Answer the question below:

Answer:

1.1x10^-4 N

Explanation:

F = G[(m_1*m_2)/r^2]

G = Gravitational constant 6.67433x10^-11 (N*m^2)/(kg^2)

m_1 = mass 1

m_2 = mass 2

r = radius between objects

F = G[(0.5kg*0.33kg)/(0.001m)^2]

F = 1.1x10^-4 N

Answer:

The unknown force will be 18.116 lb.

Explanation:

Given that,

Three forces act on a flange as shown in figure.

The net force acting on the flange is a minimum.

[tex]\dfrac{dF_{net}}{df}=0[/tex]

We need to calculate the unknown force

Using formula of net force

[tex]\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}[/tex]

Put the value into the formula

[tex]\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}[/tex]

[tex]\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}[/tex]

The magnitude of net force,

[tex]F_{net}=\sqrt{F_{x}^2+F_{y}^2}[/tex]

[tex]F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}[/tex]

[tex]F_{net}=\sqrt{F^2+(60.62)^2+121.24\times\dfrac{F}{\sqrt{2}}+(35)^2-70\times\dfrac{F}{\sqrt{2}}}[/tex]

[tex]F_{net}=\sqrt{F^2+4899.78+36.232F}[/tex]

On differentiating w.r.to F

[tex](\dfrac{dF_{net}}{dF})^2=2F+36.232[/tex]

[tex]0=2F+36.232[/tex]

[tex]F=-\dfrac{36.232}{2}[/tex]

[tex]F=-18.116\ lb[/tex]

Negative sign shows the direction of force which is downward.

Hence, The unknown force will be 18.116 lb.

Following are the solution to the given question:

Calculating the Net force on flange:

[tex]\overrightarrow{F_{net}} = ( F \cos 45^{\circ} + 70 \cos 30^{\circ} -40 ) \hat{i}+(70 \sin 30^{\circ} - F \sin 45^{\circ} )\hat{y} \\\\ \overrightarrow{F_{net}} = ( F \cos 45^{\circ} + 20.62 ) \hat{i}+(35 -F\sin 45^{\circ})\hat{y} \\\\ [/tex]

Calculating the magnitude:

[tex]= \sqrt{f_{X}^2 +f_{y}^{2}}\\\\ = \sqt{( F \cos 45^{\circ} + 20.62 )^2+(35 -F\sin 45^{\circ})^2} \\\\ = \sqt{( F^2+20.62^2+ 35^2+ 41.24 F \cos 45^{\circ} -70 F\sin 45^{\circ})}\\\\ = \sqt{( F^2+20.33F +1650.1844)} \\\\[/tex]

Differentiate the value

[tex]\to \frac{F_{net}}{df}=0 [/tex]

[tex]\to 2F-20.33=0\\\\ \to 2F=20.33\\\\ \to F=\frac{20.33}{2}\\\\ \to F= 10.165\ lb [/tex]

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Answer:

Explanation:

Crinoid fossils can be found on the shore of Lake Michigan because during the Carboniferous period, all of what is now the United States, except for a small part of the upper Midwest, and all of the states along the East Coast, except for Florida, was covered by a warm, shallow inland sea(seawater), and since crinoid fossils live in sea water, they washed up on many places like Lake Michigan.

Answer:

Explanation:

Crinoid fossils can be found on the shore of Lake Michigan because during the Carboniferous period, all of what is now the United States, except for a small part of the upper Midwest, and all of the states along the East Coast, except for Florida, was covered by a warm, shallow inland sea(seawater), and since crinoid fossils live in sea water, they washed up on many places like Lake Michigan.

Answer:

Momentum, p = 23250 kg m/s

Explanation:

Given that

Mass of a car, m = 1550 kg

Speed pf car, v = 15 m/s

We need to find the momentum of the car. The formula for the momentum of an object is given by :

p = mv

Substituting all the values in the above formula

p = 1550 kg × 15 m/s

p = 23250 kg m/s

So, the momentum of the car is 23250 kg m/s.

Answer:

5.99 m  = 6 m

Explanation:

PE = m*g*h

235.2 J = (4 kg)(9.81 m/s^2)(h)

h = (235.2 J)/(9.81*4)

h = 5.99 m

h = 6 m

Answer:

The correct answer is - the sound waves make vibration that travels through the string.

