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Anna11 [10]
3 years ago
7

What is the constraint forces

Physics
1 answer:
neonofarm [45]3 years ago
6 0

Answer:

(1) Support

(2) Tension

(3) Reaction

The constraint forces are forces that occur in pairs such that the net force (effect) is zero

The details of the constraint forces are;

(1) The support force at the top of the swing acting upwards

(2) The tension in the swing rope acting upwards

(3) The (normal) reaction between the legs of the swing frame and the ground

Explanation:

The normal reaction is the force equal in magnitude and opposite in direction to the total weight force the swing and the person on the swing posses due to gravitational attraction which prevents the downward or sideways motion of the entire swing

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Y=mx+b if y=8.18 m=1.31 b=17.2 then find x
Dafna1 [17]

Plug in the corresponding values into y = mx + b

8.18 in for y

1.31 in for m

17.2 in for b

8.18 = 1.31x + 17.2

Now bring 17.2 to the left side by subtracting 17.2 to both sides (what you do on one side you must do to the other). Since 17.2 is being added on the right side, subtraction (the opposite of addition) will cancel it out (make it zero) from the right side and bring it over to the left side.

8.18 - 17.2 = 1.31x

-9.02 = 1.31x

Then divide 1.31 to both sides to isolate x. Since 1.31 is being multiplied by x, division (the opposite of multiplication) will cancel 1.31 out (in this case it will make 1.31 one) from the right side and bring it over to the left side.

-9.02/1.31 = 1.31x/1.31

x ≈ -6.8855

x is roughly -6.89

Hope this helped!

~Just a girl in love with Shawn Mendes

8 0
3 years ago
Read 2 more answers
a hockey puck with a mass of 0.11 kg is at rest on the horizontal frictionless surface of the rink. a player applies a horizonta
stira [4]

The solution to this ques is available in the image.

Given,

Force= 1N

Mass= 0.11kg

Time= 5sec

Force= mass X accelaration

Accelaration= velocity/ time

Speed=distance/ time

Hence, the speed is 45 m/s and the distance is 225 m.

To know more about speed and distance problems the link is given below:

brainly.com/question/19610984?

#SPJ4

8 0
1 year ago
A ____________ galaxy has arms of gas and dust; where as _____________ galaxy is a category of galaxy that does not have a disti
Jet001 [13]

Answer:

spiral, an irregular

Explanation:

i took the test :)

3 0
3 years ago
The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of
Elan Coil [88]

Answer: 230.50 m

Explanation:

We have the following information:

h_{Hg-TOP}=675mmHg=0.675m the barometric reading at the top of the building

h_{Hg-BOT}=695mmHg=0.695m the barometric reading at the bottom of the building

\rho _{air}=1.18 kg/m^{3} density of air

\rho _{Hg}=13600 kg/m^{3} density of mercury

g=9.8/m^{2} gravity

And we need to find the height of the building.

In order to approach this problem, we will firstly use the following equations to find the pressure at the top of the building P_{TOP} and the perssure at the bottom P_{BOT}:

P_{TOP}=\rho _{Hg} g h_{Hg-TOP} (1)

P_{BOT}=\rho _{Hg} g h_{Hg-BOT} (2)

From (1): P_{TOP}=(13600 kg/m^{3})(9.8/m^{2})(0.675m)=89964 Pa (3)

From (2): P_{BOT}=(13600 kg/m^{3})(9.8/m^{2})(0.695m)=92629.6 Pa (4)

Having the pressures at the top and the bottom of the building, we can calculate the variation in pressure \Delta P:

\Delta P=P_{BOT} - P_{TOP} (5)

\Delta P=92629.6 Pa - 89964 Pa=2665.6 Pa (6)

On the other hand, we have a column of air with a cross-section area A and the same height of the building, lets name it h_{air}.

As pressure is defined as the force F exerted on a specific area A, we can write:

\Delta P=\frac{F}{A} (7)

If we isolate F we have:

F= A \Delta P (8)

Also, the force gravity exerts on this column of air (its weight) is:

F=m_{air} g (9)

Knowing the density of air is: \rho_{air}=\frac{m_{air}}{V_{air}} (10)

where the volume of air can be written as: V_{air}=(A)(h_{air}) (11)

Substituting (1) in (10):

\rho_{air}=\frac{m_{air}}{(A)(h_{air}} (12)

Isolating m_{air}:

m_{air}=(\rho_{air}) (A) (h_{air}) (13)

Substituting (13) in (9):

F=(\rho_{air}) (A) (h_{air}) (g) (14)

Matching (8) and (14)

A \Delta P=(\rho_{air}) (A) (h_{air}) (g) (15)

Isolating h_{air}:

h_{air}=\frac{\Delta P}{g \rho_{air}} (16)

Substituting the known and calculated values:

h_{air}=\frac{2665.6 Pa}{(9.8m/s^{2}) (1.18 kg/m^{3})} (17)

Finally:

h_{air}=230.50 m This is the height of the building

8 0
3 years ago
A substance has a volume of 10.0 cm3 and a mass of 89 grams. What is its density?
vovikov84 [41]
Density = mass / volume

D = 89 / 10.0

D = 8.9 g/cm³

hope this helps!
8 0
3 years ago
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