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Cerrena [4.2K]
4 years ago
7

If you know the amount of air pressure exerted on a tabletop, how can you calculate the force exerted on the tabletop? Multiply

the air pressure by the area of the tabletop. Divide the air pressure by the area of the tabletop. Divide the area of the tabletop by the air pressure. Add the air pressure and the area of the tabletop.
Physics
1 answer:
babymother [125]4 years ago
7 0

Answer:

Multiply the air pressure by the area of the tabletop.

Explanation:

The relationship between pressure, force and area is given by:

p=\frac{F}{A}

where in this case, p is the air pressure, F is the force exerted and A the area of the tabletop. By re-arranging the equation, we can solve for F, the force exerted:

p=\frac{F}{A}\\p\cdot A=\frac{F}{A}\cdot A\\pA=F

So, the correct answer is:

The force exerted on the tabletop can be found by multiplying the air pressure by the area of the tabletop.

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Answer:

Explanation:

Find attach the solution

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How is a frame of reference used to describe motion
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In particle kinematics, we measure the acceleration of the particle. For this, an observer is taken as a reference point. Generally we consider the observer as in origin of a co-ordinate system. The place where observer is placed is known as frame of reference. 
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4 years ago
assume that the initial speed is 25 m/s and the angle of projection is 53 degree above the hroizontal. the cannon ball leaves th
finlep [7]

Answer:

A.  xmax = 131.49 m

B.  t = 8.74 s

C.  ymax = 220.33 m

Explanation:

A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:

y=y_o+v_osin\theta-\frac{1}{2}gt^2      (1)

yo: height from the projectile is fired = 200m

vo: initial velocity of the projectile = 25m/s

g: gravitational acceleration = 9.8 m/s^2

θ: angle between the direction of the initial motion of the ball and the horizontal = 53°

t: time

You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.

When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:

0=200+(25)sin53\°t-\frac{1}{2}(9.8)t^2\\\\0=200+19.96t-4.9t^2 (2)

You use the quadratic formula to obtain the value of t:

t_{1,2}=\frac{-19.96\pm\sqrt{(19.96)^2-4(-4.9)(200)}}{2(-4.9)}\\\\t_{1,2}=\frac{-19.96\pm65.71}{-9.8}\\\\t_1=8.74s\\\\t_2=-4.66s

You use the positive value because it has physical meaning.

Now, you can calculate the horizontal range of the projectile by using the following formula:

x_{max}=v_ocos\theta t      

x_{max}=(25m/s)(cos53\°)(8.74s)=131.49m

The cannon ball travels a horizontal distance of 131.49 m

B. The cannon ball reaches the canon for t = 8.74s

C. The maximum height is obtained by using the following formula:

y_{max}=y_o+\frac{v_o^2sin^2\theta}{2g}     (3)

By replacing in the equation (3) the values of all parameters you obtain:

y_{max}=200m+\frac{(25m/s)^2(sin53\°)^2}{2(9.8m/s^2)}\\\\y_{mac}=200m+20.33m=220.33m

The maximum height reached by the cannon ball is 220.33m

3 0
3 years ago
Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of 9q. Sphere B carries a charge of -q. Sphere C
Leona [35]

Answer:

Explanation:

A and B are touched .

charge on each of them after touching = (9q - q) / 2 = 4q

when C is touched with A

charge on A and C each after touching

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When C is touched with B

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6 0
4 years ago
To review the solution to a similar problem, consult Interactive Solution 1.43. The magnitude of a force vector is 87.4 newtons
trasher [3.6K]

Answer:

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Substitute  these values into equation 1

87.4² = 72.1² + y²

y² = 87.4²-72.1²

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