Explanation:

When an individual person talks into your paper cup telephone the person on the other end can feel the bottom of their cup vibrate. The sound waves create vibration go through the string that travels through the string to the end of the cup where vibrations can feel.

The sound waves are longitudinal waves that move or travel through different mediums like air, solid, or gas. The waves create vibration in the particles.

You could create a paper cup telephone but instead of using string, test out different materials and see if those materials will allow sound vibrations to travel through them

Answer:

40 cm

hope it's help you but you can choose if your ans. is -40 cm

Explanation:

The mass of a truck = 5000 kg

Mass of a person = 50 kg

The momentum of a body is given by :

p = mv

m is mass

If momentum of the truck and the person is same. Let v₁ and v₂ are velocity of the truck and the person.

[tex]5000\times v_1=50\times v_2\\\\\dfrac{v_1}{v_2}=\dfrac{50}{5000}\\\\\dfrac{v_1}{v_2}=\dfrac{1}{100}\\\\v_2=100v_1[/tex]

So, it can only possible if the velocity of the person is 100 times of the velocity of the truck.

Answer:

Distance.  A light-year is the distance a single photon of light can travel in empty space in a year.

Answer:

D-Distance

Explanation:

Answer:

u will need 1/2 cells

Explanation:

Answer:

8

Explanation:

Answer:

rotational equilibrium

Explanation:

Answer:

I believe the answer is specific gravity

Explanation:Hope this helps :)

Answer:

.

Explanation:

Answer:

physical

Explanation:

it is a real life model that was built

Answer:

The y-axis represents position relative to the starting point, and the x-axis represents time.

Explanation:

Answer:

a = 5 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics.

[tex]v_{f}=v_{o}+a*t[/tex]

where:

Vf = final velocity = 20 [m/s]

Vo = initial velocity = 10 [m/s]

t = time = 2 [s]

a = acceleration [m/s²]

Now replacing:

[tex]20 =10 +a*2\\10=2*a\\a=5[m/s^{2} ][/tex]

Answer:

(1)Work done by snail is 382.5 N

(2)Work done by ELEPHANT is 25160 N

(3)Work done by HANS is 2420 N

(4)Work done on block is 433 N

Explanation:

The work done due to force F applied at an angle θ from the horizontal to the body is give by

Work done = Fscosθ where, s is distance traveled by body

Case 1: F= 255N , s= 3.00m and θ = 60.[tex]0^0[/tex]

Work done = Fscosθ = 255 x 3.00 x cos60.[tex]0^0[/tex] = 382.5N

thus work done by snail is 382.5 N

Case 2: F= 200N , s= 150m and θ = 33.[tex]0^0[/tex]

Work done = Fscosθ = 200 x 150 x cos33.[tex]0^0[/tex] = 25160N

thus work done by elephant is 25160 N

Case 3: F= 22.9N , s= 129m and θ = 35.[tex]0^0[/tex]

Work done = Fscosθ = 22.9 x 129 x cos35.[tex]0^0[/tex] = 2420N

thus work done by Hans is 2420 N

Case 4: F= 100N , s= 5m and θ = 30.[tex]0^0[/tex]

Work done = Fscosθ = 100 x 5.00 x cos30.[tex]0^0[/tex] = 433N

thus work done on block is 433 N

(1)  The work done by the sailor is 382.5 J.

(2)  The work done by elephant is 25160 J.

(3) The work done by Hans upon the backpack is 2420 J.

(4)  The required work done on block is 433 J.

Let us solve these questions in parts. All these questions are based on the work done. The work done due to force F applied at an angle θ from the horizontal to the body is give by,

[tex]W =F \times s \times cos \theta[/tex]

Here, s is the distance covered by the body.

(1)

Given data:

F= 255N , s= 3.00m and θ = 60.

The work done is calculated as,

W = Fscosθ

W= 255 x 3.00 x cos60 = 382.5 J.

Thus, the work done by the sailor is 382.5 J.

(2)

Given data:

F= 200N , s= 150m and θ = 33.

The work done by the elephant is calculated as,

W = Fscosθ

W= 200 x 150 x cos33. = 25160 J

Thus, the work done by elephant is 25160 J.

(3)

Given data:

F= 22.9N , s= 129m and θ = 35.

Then the work done upon the backpack is calculated as,

W= Fscosθ

W= 22.9 x 129 x cos35. = 2420 J

Thus, the work done by Hans upon the backpack is 2420 J.

