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Fynjy0 [20]
3 years ago
10

Two circles are drawn in a 12-inch by 14-inch rectangle. Each circle has a diameter of 6 inches. If the circles do not extend be

yond the rectangular region, what is the greatest possible distance (in inches) between the centers of the two circles?
Physics
1 answer:
Dennis_Churaev [7]3 years ago
6 0

Answer:

d = 10 inch

Explanation:

The farthest distance between the centers, is along the diagonal of the rectangle. Therefore, we need to calculate the diagonal of the rectangle, but counting the fact that we have both circles.

So if, one side is 12 inch, and the other is 14 inch, we can use the Pitagoras theorem which is:

d = √(a²) + (b)²

Where a and b, are the lenght of the rectangle, but without the lenght  of the diameter of both circles.

With this, the expression is this:

d = √(14 - 6)² + (12 - 6)²

d = √64+36

d = √100

d = 10 inches

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Elena L [17]

Answer:

A. 30.38°

B 5.04N

Explanation:

Using

F= ILBsin theta

2 .55N= 8.4Ax 0.5mx 1.2T x sintheta

Theta = 30.38°

B. If theta is 90°

Then

F= 8.4Ax 0.5mx 1.2x sin 90°

F= 5.04N

6 0
3 years ago
Think about the mice shown below. Using the space below numbered 1, describe what is happening from figure 1 to 3.​
UkoKoshka [18]

Answer: Natural selection is taking place.

Explanation:

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- anonymous

8 0
3 years ago
A cart loaded with bricks has a total mass of 9.13 kg and is pulled at constant speed by a rope. The rope is inclined at 24.7 ◦
blagie [28]

Answer:

W = 0.63 KJ

Explanation:

Work (W) is defined as the point product of force (F) by the distance (d)the body travels due to this force.  

W= F*d *cosα Formula (1)  

F : force (N)

d : displacement (m)

α : angle between force and displacement

Newton's second law:

∑F = m*a Formula (2)  

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the cart on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the cart

W: Weight of the cart : In vertical direction

FN : Normal force : perpendicular to the floor

f : Friction force: parallel to the floor

T : tension Force,  inclined at  θ=24.7° above the horizontal

Calculated of the W

W= m*g

W= 9.13 kg* 9.8 m/s² = 89.47 N

x-y components o the  tension force (T)

Tx = Tcosθ = T*cos 24.7° (N)

Ty = Tsin θ = T*sin 24.7°  (N)

Calculated of the FN  

We apply the formula (2)  

∑Fy = m*ay ay = 0  

FN +Ty- W = 0  

FN = W-Ty  

FN =  89.47-T*sin 24.7°

Calculated of the friction force (f)

f = μk*FN

f =(0.597)*(  89.47-T*sin 24.7° )

f= 53.41-0.249T

Calculated of the tension force of the rope (f)

We apply the formula (2) :

∑Fx = m*ax  ,  ax= 0 ,because the speed of the cart  is constant

Tx - f = 0

T*cos 24.7°-( 53.41 - 0.249T )= 0

T*cos 24.7° + 0.249T = 53.41

(1.1575)T = 53.41

T= (53.41) / (1.1575)

T= 46.14 N

Work done on the cart by the rope

We apply the formula (1)

W=T*d *cosα

W= (46.14 N)*(15.1 m) *(cos24.7)

W = 632.97 (N*m) = 632.97 (J)

W = 0.63 KJ

6 0
4 years ago
If an object accelerates from rest, with a constant acceleration of 5.4 m/s2, what will its velocity be after 28s?
aleksklad [387]
Vf = Vi + at
Vf = 0 + 5.4•28
= 151.2m/s..
not sure if its right
6 0
3 years ago
A proton in a particle accelerator is traveling at a speed of 0.99c.(a) If you use the approximate nonrelativistic equation for
____ [38]

Answer:

a)  p = 4.96 10⁻¹⁹ kg m / s , b)  p = 35 .18 10⁻¹⁹  kg m / s ,

c)  p_correst / p_approximate = 7.09

Explanation:

a) The moment is defined in classical mechanics as

                 p = m v

Let's calculate its value

               p = 1.67 10⁻²⁷ 0.99 3. 10⁸

               p = 4.96 10⁻¹⁹ kg m / s

b) in special relativity the moment is defined as

               p = m v / √(1 –v² / c²)

Let's calculate

                p = 1.67 10⁻²⁷ 0.99 10⁸/ √(1- 0.99²)

                p = 4.96 10⁻¹⁹ / 0.141

                p = 35 .18 10⁻¹⁹  kg m / s

c) the relationship between the two values ​​is

            p_correst / p_approximate = 35.18 / 4.96

            p_correst / p_approximate = 7.09

4 0
3 years ago
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