Describe what happens when divergent plates move away from each other

interval (from lowering the body to his feet taking off) lasts for 0.3 s and the mass of the player is 90 kg. Ignore air resista

steposvetlana [31]

Answer:

The force applied to the surface is 9 kilo Newton.

Explanation:

While jumping on the surface the player applies the force that is equal to its weight on the surface.

The mass of the player is given as 90 kg.

Force applied by the player = weight of the player

Force applied by the player = m × g

Where m is the mass of the player and g is acceleration due to gravity

Force applied by the player = 90 × 9.8

Force applied by the player = 882 Newton

Expressing your answer to one significant figure, we get

Force applied by the player =0. 9 kilo Newton

The force applied to the surface is 0.9 kilo Newton.

8 0

3 years ago

. Friction is a rubbing force that ___________ a spinning yo-yo.

Advocard [28]

The yo-yo speeds up when you rub it

3 0

3 years ago

Read 2 more answers

How do I solve this

lesantik [10]

multiply grav pull by mass of astro maybe with a calculator

7 0

4 years ago

The series in the He spectrum that corresponds to the set of transitions where the electron falls from a higher level to the nf

Galina-37 [17]

Ok so here is the thing. It is necessary to introduce the atomic number Z into the following equation and the reason for that is that we are not working here with hydrogen (H). It will go like this:
E=(2.18×10^-18 J)(Z^2 )|1/(ni^2 )-1/(nf^2 )| 
E=(2.18×10^-18 J)(2^2 )|1/(6 ^2 )-1/(4 ^2 )|=3.02798×10^-19 J 

After that we need to plug the E value calculated into the equation. Remember that the wavelength is always positive:

E=hc/λ 3.02798×10^-19 J=hc/λ λ=6.56×10^-7 m 

so 6.56×10^-7 m or better written 656 nm is in the visible spectrum

7 0

3 years ago

A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh

kolezko [41]

Answer:

The angular velocity is $w = 53.35 \ rounds /minute$

Explanation:

From the question we are told that

The mass of the block is $m = 61.2kg$

The of the pulley is $M = 14.2 kg$

The radius of the pulley is $R = 1.5m$

The radius of the cord around the pulley is $r = 1.5 m$

The distance of the block to the floor is $d = 8.0 m$

From the question we are told that the moment of inertia of the pulley is

$I = \frac{1}{2} MR^2 kg \cdot m^2$

Substituting value

$I = \frac{1}{2} * 14.2 * (1.5)^2$

$I = 15.975 kg \cdot m^2$

Using the Newtons law we can express the force acting on the vertical axis as

$ma = mg -T$

=> $T = mg -ma$

Now when the pulley is rotated that torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

$\tau = I \alpha$