A dog with a mass of 57.0 kg slides down a wet slope with negligible friction. The dog starts from rest and has a speed of 2.10 m/s at the bottom. What is the height of the slope (in m)?​

Answer:

C - 98j

Explanation:

Answer:

c- 98j

hope it helped

Answer:

Initial speed of train = 35.76 m/s (80 mph)

               I figured out the speed of the train (which equals 80mph) by viewing the movie. In one of the scenes, it is shown that the train reaches 80mph after Doc Ock desroys the break on the train.

                   1mph = 0.44704mps

                    80(0.44704) = 35.7632

                                      = 35.76m/s

   vf = 0m/s

   vi = 35.76m/s

Mass of train = approximately 137,100 kg (302,254 lbs)  (found in research)

Time taken to stop train = approximately 46 seconds

a = vf - vi

       t

  = 0 - 35.76

         46

  = -.7773913043

  = -.78 m/s^2

(The number is negative since the accelerarion of the train is slowing down)

According to Newton's Second Law, a = f/m. When rearranged, the formula reads f = m(a). As shown in the free body diagram, the weight of the train equals the force due to gravity.

(force = tension)

F =  m(a)

  = 137,100(-.78)

  = -106938 N

106938 N is the calculated tension of Spider-Man's webs

I also calculated the distance it took for Spider-Man to stop the train:

Approximate distance traveled by train:

d = 1/2(a)(t)^2

  = 1/2(-.78)(46)^2

  = 825.24 m

In order for Spider-Man to have been able to completely stop the train before falling off the edge of the tracks, I found that the tension of his webs must have equaled 106938N.

In order to determine the tension of Spider-Man's webs, I needed to know the mass of the train. Since the movie was unclear on what type of trian was used, I researched online to find the mass of the train. Knowing the mass of the train, i was then able to calculate the tension of Spider-Man's webs and the distance the train took to stop.

Explanation:

Answer:

s₂ = 0.475 m = 47.5 cm

Explanation:

The arc length and the angle of rotation are related through the formula:

s = rθ

where,

s = arc length

r = radius of curvature

θ = angle of rotation

First, we consider the arc length covered by the point of insertion of extensor muscles.

s₁ = r₁θ

where,

s₁ = arc length covered by insertion of extensor muscle = 5 cm

r₁ = length of insertion from knee = 4 cm

θ = Angle of Rotation = ?

Therefore,

5 cm = (4 cm)(θ)

θ = (5 cm)/(4 cm)

θ =  1.25 rad

Now, we consider the arc length covered by the foot.

s₂ = r₂θ

where,

s₂ = arc length covered by the foot = ?

r₂ = distance from knee to foot = 38 cm = 0.38 m

The angle of rotation will be the same for the foot as the insertion.

Therefore,

s₂ = (0.38 m)(1.25 rad)

s₂ = 0.475 m = 47.5 cm

Answer:

D. V/3.

Explanation:

V = rw

At point a

Linear velocity

= Va = rw

At point b

We have,

V = 3rw

So linear velocity of a

= V/Va = 3rw/re

I cross multiples and after which I got Va = v/3

Please view attachment

Answer:

Explanation:

The relation between linear and angular velocity,

[tex]V = rw[/tex]

Now, for point A the linear velocity is

[tex]V_A = rw[/tex]

And for point B

[tex]V = 3rw[/tex]

Therefore, the linear velocity of the point A

[tex]\frac{V}{V_A} = \frac{3rw}{rw}\\\\V_A = \frac{V}{3}[/tex]

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Answer:

[tex]B=3.29\ \mu T[/tex]

Explanation:

Given that,

Length of a wire is 3 m

Current in the wore, I = 10 A

The wire formed a semicircle.

We need to find the magnitude of the magnetic field (in micro-tesla) at the center of the circle along which the wire is placed.

The magnetic field at the center of semicircle is given by :

[tex]B=\dfrac{\mu_o I}{4R}[/tex]

R is radius of semicircle,

[tex]R=\dfrac{3}{\pi}=0.954\ m[/tex]

So,

[tex]B=\dfrac{4\pi \times 10^{-7}\times 10I}{4(0.954)}\\\\B=3.29\times 10^{-6}\ T\\\\B=3.29\ \mu T[/tex]

So, the magnitude of the magnetic field is [tex]3.29\ \mu T[/tex].

