If the penny is thrown horizontally at 25 m/s from the 170 meter building, how long will it take for the penny to hit the ground?

Answer:

Explanation:

1)  average velocity is

v = (21 + 39)/2 = 30m m/s

d = vt 30(4.1) = 123 = 120 m

2)  d = ½gt²

d = ½(9.8)(4.1²)

d = 82.369 = 82 m

when rounding to the two significant digits of the question numerals.

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

[tex]P=\frac{1}{2}\rho v^2\\\\v=\sqrt{\frac{2P}{\rho}}[/tex]

where,

v = flow speed = ?

P = Dynamic Pressure = 55 psi[tex](\frac{6894.76\ Pa}{1\ psi})[/tex] = 379212 Pa

[tex]\rho[/tex] = density of water = 1000 kg/m³

Therefore,

[tex]v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}[/tex]

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

[tex]\frac{V}{t} = Av\\\\t =\frac{V}{Av}[/tex]

where,

t = time to fill the pool = ?

A = Area of the mouth of hose = [tex]\frac{\pi (0.015875\ m)^2}{4}[/tex] = 1.98 x 10⁻⁴ m²

V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = [tex][\frac{\pi (9.144\ m)^2}{4}][1.524\ m][/tex] = 100.1 m³

Therefore,

[tex]t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\[/tex]

t = 18353.5 s = 305.9 min = 5.1 hours

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In 60 minutes or 3600 seconds, the tip of the minute hand traverses the circumference of a circle with radius 3.00 cm, so it moves with a tangential speed of

(3.00 cm)/(3600 s) ≈ 0.00083 cm/s = 8.3 μm/s

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:

At the highest attitude, the  velocity of  the ball is 0 m/s, so the kinetic energy is 0 as well.

Hence the answer is potential energy because it doesn't depend on velocity .

Answer:

A

Explanation:

Answer:

Hey mate.....

Explanation:

This is ur answer.....

Jelly is a colloid where colloidal particles are solids which are dispersed in the liquid. Jelly is a fluid which can not be considered as a particular one kind of state. It is in the category of 'gel' which is a colloidal form.

Hope it helps!

Brainliest pls!

Follow me! ;)

Answer:

Small particles of dust and pollution in the air can contribute to (and sometimes even enhance) these colors, but the primary cause of a blue sky and orange/red sunsets or sunrises is scattering by the gas molecules that make up our atmosphere. Large particles of pollution or dust scatter light in a way that changes much less for different colors.

Explanation:

Answer:

The data suggest that the true mean HAZ depth is larger when the current setting is higher.

Answer:

The 7 states of matter are solid, loquid, gas, fermionoc condensate, quark gluton plasm, bose einetein condensate amd ionised plasm but its usually only 3 they teach you

Answer:

7

Explanation:

solid, liquid,gas,fermionoc condensate,quark glutton plasm,bose einetein condensate amd ionised plasm.

Hi there!

We can use the following equation to find the frequency of each harmonic:

[tex]f_n = \frac{n}{2L} \sqrt{\frac{T}{\lambda}}[/tex]

n = nth harmonic

L = length of string (m)

T = Tension of string (N)

λ = linear density (kg/m)

Begin by converting the linear mass density to kg:

2.00g /m · 1 kg / 1000g = 0.002 kg/m

Now, we can use the equation to find the first three harmonics.

First harmonic:

[tex]f_1 = \frac{1}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{131.76 Hz}[/tex]

Second harmonic:

[tex]f_2 = \frac{2}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{263.52Hz}[/tex]

Third harmonic:

[tex]f_3 = \frac{3}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{395.28Hz}[/tex]

Answer:

Explanation:

Plane 2 is moving north at

8.5sin20 = 2.9 m/s

Plane 2 is moving west at

8.5cos20 = 8.0 m/s

Part A

v = √((13 - 2.9)² + 8.0²) = 12.876... 13 m/s

Part B

θ = arctan((13 - 2.9) / 8.0) = 51.617... 52° N of E

Part C

13 m/s  52° S of W

relative velocity magnitude is independent of reference frame

a.

