The 52-g arrow is launched so that it hits and embeds in a 1.50 kg block. The block hangs from strings. After the arrow joins the block, they swing up so that they are 0.47 m higher than the block's starting point. How fast was the arrow moving before it joined the block? What mechanical work must you do to lift a uniform log that is 3.1 m long and has a mass of 100 kg from the horizontal to a vertical position?

Answer:

0.02976°C

Explanation:

Heat Supplied = mcø

ø = 500/(4*4200)

ø = 0.02976 °C

Answer:

Your parents or anyone in your ancestry.

Explanation:

Answer:

Explanation:

Speed is total distance traveled over time taken to do so.

If AB is measured in kilometers, time (t) for the whole trip is

t = AB/60 + AB/40  

t = 2AB/120 + 3AB/120

t = 5AB/120 hrs

Average speed is distance over time

s = 2AB / (5AB/120)

s = 2(120)/5

s = 48 km/hr

s = 48(1000 m/km / 3600 s/hr) = 13.333333.... 13 m/s

Velocity is displacement over time.

As displacement is zero, velocity is zero

v = 0 km/hr = 0 m/s

Pretty harsh reporting answers just because they are wrong.

Newton's third law allows to find the result for the value of the reaction force during the collision is:

Newton's third law stable that the forces appear in pairs or ea that when two bodies interact, the interaction forces appear in the two bodies simultaneously, in general they are called action and reaction forces.

These furas are of the same magnitude, but in the opposite direction, each one applied to one of the bodies.

They indicate that the most lighter body collides with the one with the greatest mass with a force of F = 90 N. If we call this the action, the larger body must react with a force of equal magnitude on the lighter body.

Consequently, the reaction force is F = 90 N directed towards the lighter body.

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Answer:

There is no scientific evidence that playing specifically l badminton makes you a better person, but sport and exercise in general release hormones which can make you feel more happy therefore making you nicer to the people around you and 'a better person'.

Answer:

It's true because playing any sport makes a person happy. So a happy person is a better person.

Please Mark as brainliest.

Answer:

Velocity of the stone after 1.5 seconds has passed = 37 m/s

Explanation:

Initial velocity (u) = 22 m/s

Time (t) = 1.5 sec

Acceleration due to gravity (g) = 10 m/s²

By using kinematics equation:

v = u + gt

v = 22 + 10 × 1.5

v = 22 + 15

v = 37 m/s

Final velocity (v) = 37 m/s

The density of the liquid will be equal to   [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]

The density of an object is defined as the ratio of the mass of an object to the volume of the object.

Volume of tube = 2^2 * pi * 30 = 377 cm^3

Volume of tube submerged = 25* 377 / 30 = 314 cm^3

Buoyancy = weight of liquid displaced

Volume of liquid displaced = 314 cm^3

Mass of tube and lead = 250 + 30 = 280 g

Now from the mass density by definition

[tex]\rho = \dfrac{m}{v}[/tex]

[tex]m=\rho \times v[/tex]

Mass of liquid displaced = Mass being supported

[tex]314 \times \rho = 280 g[/tex]

[tex]\rho= \dfrac{280}{ 314 } = .892 \frac{g}{cm^3}[/tex]

Thus the density of the liquid will be equal to   [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]

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The density of the liquid is 1.67 g/[tex]cm^3[/tex].

The volume of the tube is

30 * 4 * 3.14 * 0.25 = 94.2 [tex]cm^3[/tex].

The mass of the lead pellets and the plastic tube is

30 + 250 = 280 g.

The volume of the lead pellets is

250 / 11.34 = 22 [tex]cm^3[/tex].

The volume of the liquid that the tube displaces is

94.2 - 22 = 72 [tex]cm^3[/tex].

The density of the liquid is

280 / 72 = 1.67 g/ [tex]cm^3[/tex].

Therefore, the density of the liquid is 1.67 g/ [tex]cm^3[/tex].

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#SPJ3

Answer: An answer on what? I’ll never ignore you!

Explanation:

Answer:

an answer on what?

Explanation: Im here to help!!

Answer:

They are...if I'm correct Chemically combined, sorry if I'm wrong.

Answer:

[tex]a[/tex]

Explanation:

is a corect anser

Answer:

a = 0.01m/s²

Explanation:

V_f = V_0+a*t

V_f = Velocity final

V_0 = Velocity initial

a = acceleration

t = time

a = (V_f-V_0)/t

a = (540m/s-240m/s)/((8hr)*(60min/1hr)*(60s/1min))

a = 0.01m/s²

Answer:

P = 2 pi (L / g)^1/2

P2 / P1 = (8 / 2)^1/2 = 2

The period would be twice as long or 5.6 sec.

Answer:

Explanation:

30 = e^5k

ln30 = lne^5k

ln30 = 5k

k = ln30/5

k = 0.68023947...

round to your heart's content.

Hi there!

We can use the following:

W = ΔKE = F · d

Find the work done on the cart:

W = 200 · 10 = 2000 J

Now, this is equal to the change in kinetic energy of the object. Its initial kinetic energy is 0 J since it starts from rest, so:

2000J = KEf - KEi

KE is given as:

[tex]KE = \frac{1}{2}mv^2[/tex]

2000J = 1/2(55)v²

4000 = 55v²

√(4000/55) = 8.53 m/s

Answer:

Explanation:

How long does the football need to rise?

4.70/3 = 2.35 s

What height will the football reach?

h = ½(9.81)2.35² = 27.1 m

With what speed does the punter need to kick the football?

vy = g•t = 9.81(2.35) = 23.1 m/s

vx = d/t = 56.0/4.70 = 11.9 m/s

v = √(vx²+vy²) = 26.0 m/s

At what angle (θ), with the horizontal, does the punter need to kick the football?