Answer:

The final speed of the electron is 6.626 x 10⁷ m/s

Explanation:

force applied to the electron, f = 4 x 10⁻¹⁵ N

distance traveled by the electron, d = 50 cm = 0.5 m

The work done on the electron;

W = F x d

W = (4 x 10⁻¹⁵ N) x (0.5 m)

W = 2 x 10⁻¹⁵ J

The kinetic energy of the electron is given by;

[tex]K.E = \frac{1}{2}m_ev^2[/tex]

Apply work - energy theorem;

K.E = W

[tex]\frac{1}{2}m_ev^2 = 2 *10^{-15}\\\\ v^2 = \frac{2( 2 *10^{-15})}{m_e}\\\\v= \sqrt{\frac{2( 2 *10^{-15})}{m_e}}\\\\ v = \sqrt{\frac{2( 2 *10^{-15})}{9.11*10^{-31}}}\\\\v = 6.626*10^7 \ m/s[/tex]

Therefore, the final speed of the electron is 6.626 x 10⁷ m/s

Answer:

That the universe is at least 13.2 billion years old

Explanation:

The most distant galaxy we have observed is more than 13.2 billion light-years away. this evidence indicates. That the universe is at least 13.2 billion years old

The universe is the sum total of all existent matter and space. The universe is estimated to be at least 10 billion light-years across and contains an enormous number of galaxies;

It has been expanding since the Big Bang some 13 billion years ago.

The most distant galaxy we have observed is more than 13.2 billion light-years away. this evidence indicates. That the universe is at least 13.2 billion years old

Hence option c is correct.

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Answer:

[tex]P=V^2/R[/tex]

[tex]R=V^2/P[/tex]

Upon substituting the values

a) R= 75*75/7800 = 0.72ohm

   Also

   R=ρ*L/A

   L=R*A/ρ

   L=[tex]\frac{0.72*3.73*10^-6 }{5*10^-7}[/tex] = 5.44m

b) when the voltage 107 is used

    R=107*107/7800 = 1.468OHM

     L=[tex]\frac{1.468*3.73*10^-6 }{5*10^-7}[/tex] =11.1m

Answer:

v = 38.34 m

Explanation:

Given that,

Mass of a skier, m = 75 kg

It falls down a 75.0 m high slope without friction, h = 75 m

We need to find the velocity of the skier at the bottom of the slope. It is based on the concept of conservation of energy. So, v is the required velocity. So,

[tex]mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 75} \\\\v=38.34\ m/s[/tex]

So, the velocity of the skier at the bottom of the slope is 38.34 m.

Answer:

change of momentum

= mass * change in velocity

= 0.14 * (36 - 0)

= 5.04 kg m/s ---answer

Explanation:

Answer:

Explanation:

Let force P = 193 pounds

second force = Q

Angle that resultant makes with force P is θ

Tanθ = Q sin86.27 / (P + Q cos86.27 )

Tan 31.4 = Q sin86.27 / (193 + Q cos86.27 )

.61 = .99 Q / (193 + .065Q )

117.73 + .04 Q = .99Q

117.73 = .95 Q

Q = 123.93  pounds .

Answer:

The minimum uncertainty in the position of the proton is 1.578 x 10⁻¹⁴ m

Explanation:

Given;

velocity of the proton, v = 2 x 10⁸ m/s

uncertainty in the velocity, = 1 % = 0.01 v = 0.01 x 2 x 10⁸ m/s = 2 x 10⁶ m/s

The momentum of the proton is given by;

ΔP = mv = (1.67 x 10⁻²⁷)(2 x 10⁶ ) = 3.34 x 10⁻²¹ kgm/s

The minimum uncertainty in the position of the proton is given by;

[tex]\delta x = \frac{h}{4\pi \delta P}\\\\ \delta x =\frac{6.626*10^{-34}}{4\pi *3.34*10^{-21}}\\\\ \delta x = 1.578*10^{-14} \ m[/tex]

Therefore, the minimum uncertainty in the position of the proton is 1.578 x 10⁻¹⁴ m

Answer:

Explanation:

Mars does not appear to have any water, while Earth is mostly water. Mars also has yellow-brown sky (pink-brown at sunrise and sunset), which suggests that it has a different atmospheric makeup.

