A5 pound object is moving horizontally with a speed of 15m/s. How much force is needed to keep the object

A solid object has a density of 2.85 g/cm3. According to Archimedes, what can be done to make this object float in water?

Juli2301 [7.4K]

2: It's not just the capillary action, but the pull from transpiration (the evaporation of water from the tree) that is used to pull water up from the roots.

The second question needs context. Strong bonds alone won't cause tension. I don't see how adhesion is different. High vapour pressure could do it, but it's the difference in pressures that'd cause tension (and the resistance of that pressure by the surface). So, a low and high pressure would be needed. Poorly worded question :( 

1: "Adhesion is the tendency of certain dissimilar molecules to cling together due to attractive forces." [1] 

3: The other three answere would not work. Think of a boat. 

3: If you push gas, it will be compressed(get smaller). If you push liquid it will push something else. Thus, liquids are good for transferring force. This is a hydraulic system.

4 0

3 years ago

The two uniform, slender rods B1and B2, each of mass 2kg, are pinned together at P, and then B1is suspended from a pin at O. (Th

Bezzdna [24]

Answer:

hello the diagram relating to this question is attached below

a) angular accelerations : B1 = 180 rad/sec, B2 = 1080 rad/sec

b) Force exerted on B2 at P = 39.2 N

Explanation:

Given data:

Co = 150 N-m ,

a) Determine the angular accelerations of B1 and B2 when couple is applied

at point P ; Co = I* ∝B2'

150 = ( (2*0.5^2) / 3 ) * ∝B2

∴ ∝B2' = 900 rad/sec

hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec

at point 0 ; Co = Inet * ∝B1

150 = [ (2*0.5^2) / 3 + (2*0.5^2) / 3 + (2*0.5^2) ] * ∝B1

∴ ∝B1 = 180 rad/sec

hence angular acceleration of B1 = 180 rad/sec

b) Determine the force exerted on B2 at P

T2 = mB1g + T1 -------- ( 1 )

where ; T1 = mB2g ( at point p )

= 2 * 9.81 = 19.6 N

back to equation 1

T2 = (2 * 9.8 ) + 19.6 = 39.2 N

3 0

3 years ago

Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

Strike441 [17]

Answer:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$

Explanation:

The electric field equation of a electromagnetic wave is given by:

$E=E_{max}(kx-\omega t)$ (1)

E(max) is the maximun value of E, it means the amplitude of the wave.
k is the wave number
ω is the angular frequency

We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).

By definition:

$k=\frac{2\pi}{\lambda}$

$k=8.98*10^{6} [rad/m]$

And the relation between λ and f is:

$c=\lambda f$

$f=\frac{c}{\lambda}$

$f=\frac{3*10^{8}}{700*10^{-9}}$

$f=4.28*10^{14}$

The angular frequency equation is:

$\omega=2\pi f$

$\omega=2\pi*4.28*10^{14}$

$\omega=2.69*10^{15} [rad/s]$

Therefore, the E equation, suing (1), will be:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$ (2)

For the magnetic field we have the next equation:

$B=B_{max}(kx-\omega t)$ (3)

It is the same as E. Here we just need to find B(max).

We can use this equation:

$E_{max}=cB_{max}$

$B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}$

$B_{max}=1.17*10^{-8}T$

Putting this in (3), finally we will have:

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$ (4)

I hope it helps you!

8 0

3 years ago

In which direction does a bag at rest move when a force of 20 newtons is applied from the right?

worty [1.4K]

Answer:

in the direction of the applied force

Explanation:

8 0

4 years ago

A person jumps off a diving board 5.0 m above the water's surface into a deep pool. The person's downward motion stops 2.2 m bel

shutvik [7]

Answer:

9.9 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 5+0^2}\\\Rightarrow v=9.9\ m/s$

If the body has started from rest then the initial velocity is 0. In order to find the velocity just before hitting the water then the distance at which the downward motion stops is irrelevant.

Hence, the speed of the diver just before striking the water is 9.9 m/s

7 0

3 years ago