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dimulka [17.4K]
3 years ago
13

3. A force of 200 N is applied to an object of area 2 sq.m . the pressure will be​

Physics
2 answers:
Vladimir79 [104]3 years ago
8 0

Answer:

100 Pa

Explanation:

Pressure = Force/Area

P = 200/2

P = 100

The SI unit is Pa for pressure, one newton per square meter N/m².

MAVERICK [17]3 years ago
3 0

Answer: 100 Pa

Explanation:

P = F / A = 200/ 2 = 100 Pa

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Holding onto a tow rope moving parallel to a frictionless ski slope, a 61.8 kg skier is pulled up the slope, which is at an angl
nata0808 [166]

Answer:

a) Frope= 71.7 N

b) Frope=6.7 N

Explanation:

In the figure the skier is simulated as an object, "a box".

a) At constant velocity we can say that the object is in equilibrium, so we apply the Newton's first law:

∑F=0

Frope=w*sen6.8°

Frope=71.71N

Take into account that w is the weight that is calculated as mass per gravitiy constant:

w=m*g

w=61.8Kg*9.8\frac{m}{s^{2} }

w=605.64N

b) In this case the system has an acceleration of 0.109m/s2.  Then, we apply Newton's second law of motion:

F=m*a

F=61.8Kg*0.109m/s2

Frope=6.73N

8 0
3 years ago
A projectile is launched at ground level with an initial speed of 54.5 m/s at an angle of 35.0° above the horizontal. It strikes
Alchen [17]
<h2>Answer: x=125m, y=48.308m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:

x-component:

x=V_{o}cos\theta t   (1)

Where:

V_{o}=54.5m/s is the projectile's initial speed

\theta=35\° is the angle

t=2.80s is the time since the projectile is launched until it strikes the target

x  is the final horizontal position of the projectile (the value we want to find)

y-component:

y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}   (2)

Where:

y_{o}=0  is the initial height of the projectile (we are told it  was launched at ground level)

y  is the final height of the projectile (the value we want to find)

g=9.8m/s^{2}  is the acceleration due gravity

Having this clear, let's begin with x (1):

x=(54.5m/s)cos(35\°)(2.8s)   (3)

x=125m   (4)  This is the horizontal final position of the projectile

For y (2):

y=0+(54.5m/s)sin(35\°)(2.8s)-\frac{(9.8m/s^{2})(2.8s)^{2}}{2}   (5)

y=48.308m   (6)  This is the vertical final position of the projectile

4 0
3 years ago
What must happen to an atom of magnesium in order to become a magnesium ion Mg+2?
igomit [66]

Answer:

Answer is: c. It must lose two electrons and become an ion.

Magnesium (Mg) is metal from 2. group of Periodic table of elements and has low ionisation energy and electronegativity, which means it easily lose valence electons (two valence electrons).

Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.

Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.

Explanation:

4 0
3 years ago
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choli [55]
Well I think B hope this helps
8 0
2 years ago
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Calculate the pressure the fluid exerted on your diver. The acceleration of gravity is 9.8 m/s2 and the density of the water is
erica [24]

Complete question:

A diver is 10 m below the surface of water. Calculate the pressure the fluid exerted on the diver. The acceleration of gravity is 9.8 m/s2 and the density of the water is 1000 kg/m3. Answer in units of Pa. Show your work.

Answer:

Tthe pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa

Explanation:

Given;

density of water, ρ = 1000 kg/m³

diver's position below the surface of the water, h = 10 m

acceleration due to gravity, g = 9.8 m/s²

Let the atmospheric pressure, P₀ = 101325 Pa

The pressure 10 m below the surface of the water is calculated as;

P =  P₀  + ρgh

P = 101325 Pa  +  (1000 x 9.8 x 10)Pa

P = 199325 Pa

P = 1.99 x 10⁵ Pa.

Therefore, the pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa

5 0
2 years ago
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