Answer:

Mars does not appear to have any water, while Earth is mostly water. Mars also has yellow-brown sky (pink-brown at sunrise and sunset), which suggests that it has a different atmospheric makeup.

Explanation:

Angular velocity of the coil :

[tex]\omega=\dfrac{2\pi n}{60}[/tex]

[tex]\omega=\dfrac{2\times \pi\times 3500}{60}\\\\\omega =366.52\ rad/s[/tex]

Now,

Peak voltage is given by :

[tex]V=NAB\omega\\\\V= 200\times \dfrac{\pi (0.1)^2}{4}\times 0.7\times 366.52\\\\V=403.01\ V[/tex]

Hence, this is the required solution.

Answer:

0.04558

Explanation:

Luminosity is defined as the total amount of energy a star can produce in a second. Luminosity is given by the formula

σ * A * T⁴

where;

σ = Stefan-Boltzmann constant

= 5.670367 * 10⁻⁸

A= Surface Area

=4πr²

T= Temperature in Kelvin

In the question,

Temperature= 4000K

Radius= 2.0 A.U

Substituting the variables into the equation,

0.000000056704 * 4(3.14)(2.0²) * 4000⁴

= 0.000000056704 * 50.24 * 16000

= 0.04558

Answer:

a) They both hit the ground at the same time

b) Watermelon will hit first, since its speed is faster than the pumpkin speed

c) h = 29,6 m

d)t = 3,02 sec

Explanation:

Equations for fall free movement are:

vf = v₀ + g*t       when   v₀ = 0   (dropped case)  vf = g*t

h = v₀*t + 1/2*g*t²

a) For both ( watermelon and pumpkin) the equation of speed is the same:

vf = g*t²    Both will have the same speed second through second

They both hit the ground at the same time

b) Now is watermelon is thrown with v₀ = 10 m/s

Watermelon will hit first since its speed is faster than the pumpkin speed

vf(watermelon) =  10 + g*t

vf₂ (pumpkin)   =  g*t

c) h = v₀*t + (1/2)*g*t²

h = (10)*1 + (1/2)*9,8*1

h = 10 + 19,6

h = 29,6 m

d)  h = g*t

t = 29,6/9,8

t = 3,02 sec

Answer:

The fundamental frequency is 10 Hz

Explanation:

The frequency of the fundamental mode of the string is given by;

[tex]f_o =\frac{1}{2l}\sqrt{\frac{T}{\mu }[/tex]

where;

L is length of the string = 2.5 m

T is tension of the string, = 10 N

μ is mass per unit length = m / L = 0.01 / 2.5 = 0.004 kg/m

Substitute the given values and solve for the frequency;

[tex]f_o =\frac{1}{2l}\sqrt{\frac{T}{\mu }}\\\\ f_o =\frac{1}{2*2.5}\sqrt{\frac{10}{0.004}}\\\\ f_o = 0.2\sqrt{2500}\\\\ f_o = 0.2 *50\\\\f_o = 10 \ Hz[/tex]

Therefore, the fundamental frequency is 10 Hz

Answer:

Explanation:

A 52.0 kg skateboarder wants to just make it to the upper edge of a "quarter pipe," a track that is one-quarter of a circle with a radius of 2.60 m .

