leaps upward, rising 0.532 m in the first 0.202 s. (a) what is its initial speed as it leaves the ground? (b) what is its speed at the height of 0.532 m? (c) how much higher does it go? use g=9.81 m/s2.

Astartled armadillo leaps upward, rising 0.532 - id05322795108, vbodbodlhviprw2954

Ответ:

nicpinela1234

04.01.2022

A) $d=v_0t+\frac{1}{2}at^2$
0.532 m = 0.202v_0 + ½(-9.81 m/s²)(0.202 s)²
v_0 = 3.62 m/s

B) $d = \frac{1}{2}(v+v_0)t$
0.532 m = ½(v + 3.62 m/s)(0.202) s
v = 1.65 m/s

C) v^2=v_0^2+2ad

At the highest point v = 0, so
0 = (1.65 m/s)^2 + 2(-9.81 m/s²)d
d = 0.139 m

0,0(0 оценок)

Ответ:

juniorgutierrez997

15.08.2021

Physics

Explanation:

By definition, the density of the substance is given by:

D = \frac{m}{V}D=

Where,

m: mass of the substance

V: volume of the substance

Calculating the volume of the substance we have:

V = (50) * (10) * (30) V = 15000 cm ^ 3V=(50)∗(10)∗(30)V=15000cm

Then, replacing values we have:

D = \frac{3300}{15000}D=

15000

3300

D = 0.22 \frac{g}{cm^3}D=0.22

0,0(0 оценок)

Astartled armadillo leaps upward, rising 0.532 m in the first 0.202 s. (a) what is its initial speed as it leaves the ground? (b) what is its speed at the height of 0.532 m? (c) how much higher does it go? use g=9.81 m/s2.

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Physics

Ask an AI advisor a question