? What is the difference between the Primitive cell and convectional cell

A pendulum is formed by taking a 2kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. It is ob

Mekhanik [1.2K]

Please elaborate more on your question so I can help you

5 0

3 years ago

A nova occurs when

eimsori [14]

C. hydrogen accreted onto a white dwarf from a close companion rapidly fuses to helium, releasing a large amount of energy.

The accreted material, composed mainly of hydrogen, is compacted on the surface of the white dwarf due to the intense gravitational force on that place. As material accumulates, The white dwarf becomes increasingly hot, until it reaches the critical temperature for ignition of nuclear fusion.

5 0

3 years ago

A campus bird spots a member of an opposing football team in an amusement park. The football player is on a ride where he goes a

kenny6666 [7]

Answer:

No..

Explanation:

As the bird releases the drop there is no internal force which will drive it into circular path but it will fall on tangent of the arc at the point of release because it has a tangential velocity same as bird. Path will be parabola in vertical plane.

As the person is on circular arc constantly moving it will never meet that drop.

6 0

3 years ago

Starting at 1.0 m/s, a cheetah runs with a constant acceleration for 4.8 s reaching a speed of 28 m/s. What is the acceleration

LenaWriter [7]

Answer:

c. $5.6m/s^2$

Explanation:

Initial velocity of cheetah,u=1 m/s

Time taken by cheetah =4.8 s

Final velocity of cheetah,v=28 m/s

We have to find the acceleration of this cheetah.

We know that

Acceleration, $a=\frac{v-u}{t}$

Where v=Final velocity of object

u=Initial velocity of object

t=Time taken by object

Using the formula

Then, we get

Acceleration, a= $\frac{28-1}{4.8}=\frac{27}{4.8} m/s^2$

Acceleration= $a=5.6 m/s^2$

Hence, the acceleration of cheetah= $5.6m/s^2$

5 0

3 years ago

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=\frac{F}{m}$ . Now $a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2})$ and $a_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2})$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=\frac{1}{2}*a_{girl}*t^{2}$ and $15.0-x=\frac{1}{2}*a_{sled}*t^{2}$ , solving for the time we get: $t=\sqrt{\frac{2x}{a_{girl} } }$ and $t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }$ . Finally we get: $\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} }$ and replacing the values we have got: $\frac{x}{0.14} =\frac{(15.0-x)}{0.73}$ so $5.33*x=15-x$ so x=2.37 (meters).

5 0

3 years ago