θ = arctan(vy/vx) = 62.7°

Answer:

Explanation:

a) the acceleration of the puck on the rough ice.

a = μg = 0.550(9.81) = 5.3955 = 5.40 m/s²

 (comes from μ = F/N = ma/mg = a/g)

b) the distance from the end boards the puck is when it comes to a stop.

v² = u² + 2as

0² = 12.0² + 2(-5.40)s

s = 13.3 ft

so distance from the boards is

15.7 - 13.3 = 2.4 m

by the way...that's some VERY rough ice...more like sand.

7.4 s

Explanation:

Given:

[tex]v_0 = 73\:\text{m/s}[/tex]

[tex]v = 0[/tex]

[tex]a = -9.8\:\text{m/s}^2[/tex]

[tex]t = ?[/tex]

To solve the time it takes for the object to come to a stop, we are going to use the equation below:

[tex]v = v_0 + at \Rightarrow t = \dfrac{v - v_0}{a}[/tex]

Using the given values above, we get

[tex]t = \dfrac{0 - 73\:\text{m/s}}{-9.8\:\text{m/s}^2}[/tex]

[tex]\;\;\;\;= 7.4\:\text{s}[/tex]

The wavelength will remain unchanged.

Explanation:

The velocity [tex]v[/tex] of a wave in terms of its wavelength [tex]\lambda[/tex] and frequency [tex]\nu[/tex] is

[tex]v = \lambda\nu[/tex] (1)

so if we double both the velocity and the frequency, the equation above becomes

[tex]2v = \lambda(2\nu)[/tex] (2)

Solving for the wavelength from Eqn(2), we get

[tex]\lambda = \dfrac{2v}{2\nu} = \dfrac{v}{\nu}[/tex]

We would have gotten the same result had we used Eqn(1) instead.

Answer:

the wavelength increases

Explanation:

Answer:

i think it's C thx correct me if wrong

The statement that describes the motion of a block while it is in equilibrium is: The block moves at a constant speed.

A state of equilibrium in physics refers to a state of rest or the forces exerted on the object is in a balanced state.

In dynamic equilibrium, the acceleration of a body is zero. This means that the body is moving at a uniform speed.

Therefore, the statement that describes the motion of a block while it is in equilibrium is: The block moves at a constant speed.

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Answer:120.3676

Explanation: using the molecular calculator and molar mass of MgSO4. hope this helps!

Answer:

λ/2 = length of pipe

The pipe is open at both ends having wavelength of A-N-A or antinode-node-antinode which is 1/2 wavelength

λ = 331 m/s / (65.4 / s) = 5.06 m     wavelength of sound

L = 5.06 m / 2 = 2.53 m      length of pipe

Answer:

Explanation:

f = [tex]\sqrt{T/(m/L)} / 2L[/tex]

T = 120 N

L = 3.00 m

(m/L) = 120 g/cm(100 cm/m / 1000 g/kg) = 12 kg/m

                                                  (wow that's massive for a "rope")

f = [tex]\sqrt{120/12} /(2(3))[/tex])

f = [tex]\sqrt{10\\}[/tex]/6 = 0.527 Hz

This is a completely silly exercise unless this "rope" is in space somewhere as the weight of the rope (353 N on earth) far exceeds the tension applied.

A much more reasonable linear density would be 120 g/m resulting in a frequency of √1000/6 = 5.27 Hz on a rope that weighs only 3.5 N

Answer:

Explanation:

momentum is mass times velocity

p = mv

so take the momentum of the truck in question 17 and divide by the mass of this car

v = p/m = p / 1400

Hi there!

Recall Newton's Law of Universal Gravitation:

[tex]\large\boxed{F_g = G\frac{m_1m_2}{r^2}}[/tex]

Where:

Fg = Force of gravity (N)

G = Gravitational Constant

m1, m2 = masses of objects (kg)

r = distance between objects (m)

Plug in the given values stated in the problem:

[tex]F_g = (6.673*10^{-11})\frac{50 * 50}{25^2} = \boxed{2.669 * 10^{-10} N}[/tex]

Answer:

Dynamic stretching

Explanation:

Dynamic stretching is a form of stretching beneficial in sports utilizing momentum from form.

Answer:

Let’s substitute in our givens.

(0.002 kg)(700 m/s) + (5 kg)(0 m/s) = vf(0.002 kg + 5 kg)

I assume you are proficient in algebra I, so I will not include the steps to simplify this equation.

Note that I have considered the bullet’s velocity to be in the positive direction,

The answer is vf = 0.280 m/s

The tangential speed of a wad of chewing gum to the rim of the wheel is approximately 1337.659 centimeters per second.

Let suppose that the wheel rotates at constant angular speed ([tex]\omega[/tex]), in radians per second, the tangential speed of a wad of chewing gum to the rim of the wheel ([tex]v[/tex]), in centimeters per second, is:

[tex]v = 2\pi\cdot r\cdot f[/tex] (1)

Where:

If we know that [tex]f = 5.13\,hz[/tex] and [tex]r = 41.5\,cm[/tex], then the tangential speed of the chewing gum is:

[tex]v = 2\pi\cdot (41.5\,cm)\cdot (5.13\,hz)[/tex]

[tex]v \approx 1337.659\,\frac{cm}{s}[/tex]

The tangential speed of a wad of chewing gum to the rim of the wheel is approximately 1337.659 centimeters per second.

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