(4)

Given data:

F= 100N , s= 5m and θ = 30.

The work done is calculated as,

W = Fscosθ

W= 100 x 5.00 x cos30. = 433 J

Thus, the required work done on block is 433 J.

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Answer:

c

Explanation:

Answer:

a, 22 kg and 22 kg

b, 215.8 N and 440 N

Explanation:

a

The mass of the ball remains constant and unchanged irrespective of where it has been to, need to go or is going. So, basically the mass of the ball on earth is as the same mass of the ball on the said planet, 22 kg

b

The weight of any object factors in the acceleration due to gravity of the said area(or planet).

W = mg, with m being the mass and g being the acceleration due to gravity.

On earth

W = 22 * 9.81 = 215.8 N

On the said planet,

W = 22 * 20 = 440 N

Do therefore, the weight is 215.8 N on earth and 440 N on the planet

Answer:

[tex]P_2 - P_1 = 1.8 * 10^4\ Pa[/tex]

Explanation:

Given

[tex]Height (h) = 1.70m[/tex]

Required

Determine the difference in the blood pressure from feet to top

This is calculated using Pascal's second law.

The second law is represented as:

[tex]P_2 = P_1 + pgd[/tex]

Subtract P1 from both sides

[tex]P_2 - P_1 = pgd[/tex]

Where

[tex]p = blood\ density = 1.06 * 10^3kg/m^3[/tex]

[tex]g = acceleration\ of\ gravity = 9.8N/kg[/tex]

[tex]d =height = 1.70m[/tex]

P2 - P1 = Blood Pressure Difference

So, the expression becomes:

[tex]P_2 - P_1 = 1.06 * 10^3 * 9.8 * 1.70[/tex]

[tex]P_2 - P_1 = 17659.6Pa[/tex]

[tex]P_2 - P_1 = 1.8 * 10^4\ Pa[/tex]

Hence, the difference in blood pressure is approximately [tex]1.8 * 10^4\ Pa[/tex]

Answer:

Use equation 1/2 * m * v²

Explanation:

where m = mass in kg

and v = velocity in m/s

plug accordingly.

Answer:

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 5000 × 15

We have the final answer as

Hope this helps you

When the object is at rest, there is a zero net force due the cancellation of the object's weight w with the normal force n of the table pushing up on the object, so that by Newton's second law,

∑ F = n - w = 0   →   n = w = mg = 112.5 N ≈ 113 N

where m = 12.5 kg and g = 9.80 m/s².

The minimum force F needed to overcome maximum static friction f and get the object moving is

F > f = 0.50 n = 61.25 N ≈ 61.3 N

which means a push of F = 15 N is not enough the get object moving and so it stays at rest in equilibrium. While the push is being done, the net force on the object is still zero, but now the horizontal push and static friction cancel each other.

So:

(a) Your free body diagram should show the object with 4 forces acting on it as described above. You have to draw it to scale, so whatever length you use for the normal force and weight vectors, the length of the push and static friction vectors should be about 61.3/112.5 ≈ 0.545 ≈ 54.5% as long.

(b) Friction has a magnitude of 15 N because it balances the pushing force.

(c) The object is in equilibrium and not moving, so the acceleration is zero.

Answer:

60n left

Explanation:

Do

Answer:

True

Explanation:

Answer:

Explanation:

From the given information; the diagram below shows a clearer understanding.

The blood pressure in the brain [tex]P_{brain} = P_{heart} - \delta ( a - g ) h[/tex]

= 13300 - 1060 (12-9.81) 0.35

=  13300 -  1060 (2.19) 0.35

= 13300 - 812.49

= 12487.51  Pa

The blood pressure in the feet [tex]P_{feet} = P_{heart} + \delta (a + g) h[/tex]

= 13300 + 1060 (12 + 9.81) 1.45

= 13300 + 1060( 21.81 ) 1.45

= 13300 + 33521.97

= 46821.97 Pa

Answer:

Explanation:

[tex]P_brain = P_heart - y(a - g)h\\\\P_brain = 13300 - 1060 (12-9.81)0.35\\\\P_brain = 12487.51 pa\\\\[/tex]

[tex]P_feet = P_heart + y(a+g)*h_r\\\\P_feet = 13300 + 1060(12+9.81)*(1.8-0.35)\\\\P_feet = 46821.97pa[/tex]

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