The components of the force are Fx = 2.2163 N, Fy = -0.5445 N and Fz = 0 N

The force on a current carrying conductor in a magnetic field is given by F = iL × B where i = current = 9.00 A, L = 25.0 cmk = 0.25 mk (since the conductor is along the z-direction). B = magnetic field. Since B has component Bx = -0.242T, By= -0.985, and Bz = -0.336, B = -0.242i + (-0.985j) + (-0.336)k = -0.242i - 0.985j - 0.336)k.

So, F = iL × B

F = 9.00 A{(0.25 m)k × [-0.242Ti + (-0.985Tj) + (-0.336T)k]T}

F = 9.00 A{(0.25 m)k × (-0.242T)i + (0.25 m)k × (-0.985Tj) + (0.25 m)k × (-0.336T)k]}

F = 9.00 A{-0.0605mT)k × i + (-0.24625 mT)k × j + (-0.084 m)k × k]}

F = 9.00 A{-0.0605mT)j + (-0.24625 mT) × -i + (-0.084 mT) × 0]}

F = 9.00 A{-0.0605mT)j + (0.24625 mT)i + 0 mT]}

F = -0.5445 AmT)j + (2.21625 AmT)i + 0 AmT]}

F = -0.5445j + 2.21625i + 0 k

F = (2.2163i - 0.5445j + 0 k) N

So, the components of the force are Fx = 2.2163 N, Fy = -0.5445 N and Fz = 0 N

b.

The magnitude of the net force on the wire is 2.282 N

The net force F = √(Fx² + Fy² + Fz²)

F = √[(2.2163 N)² + (-0.5445 N)² + (0 N)²)

F = √[(4.912 N)² + 0.2964 N)² + (0 N)²)

F = √[5.2084 N)²

F = 2.2822 N

F ≅ 2.282 N

So, the magnitude of the net force on the wire is 2.282 N

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According to above question ~

Let's find the charge (q) by using formula ~

Hence, 12 coulombs of charge flow past any point in the wire in 3 seconds

Answer:

Planck postulated that the energy of light is proportional to the frequency, and the constant that relates them is known as Planck's constant (h). His work led to Albert Einstein determining that light exists in discrete quanta of energy, or photons.

Explanation:

Answer:

Energy does not occur in continuous amounts but in discrete amounts described by:

E = N h ∨   where N is the number of quanta (energy units), ∨ the frequency of the energy, and h Planck's constant (6.63E-34 J-sec)

Answer:

im pretty sure the answer is C

Answer:

Instrument Use

Accelerometer Measures acceleration

Altimeter Measures altitude of an aircraft

Ammeter Measures electric current in ampere

Anemometer Measures wind speed

Answer: Contagion theory states that crowds cause people to act in a certain way; convergence theory says the opposite, claiming that people who wish to act in a certain way come together to form crowds.

Explanation: Its C

Answer:

im here\

Explanation:

Answer:

C

Explanation:

Definition of drug: a medicine or other substance which has a physiological effect when ingested or otherwise introduced into the body

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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Answer:

5 m/s

Explanation:

5000/1000=5 m/s

:))

Answer:

Explanation:

The answer:

https://www.chegg.com/homework-help/questions-and-answers/race-car-starting-rest-travels-around-circular-turn-radius-247-m-certain-instant-car-still-q402991

Total acceleration of car is 148.31 m/s².

Centripetal acceleration is a characteristic of an object's motion along a circular path. Centripetal acceleration applies to any item travelling in a circle with an acceleration vector pointing in the direction of the circle's center.

Given parameters:

Radius of the circular path: r = 22.5 m.

Angular speed: ω = 0.541 rad/s.

So, centripetal acceleration; α = ω²r = (0.541)² × 22.5 m/s² = 6.58 m/s².

Tangential acceleration: [tex]\alpha_t[/tex] = αr = 6.58 × 22.5 = 148.16 m/s².

Hence, total acceleration = √(α² + [tex]\alpha_t[/tex]²) = √(6.58² +148.16²) = 148.31 m/s².

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Answer:

When you are wet or you are sweating current passes easily.

Explanation:

wetness increases conductivity. therefore 'when it is wet'.

