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3 years ago

7

To identify whether the forces are balanced or not, you need to whet

1 answer:

jok3333 [9.3K]3 years ago

7 0

Answer:

what are you asking?

Explanation:

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He lungs and respiratory system work closely with the ___________ system to make sure oxygen reaches all the cells of our body.

vlada-n [284]

Im pretty sure it is answer a

8 0

3 years ago

Read 2 more answers

A bug of mass 2.2 g is sitting at the edge of a cd of radius 3.0 cm. if the cd is spinning at 280 rpm, what is the angular momen

m_a_m_a [10]

M = 2.2 g = 2.2 x 10⁻³ kg, the mass of the bug.
r = 3.0 cm = 0.03 m, the radial distance from the center.

The angular speed is
ω = 280 rpm
= (280 rev/min)*(2π rad/rev)*(1/60 min/s)
= 29.3215 rad/s

The moment of inertia of the bug is
I = mr²
= (2.2 x 10⁻³ kg)*(0.03 m)²
= 1.98 x 10⁻⁶ kg-m²

Calculate the angular momentum of the bug.
J = Iω
= (1.98 x 10⁻⁶ kg-m²)*(29.3215 rad/s)
= 5.806 x 10⁻⁵ (kg-m²)/s

Answer: 5.806 x 10⁻⁵ (kg-m²)/s

5 0

4 years ago

In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-l

Ksenya-84 [330]

Answer:

10.4 m/s

Explanation:

The problem can be solved by using the following SUVAT equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the diver in the problem, we have:

$u=+6.3 m/s$ is the initial velocity (positive because it is upward)

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

By substituting t = 1.7 s, we find the velocity when the diver reaches the water:

$v=+6.3 + (-9.8)(1.7)=-10.4 m/s$

And the negative sign means that the direction is downward: so, the speed is 10.4 m/s.

3 0

4 years ago

Read 2 more answers

Two conducting spheres are each given a charge Q. The radius of the larger sphere is three times greater than that of the smalle

Zepler [3.9K]

Answer:

1/9 E

Explanation:

The electric field produced by a charged sphere outside the sphere is equal to that produced by a single point charge:

$E=k\frac{Q}{r^2}$

where

k is the Coulomb's constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

In this problem, we have a sphere of radius R (smaller sphere) with a charge Q that produces an electric field of magnitude E at r=R.

For the larger sphere,

R' = 3R

Therefore, the electric field at r=R' will be

$E'=k\frac{Q}{R'^2}=k\frac{Q}{(3R)^2}=\frac{1}{9}(k\frac{Q}{R^2})=\frac{E}{9}$

So, the electric field just outside the larger sphere will be 1/9 E.

8 0

3 years ago

A body on the circular orbit makes an angular displacement given by ∅(t)=2t^2+5t+5. If time t is in seconds, calculate the angul

Talja [164]

Answer:

Angular Velocity at 2 s= 13 rad/s

Explanation:

∅(t)=2 $t^{2}$ + 5t + 5

where represents angular displacement at any given time t

Angular Velocity (ω)= $\frac{\textrm{d ∅(t) }}{\textrm{dt}}$

ω=4t+ 5

Putting in t=2

ω=8+5=13 rad/s

4 0

3 years ago