Types of Bone Tissue

A small rocket with mass 20.0 kg is moving in free fall toward the earth. Air resistance can be neglected. When the rocket is 80

Mekhanik [1.2K]

Answer:

= 308.5 N

Explanation:

acceleration of rocket for safe landing

$v_{y} ^{2} = v_{0} ^{2} +2ay$

$a = \frac{v_{y}^{2} - v_{0}^{2} }{2y}$

$v_{0}$ = initial velocity

$v_{y}$ =final Velocity

m = mass of rocket

$\frac{0^{2} - 30^{2} }{2\times80}$

-5.625 $m/s^{2}$

Upward force

F - mg = ma

F = ma+ mg

F = m(a+g)

m= mass

a = 5.625 $m/s^{2}$

g = 9.8 $m/s^{2}$

= 20(5.625 $m/s^{2}$ + 9.8 $m/s^{2}$ )

= 308.5 N

7 0

3 years ago

Because communication involves two or more people acting in both sender and receiver roles and because their messages are depend

djyliett [7]

Answer:

Transactional process

Explanation:

Because communication involves two or more people acting in both sender and receiver roles and because their messages are dependent on and influenced by those of their partner means that communication is a transactional process i.e. a give and take process.

If the sender relays a message and the receiver does not reply or reply not taking cognizance of the sender's message, communication has not taken place.

For example, if consider this.

Sender: How old are you?

Reciever: My Cat is Red.

The response of the receiver is not in tune with the sender and so communication has not taken place.

4 0

3 years ago

A baton twirler has a baton of length 0.36 m with masses of 0.48 kg at each end. Assume the rod itself is massless. The rod is f

AleksAgata [21]

Answer:

Part a)

When rotated about the mid point

$\tau = 0.021Nm$

Part b)

When rotated about its one end

$\tau = 0.042 Nm$

Explanation:

As we know that the angular acceleration of the rod is rate of change in angular speed

so we will have

$\alpha = \frac{\Delta \omega}{\Delta t}$

$\alpha = \frac{2.4 - 0}{3.6}$

$\alpha = 0.67 rad/s^2$

Part a)

When rotated about the mid point

$I = 2mr^2$

$I = 2(0.48)(0.18)^2$

$I = 0.0311 kg m^2$

now torque is given as

$\tau = 0.0311 (0.67)$

$\tau = 0.021Nm$

Part b)

When rotated about its one end

$I = m(2r)^2$

$I = (0.48)(0.36)^2$

$I = 0.0622 kg m^2$

now torque is given as

$\tau = (0.0622)(0.67)$

$\tau = 0.042 Nm$

3 0

3 years ago

What is the resultant of the vectors shown?

ad-work [718]

Answer:

Option B

Explanation:

Option A is the wrong answer because the horizontal vector is in the opposite direction.

Option C is the wrong answer as the horizontal vector is in the opposite direction and all the vectors are connected head to tail [of the arrows] [Triangle law of vector addition]

Option D is the wrong answer as the horizontal vector is in the opposite direction.

5 0

2 years ago

Explain why a 5eV photon of light cannot be absorbed or emitted by the hydrogen atom

sweet [91]

Answer:

None of the transitions in the hydrogen atom corresponds to a photon energy of 5eV hence no photon of this energy is absorbed or emitted by the hydrogen atom.

Explanation:

Electrons in a hydrogen atom must be in one of the allowed energy levels. If an electron is in the first energy level, it must have exactly -13.6 eV of energy. If it is in the second energy level, it must have -3.4 eV of energy and so on.

If the electron wants to jump from the first energy level, n = 1, to the second energy level n = 2. The second energy level has higher energy than the first, so to move from n = 1 to n = 2, the electron needs to gain energy. It needs to gain (-3.4) - (-13.6) = 10.2 eV of energy to be excited to the second energy level.

The step from the second energy level to the third is much smaller. It takes only 1.89 eV of energy for this excitation to take place. It takes even less energy to excite electrons in hydrogen from the third energy level to the fourth, and even less from the fourth to the fifth.

None of these transitions in the hydrogen atom corresponds to a photon energy of 5eV hence no photon of this energy is absorbed or emitted by the hydrogen atom.

5 0

3 years ago