What speed does he need at the bottom

Let the lowest point of the circle = (0,0)

Center of circle = (0, 2.6)

The height of the circle = 2 * radius = 2 * 2.60 = 5.20 m

Highest point = (0, 5.2)

As the skateboarder moves around ¼ of a vertical circle, the skateboarder moves from the lowest position to a position that is ½ the way up to the highest position. This is the point that is 2.6 meters directly to the right of left of the center = (2.6, 2.6)

As the skateboarder has moved 2.6 m upward, the potential energy increase = m * g * ∆h = 52.0 * 9.8 * 2.6

During this same time, the kinetic energy decreased from the maximum to 0. The decrease of KE = the increase of PE

½ * m * ∆v^2 = m * g * ∆h

½ * 52 * ∆v^2 = 52.0 * 9.8 * 2.6

½ * ∆v^2 = 9.8 * 2.6

∆v = (2 * 9.8 * 2.6)^0.5 = 7.14 m/s

The velocity at highest point, (2.6,2.6) is 0 m/s

So the velocity at the lowest point must be 7.14 m/s

Let’s see if the skateboarder has enough KE to move upward 2.6 m.

KE at bottom = ½ * 52 * 7.14^2 = 1325.5

PE = 52 * 9.8 * h

52 * 9.8 * h = 1325.5

h = 2.6 m

The answer is OK

The speed needed by the skateboarder at the bottom is 7.8 m/s.

The given parameters;

Apply the principle of conservation of mechanical energy to determine the sped of the skateboarder at the bottom;

[tex]P.E_{top} = K.E_{bottom}\\\\mgh = \frac{1}{2}mv^2\\\\gh = \frac{1}{2} v^2\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8 \times 3.1} \\\\v = 7.8 \ m/s[/tex]

Thus, the speed needed by the skateboarder at the bottom is 7.8 m/s.

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Answer:

The answer is "[tex]\bold{MB= \frac{L^2 \ WB}{12}}[/tex]"

Explanation:

please find the complete question in the attached file:

From A to B  

[tex]reqAB = \int^{\frac{1}{2}}_{0} \frac{WB}{\frac{L}{2}} x \ dx = \frac{L\ WB}{4}[/tex]

From symmetry [tex]Ay = Cy = FeqAB[/tex]  

The FeqAB position is found on a table of common centres:

[tex]\frac{2}{3}( \frac{L}{2})[/tex]

[tex]\to \Sigma_{NB=0} FX = NB=0\\\\\to \Sigma_{VB=0} Fy = \frac{LWB}{4} - \frac{LWB}{4} -VB =0 \\\\\to \Sigma M_{.CCW \ A} = \frac{L}{3} - \frac{LWB}{4} - \frac{L}{2}VB+MB =0 \\\\MB= \frac{L^2 \ WB}{12}[/tex]

Answer:

v = 11.6 m/s

Explanation:

       [tex]v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{((7.5m/s)^{2} + (-8.9m/s)^{2}[/tex]

       ⇒ v = 11.6 m/s

Answer:

the vibration of the glass tube creating sound waves within itself.  reason is the rising pitch is that the liquid rise shortens the length of the vibrating area within the tube.

Explanation:

Answer:

mk7000000.0000000000

Answer:

1,642.8 coulombs and 1.39*10²¹ electrons pass through any cross section the width of the wire.

Explanation:

Electric current is the circulation of electric charges in an electric circuit, while electric current intensity (I) is the amount of electricity or electric charge (Q) that circulates through a circuit in unit time (t).

Then, the intensity of electric current is expressed as:

[tex]I=\frac{Q}{t}[/tex]

Where:

In this case:

Replacing:

[tex]7.4 A=\frac{Q}{222 s}[/tex]

Solving:

Q= 7.4 A* 222 s

Q= 1,642.8 C

A Coulomb represents about 6.24*10¹⁸ electrons, so you can apply the following rule of three: if 1 C represents 6.24*10¹⁸ electrons, 222 C how many electrons does it represent?

[tex]electrons=\frac{222 C*6.24*10^{18} }{1C}[/tex]

electrons= 1.39*10²¹

1,642.8 coulombs and 1.39*10²¹ electrons pass through any cross section the width of the wire.

Answer:

Locating misplaced electrical equipment.

Explanation:

Ultrasound is primarily used by doctors to locate fractured bones or damaged organs, so I assume it can be used to locate other things too.

Answer:

It is increasing .60 m/s

Explanation:

Answer:

900J

Explanation:

The Work done can be calculated using

Work done = force × distance

According to this question, force = 300N, distance = 3m

Work done = 300 × 3

Work done = 900J

Gray exerted 900J of work