Answer:

200 feet

Explanation:

20 seconds

Explanation:

Let [tex]x_a[/tex] be the distance traveled by the accelerating car and [tex]x_c[/tex] be the distance traveled by the car moving with a constant velocity. When they cover the same distance, we can write

[tex]x_a = x_c \Rightarrow v_{0a}t + \frac{1}{2}at^2 = v_ct[/tex]

where [tex]v_c[/tex] is the velocity of car moving at a constant rate and a is the acceleration of the accelerating car. Since the accelerating car started from rest, then [tex]v_{0a}[/tex] is zero so our equation above simplifies to

[tex]\frac{1}{2}at^2 = v_ct[/tex]

Note that the variable t cancels out so solving for t, we get

[tex]\frac{1}{2}at = v_c \Rightarrow t = \dfrac{2v_c}{a}[/tex]

Plugging in the given values,

[tex]t = \dfrac{2(20\:\text{m/s})}{2\:\text{m/s}^2} = 20\:\text{s}[/tex]

Answer:

Approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex], assuming that no external force (e.g., gravitational pull) was acting on this rocket.

Explanation:

Assume that no external force is acting on this rocket. The system of the rocket and the fuel on the rocket would be isolated (an isolated system.) The momentum within this system would be conserved.

Let [tex]v_{0}\; \rm m\cdot s^{-1}[/tex] be the initial velocity of the rocket.

The velocity of the exhaust gas would be [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] since the gas is ejected away from the rocket.

Let [tex]\Delta v\; \rm m\cdot s^{-1}[/tex] denote the increase in the velocity of the rocket. The velocity of the rocket after ejecting the gas would be [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex].

The momentum [tex]p[/tex] of an object of velocity [tex]v[/tex] and mass [tex]m[/tex] is [tex]p = m \cdot v[/tex].

The combined mass of the rocket and the fuel was [tex]10000\; \rm kg[/tex]. The initial momentum of this rocket-fuel system would be:

[tex]\begin{aligned}p_{0} &= m \cdot v\\ &= 10000\; {\rm kg} \times v_{0}\; {\rm m \cdot s^{-1}} \\ &= (10000\; v_{0})\; \rm {kg \cdot m\cdot s^{-1}}\end{aligned}[/tex].

The momentum of the [tex]5.0\; \rm kg[/tex] of fuel ejected at [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] would be:

[tex]\begin{aligned} & 5.0 \; {\rm kg} \times (v_{0} - 5000)\; {\rm m\cdot s^{-1}}\\ =\; & (5.0\, v_{0} - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

After ejecting the [tex]5.0\; \rm kg[/tex] of the fuel, the mass of the rocket would be [tex]10000\; \rm kg - 5.0\; \rm kg = 9995\; \rm kg[/tex]. At a velocity of [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex], the momentum of the rocket would be:

[tex]\begin{aligned} & 9995 \; {\rm kg} \times (v_{0} + \Delta v)\; {\rm m\cdot s^{-1}}\\ =\; & (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

Take the sum of these two quantities to find the momentum of the rocket-fuel system after the fuel was ejected:

[tex]\begin{aligned}p_{1} &= (5.0\, v_{0} - 25000)\; {\rm kg \cdot m\cdot s^{-1}\\ &\quad\quad + (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}} \\ &= (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

The momentum of the rocket-fuel system would be conserved. Thus [tex]p_{0} = p_{1}[/tex].

[tex](10000\, v_{0})\; {\rm kg \cdot m\cdot s^{-1}} = (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

Solve this equation for [tex]\Delta v[/tex], the increase in the velocity of the rocket.

[tex]10000\, v_{0} = 10000\, v_{0} + 9995\, \Delta v - 25000[/tex].

[tex]9995\, \Delta v = 25000[/tex].

[tex]\begin{aligned}\Delta v &= \frac{25000}{9995} \approx 2.5\end{aligned}[/tex].

Thus, the velocity of the rocket would increase by approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex] after ejecting the [tex]5.0\; \rm kg[/tex] of fuel.

The mechanical advantage of ramp A is greater than others and it will require the least force to move the load to greater distance.

The mechanical advantage of the ramp is calculated as follows;

[tex]M.A = \frac{L}{h}[/tex]

The mechanical advantage of the ramp B, C, D and E is calculated as;

[tex]M.A = \frac{L}{h} \\\\[/tex]

The mechanical advantage of the ramp A is calculated as follows;

[tex]M.A = \frac{2L}{h} \\\\M.A = 2(\frac{L}{h} )[/tex]

Since the length of the ramp A is greater than other ramps, the mechanical advantage will be greater and it will require the least force to move the load to greater distance.

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