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Stolb23 [73]
1 year ago
5

Two forces F1=20N and F2=30N are acting on an object as shown in figure: F1=20N F2=30N (i) Find the net force acting on the obje

ct? (ii) State the direction of the net force acting on the object? (iii) If the body still does not move under the application of these forces, what can be the possible reason for this? Identify the name of this extra force and its direction. Two forces F1 = 20N and F2 = 30N are acting on an object as shown in figure : F1 = 20N F2 = 30N ( i ) Find the net force acting on the object ? ( ii ) State the direction of the net force acting on the object ? ( iii ) If the body still does not move under the application of these forces , what can be the possible reason for this ? Identify the name of this extra force and its direction .
​
Physics
1 answer:
Harman [31]1 year ago
5 0

The net force is the 10N force and its direction is towards F2 force.

<h3>What is resultant force?</h3>

The resultant force or the net force is the actual force that acts on a body. We are told that the two forces act in opposite directions hence the net force is F2 - F1  = 30 N - 20 N = 10 N.

ii) The direction of this net force is going to be towards the 30N force.

iii) If the body does not move under the application of these forces then the F1 force must be reinforced by another force which balances it against the F2 force.

Learn more about net force:brainly.com/question/18031889?

#SPJ1

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The energy released by the exploding gunpowder in a cannon propels the cannonball forward. Simultaneously, the cannon recoils. T
Lelechka [254]

Answer:

the launched cannonball

Explanation:

4 0
3 years ago
The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu
Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

E=\phi + K

where

E=\frac{hc}{\lambda} is the energy of the incident light, with h being the Planck constant, c being the speed of light, and \lambda being the wavelength

\phi is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

\lambda=190 nm=1.9\cdot 10^{-7}m, so the energy of the incident light is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J

Converting in electronvolts,

E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

\phi = E-K=6.5 eV-4.0 eV=2.5 eV

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

\lambda=510 nm=5.1\cdot 10^{-7} m

So its energy is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J

Converting in electronvolts,

E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: \phi = 4.5 eV

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

E=\phi+K=4.5 eV+2.7 eV=7.2 eV

Then the copper is replaced with sodium, which has work function of

\phi = 2.3 eV

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

K=E-\phi = 7.2 eV-2.3 eV=4.9 eV

7 0
3 years ago
This should be correct
sweet-ann [11.9K]

Maybe it is, maybe it isn't.  We can't tell, until we see what "this" is. 

Show us a drawing, an equation, an expression, a statement ... something !

5 0
3 years ago
A black stone was discovered to have magnetic properties.
Jobisdone [24]

Answer:

a) this stone was first called Schorl stone

b)the first modern name for this mineral was Black tourmaline

Explanation:

hope this helps you!

:)

5 0
3 years ago
If a force of 3000 N is applied to a large rock, but the rock does not move, how much work is done on the rock?
kkurt [141]

Answer:

No work was done.

W = 0

Explanation:

Work is said to be done whenever a force of one newton moves a body of one kilogram through a distance of one meter. Meaning the applied force has to move the body from a point of rest through certain distance.

Work = force × distance

So, in the case of this question, we only have the force been applied, but no distance was covered. Hence, no work was done.

W = 3000× 0 meter

W = 0

8 0
3 years ago
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