help :”)
a skydiver jumps out of a plane and falls for 45 seconds before deploying his parachute. how far did he fall?

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Distance travelled by the truck is ~

And it's displacement is ~

[tex] \large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}[/tex]

See the diagram in attachment for reference ~

Let O be the initial point, It travels 60 km towards west till point B and then 80 km towards north till point P and returns to initial point O in a straight line, now as we can observe here, it forms a right angled Triangle.

The measure of two legs is 60 km and 80 km, let's find the hypotenuse ~

According to Pythagoras theorem ~

hypotenuse² = sum of squares of other two legs

that is ~

So, the distance between the point A and O is 100 km

Now, The total distance is equal to the distance covered through actual path that is ~

And displacement is the distance between the final point and initial point, but since the truck returns to the point from where it started the journey, so the final and initial point is same therefore displacement is equal to 0.

Explanation:

The torque exerted on the wrench is

[tex]\tau = Fd\sin{\theta} = Fd[/tex]

since the force is applied perpendicular to the wrench, in which sin90 = 1. The torque then is

[tex]\tau = (185\:\text{N})(0.11\:\text{m}) = 20.35\:\text{N-m}[/tex]

Answer:

●Bx=Bcos40

Bx=10 × 0.76

Bx=7.6

●By=Bsin40

By=10×0.64

By=6.4

Hope it will help you.

Answer:

Explanation:

Conservation of momentum

115v + 133(0) = (115 + 133)1.35

v = 2.911304...

v= 2.91 m/s east

Answer:

The velocity east is 2.91

Explanation:

Fill in the box

2.91

The distance traveled by the body in the given time is 1,250 m.

The given parameters;

The distance traveled by the body in the given time is calculated as follows;

[tex]s =ut + \frac{1}{2} at^2\\\\s = 0 \ + \ \frac{1}{2} (4)(25^2)\\\\s =1,250 \ m[/tex]

Thus, the distance traveled by the body in the given time is 1,250 m.

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Answer:

3i - 2j + 8k

Explanation:

p + q = (2i + i) + (2j - 4j ) + (4k + 4k )

= 3i -2j + 8k

Use the meter stick to measure your height to the level of your eyes, then use trigonometry ratio formula to calculate the height from your eyes' level and above of the statue of liberty with angle of elevation, then, add the two heights.

If you are standing at a known distance from the statue of liberty, a meter stick can be used to measure your known distance away from the statue of liberty.

To determine its height using only a meter stick, the angle at which you look at the peak of statue of liberty must be measured or known. The height of the statue of liberty can be calculated if you know the angle of elevation at which you look at the peak of the statue, and the availability of the meter stick.

Use the meter stick to measure your height to the level of your eyes, then use trigonometry ratio formula to calculate the height from your eyes' level and above of the statue of liberty. That is,

Tan Ф = opposite / adjacent

Tan Ф = H/d

H = d x TanФ

Where

H = calculated height from the eyes level and above

Ф = angle of elevation

d = known distance away from the statue

Let h = Your measured height of your body to the eyes level

Then,

The height of the statue of liberty = H + h

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Answer:

Explanation:

s = s₀ + v₀t + ½at²

s = 0 + 5.4(12) + ½(0)12²

s = 64.8 m

A similar theme in both "The Stag at the Pool" and "Hans in Luck" is A. The things we think are valuable sometimes turn out not to be valuable after all.

In "The stag at the pool", a thirsty stag went to the pool to drink. As he bent over, he admired his antlers but despised his legs. He saw a lion and ran but his antlers got stuck in a tree and he was caught by the lion.

The theme in "Hans in Luck" shows that one doesn't need possessions to be happy. The character in the story had several possessions but was still unhappy. Both themes showed that the things we think are valuable sometimes turn out not to be valuable after all.

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The lead ball will float with about 17% of its volume above the surface of the mercury.

We know that density is defined as mass per unit volume of a substance. The density of a substance is an intrinsic property which can be used to identify a substance.

Given that Lead is less dense that mercury, we know that lead will float on mercury. Since the density of mercury is 13.6 g/cm3 and that of lead is 11.3 g/cm3, lead ball will float with about 17% of its volume above the surface of the mercury.

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Missing parts;

A spherical ball of lead (density 11.3 g/cm3) is placed in a tub of mercury (density 13.6 g/cm3). Which answer best describes the result?

A.The lead ball will float with about 83% of its volume above the surface of the mercury.

B.The lead ball will float with about 17% of its volume above the surface of the mercury.

C.The lead ball will float with its top exactly even with the surface of the mercury.

D.The lead will sink to the bottom of the mercury.

E.none of the above

Its relative speed compared to Earth is 0.921

The speed of the object v = 2πr/T where r = radius of orbit and T = period of orbit.

Let v = speed of earth, r = radius of earth orbit = 1 AU and T = period of earth orbit.

So, v = 2πr/T

Also, v' = speed of NEO, r' = radius of NEO orbit = distance of NEO from sun = 1.18 AU and T' = period of NEO orbit.

So, v' = 2πr'/T'

v'/v = 2πr'/T' ÷ 2πr/T

v'/v = r'/r × T/T'

From Kepler's law, T² ∝ r³

So, T'²/T² = r'³/r³

(T'/T)² = (r'/r)³

T'/T =  √[(r'/r)]³

T/T' = √[(r'/r)]⁻³

So, substituting this into the equation, we have

v'/v = r'/r × T/T'

v'/v = r'/r × √[(r'/r)]⁻³

v'/v = √[(r'/r)]⁻¹

Since r' = 1.18 AU and r = 1 AU, r'/r = 1.18

So, v'/v = √[(r'/r)]⁻¹

v'/v = √[(1.18)]⁻¹

v'/v = [1.0863]⁻¹

v'/v = 0.921

So, its relative speed compared to Earth is 0.921

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Answer: Im not entirly sure but I think It's D all the above. I think all but B because I never really heard of that but if you look in our history I think that happen im not sure I would wait untill you know that somone knows for sure.

Explanation:

Answer: 38 weeks (266 days) from the date of conception.

Explanation: www.momjunction.com/pregnancy-due-date-calculator/

Explanation:

These prefixes are very commonly used in naming chemical compounds.

Di- means two.

For example, carbon dioxide's formula is be written as [tex]\text{CO}_2,[/tex] and it has 2 oxygen atoms, hence "di-oxide."

Tetra - means four.

For example, carbon tetrachloride's chemical formula is written as [tex]\text{CCl}_4[/tex], and there are four chlorine atoms

Deca- means ten

For example, lanthanum decahydride's chemical formula is written as [tex]\text{LaH}_{10}.[/tex] In this case there are 10 hydrogen atoms for every lanthanum atom.

Hepta - means seven

For example, iodine heptafluoride is written as [tex]\text{IF}_7[/tex]. Note the seven fluorine atoms attached to the iodine atom, hence the name "hepta-fluoride."

Answer:

D

Explanation:

Process of elimnination + it's the only one that makes sense

An object will begin moving from rest when acted upon by forces that are greater in one direction than in any other direction. Hence, Option (D) is correct.

The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

When  forces that are greater in one direction than in any other direction, resultant will be  unbalanced forces.  Unbalanced forces are those acting on a body when the net force acting on the body is greater than zero. The body alters its state of motion when unbalanced forces act on it.

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#SPJ2

Answer:

4600 Kg/m³

Explanation:

Volume of block=5m³

Mass of block= 920 kg/m³

Density=mass × volume

=920 × 5³

=4600 /m³

The density of ice is 4600 Kg/m³

___________________________________

(Hope this helps can I pls have brainlist (crown)☺️)

Explanation:

Solution:

Here

volume=5

Density=920

Density =Mass/Volume

or,Mass=Density*Volume

or,M=920*5

so,M=4600kg

Answer:

I think false I don't know if I right but I hoped this help

Answer:

D. Greater than mg​

Explanation:

According to Newton’s second law of motion, the net force equals mass times acceleration. We are going to use a free body diagram (force diagram) to show that the equation of the motion is given by

T – mg = – ma

Thereby,

T = mg – ma

and the answer is: (d)

D. Greater than mg​

_________________________________

(hopet his helps can I pls have brainlist  (crown)☺️)

Answer:

I don't know he he.

just joking

Answer:

Explanation:

A CD has an OD of 120 mm and an ID of 15 mm and has a mass between 14 and 33 grams. Let's call it m

Lets call the outer and inner radii R and r respectively

Find the moment of inertia about a line perpendicular to the surface of the disc through its center. We can integrate or look up the result from standard tables

I = ½m(R² + r²)

then use the parallel axis theorem to shift the position of the axis

I = ½m(R² + r²) + md²

where d is the distance of the shift. In this case d = R

I = ½m(R² + r²) + mR²

I = m(1.5R² + 0.5r²)

If we select a mass of say 20 grams

I = 0.020(1.5(0.060²) + 0.5(0.0075²))

I = 0.0001085625 kg•m²

Answer:

36 C= 96.8 F

373 K= 99.85

Explanation:

C to F: (36 x 1.8) + 32

         = 64.8 +32

         = 96.8 F

K to C: C= K- 273.15

           C= 373-273.15

           C= 99.85

____

= 36°C

=( 36 × 9/5 ) + 32

=(36 ÷ 5 × 9) + 32

=(7,2 × 9) + 32

= 64,8 + 32

______

______

= 373 K

= 373 - 273

[tex] \boxed { \sf semoga \: membantu \: :v }[/tex]

Answer:

[tex]k.e. = \frac{1}{2}mv {}^{2} \\ 800= \frac{1}{2} \times 40 \times v {}^{2} \\ 800 = 20v {}^{2} \\ v {}^{2} = 400 \\ v {}^{2} = 20ms {}^{ - 1} [/tex]

k.e. = kinetic energy

hope helpful <3

Answer:

Most importantly metals can pass an electric current without being affected and changed by the electricity. Electrical conductivity combined with ductility makes metals the most suitable materials for electrical transmission wires.

Answer:

This happens when the mass of one of the objects is very large and does not move.

Explanation:

11.4 m/s

Explanation:

The cord will break when the centripetal force exerted on it meets or exceeds the maximum tension [tex]T_{max}[/tex] that it can handle.

[tex]T_{max} = m\dfrac{v_{max}^2}{r}[/tex]

Solving for [tex]v_{max},[/tex] we get

[tex]v_{max}^2 = \dfrac{rT_{max}}{m}[/tex]

or

[tex]v_{max} = \sqrt{\dfrac{rT_{max}}{m}} =\sqrt{\dfrac{(1.3\:\text{m})(60\:\text{N})}{(0.6\:\text{kg})}}[/tex]

[tex]\:\:\:\:\:= 11.4\:\text{m/s}[/tex]

We have that for the Question "" it can be said that Calculate the moment which must be applied to the handle of the screw to raise the block is

From the question we are told

The vertical position of the 100-kg block is adjusted by the screw activated wedge. Calculate the moment which must be applied to the handle of the screw to raise the block. The single thread screw has square threads with a mean diameter of 30 mm and advances 10 mm each complete turn. The coefficient of friction for the screw threads is 0.24, and the coefficient of friction for all the mating surfaces of the block and the wedge is 0.40. Neglect friction at the ball joint A

Generally the equation for the Block is mathematically given as

[tex]\sum Fy=0[/tex]

[tex]981cos21.80 = R_2cos53.6\\\\R_2=1535N[/tex]

the equation for the Wedge is mathematically given as

[tex]\sum Fx=0\\\\1535cos36.4=Pcos21.8\\\\P=1331N[/tex]

the equation for the Screw is mathematically given as

[tex]\beta = tan^{-1}*\frac{L}{2*\pi*r} \\\\\beta = tan^{-1}*\frac{10}{2*\pi*(15)} \\\\\\beta = 6.06\\\\\theta = tan^{-1}*0.25 \\\\\theta = 14.04\\\\\\Therefore\\\\\theta + \beta = 20.1\\\\[/tex]

Therefore

[tex]M = Pr tan (\theta + \beta)\\\\M = 1331(0.015) tan20.09\\\\M = 7.30 N.m[/tex]

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Answer:

A 5.00- kg block is placed on top of a 10.0 -kg block (Fig. P5.68). A horizontal force of 45.0 N is applied to the 10.0-kg block, and the 5.00- kg block is tied to the wall. The coefficient of kinetic friction between all surfaces is 0.200. (a) Draw a free-body diagram for each block and identify the action-reaction forces between the blocks.

(b) Determine the tension in the string and the magnitude of the acceleration of the 10.0-kg block.

The horizontal acceleration of the block is 5 m/s².

To calculate the horizontal acceleration on the block, we use the formula below.

Formula:

Where:

Make "a" the subject of the equation.

Substitute these values into equation 2

Substitute these values into equation 2

Hence the horizontal acceleration of the block is 5 m/s².

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