Two Atoms are isotopes of each other if they have the same number of_______ but a different number ________

An object is dropped from a height H. During the final second of its fall, it traverses a distance of 53.2 m. What was H? An obj

Serhud [2]

Answer:

H = 171.90 m

Explanation:

given data

distance = 53.2 m

height = H

to find out

height H

solution

we know height is here H = $\frac{1}{2} gt^2$ ......................1

here t is time and a is acceleration

so

we find t first

we know during time (t -1) s , it fall distance (H - 53.2) m

so equation of distance

$H - 53.2 = \frac{1}{2} g (t-1)^2$

$H - 53.2 = \frac{1}{2} g (t^2-2t+1)$

$H - 53.2 = \frac{1}{2} gt^2-gt+\frac{1}{2} g$ ................2

now subtract equation 2 from equation 1 so we get

$H - (H - 53.2) =\frac{1}{2} gt^2- (\frac{1}{2} gt^2-gt+\frac{1}{2} g)$

53.2 = gt - $\frac{1}{2} g$

53.2 = 9.81 t - $\frac{1}{2} 9.8$

t = 5.92 s

so from equation 1

H = $\frac{1}{2} (9.81)5.92^2$

H = 171.90 m

5 0

3 years ago

2×3.14√(1.0m/(9.8〖ms〗^(-1) )=)

il63 [147K]

This is the period in a simple harmonic motion which is 2 seconds in this question.

<h3>What is Period ?</h3>

The period of an oscillatory object can be defined as the total time taken by a vibrating body to make one complete revolution about a reference point.

We are given the below question

2×3.14√(1.0m/(9.8〖ms〗^(2) )= T

This question can as well be expressed as

2π√(L/g) which is equal to period T.

In a nut shell, Period T = 2×3.14√(1.0m/9.8)

T = 6.28√0.102

T = 6.28 × 0.32

T = 2.006 s

Therefore, the period T of the oscillation is 2 seconds approximately.

Learn more about Period here: brainly.com/question/12588483

#SPJ1

8 0

1 year ago

In order to rendezvous with an asteroid passing close to the earth, a spacecraft must be moving at 8.50×103m/s relative to the e

Mariulka [41]

Answer:

v₀ = 13.9 10³ m / s

Explanation:

Let's analyze this exercise we can use the basic kinematics relationships to love the initial velocity and the acceleration we can look for from Newton's second law where force is gravitational attraction.

F = m a

G m M / x² = m dv / dt = m dv/dx dx/dt

G M / x² = dv/dx v

GM dx / x² = v dv

We integrate

v² / 2 = GM (-1 / x)

We evaluate between the lower limits where x = Re = 6.37 10⁶m and the velocity v = vo and the upper limit x = 2.50 10⁸m with a velocity of v = 8.50 10³ m/s

½ ((8.5 10³)² - v₀²) = GM (-1 /(2.50 10⁸) + 1 / (6.37 10⁶))

72.25 10⁶ - v₀² = 2 G M (+0.4 10⁻⁸ - 1.57 10⁻⁷)

72.25 10⁶ - v₀² = 2 6.63 10⁻¹¹ 5.98 10²⁴ (-15.3 10⁻⁸)

72.25 10⁶ - v₀² = -1.213 10⁸

v₀² = 72.25 10⁶ + 1,213 10⁸

v₀² = 193.6 10⁶

v₀ = 13.9 10³ m / s

6 0

4 years ago

A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res

Natalija [7]

Answer: