If a leaf falls from a tree, has work been done on the leaf? Explain.

Where is the centre of mass of a system of two particles is situated?

Sever21 [200]

Answer:

In a two particle system, the center of mass lies on the center of the line joining the two particles.

4 0

3 years ago

A rescue pilot drops a survival kit while her plane is flying at an ultitude of 2500m with a forward velocity of 95m. If the air

never [62]

Answer:

Approximately $2.1\; \rm km$ , assuming that $g = -9.8\; \rm m \cdot s^{-2}$ .

Explanation:

Let $t$ denote the time required for the package to reach the ground. Let $h(\text{initial})$ and $h(\text{final})$ denote the initial and final height of this package.

$\displaystyle h(\text{final}) = \frac{1}{2}\, g\, t^2 + h(\text{initial})$ .

For this package:

Initial height: $h(\text{initial}) = 2500\; \rm m$ .
Final height: $h(\text{final}) = 0\; \rm m$ (the package would be on the ground.)

Solve for $t$ , the time required for the package to reach the ground after being released.

$\displaystyle t^{2} = \frac{2\, (h(\text{final}) - h(\text{initial}))}{g}$ .

$\begin{aligned} t &= \sqrt{\frac{2\, (h(\text{final}) - h(\text{initial}))}{g} \\ &\approx \sqrt{\frac{2\times (0\; \rm m - 2500\; \rm m)}{(-9.8\; \rm m \cdot s^{-2})}} \approx 22.588\; \rm s\end{aligned}$ .

Assume that the air resistance on this package is negligible. The horizontal ("forward") velocity of this package would be constant (supposedly at $95\; \rm m \cdot s^{-1}$ .) From calculations above, the package would travel forward at that speed for about $22.588\; \rm s$ . That corresponds to approximately: $95\; \rm m \cdot s^{-1} \times 22.588\; \rm s \approx 2.1 \times 10^{3}\; \rm m = 2.1\; \rm km$ .

Hence, the package would land approximately $2.1\; \rm km$ in front of where the plane released the package.

5 0

3 years ago

Suppose a rocket launches with an acceleration of 34.0 m/s2 what is the apparent weight of an 85-kg astronaut aboard this rocket

docker41 [41]

Answer:

the apparent weight of the astronaut in the rocket is 3723 N.

Given:

acceleration = 34 $\frac{m}{s^{2} }$

mass of astronaut = 85 kg

To find:

Apparent weight of the astronaut = ?

Solution:

Total weight of the astronaut in a rocket is given by,

W = w + F

W = apparent weight of the astronaut

w = weight of astronaut on earth surface = mg

F = force acting on the astronaut = ma

W = mg+ma

W = m (g+a)

W = 85 (9.8 + 34)

W = 3723 N

Thus, the apparent weight of the astronaut in the rocket is 3723 N.

7 0

3 years ago

Read 2 more answers

A sphere of radius R has total charge Q. The volume charge density (C/m3) within the sphere is rho(r)=C/r2, where C is a constan

san4es73 [151]

Answer: C = Q/4πR

Explanation:

Volume(V) of a sphere = 4πr^3

Charge within a small volume 'dV' is given by:

dq = ρ(r)dV

ρ(r) = C/r^2

Volume(V) of a sphere = 4/3(πr^3)

dV/dr = (4/3)×3πr^2

dV = 4πr^2dr

Therefore,

dq = ρ(r)dV ; dq =ρ(r)4πr^2dr

dq = C/r^2[4πr^2dr]

dq = 4Cπdr

FOR TOTAL CHANGE 'Q', we integrate dq

∫dq = ∫4Cπdr at r = R and r = 0

∫4Cπdr = 4Cπr

Q = 4Cπ(R - 0)

Q = 4CπR - 0

Q = 4CπR

C = Q/4πR

The value of C in terms of Q and R is [Q/4πR]

7 0

3 years ago

A heliocentric system is _____-centered.

aniked [119]

A heliocentric system is a sun-centered

3 0

4 years ago

Read 2 more answers