Please help with this ski question to find Delta y ​

Answer:

m_2 = 2600kg

Explanation:

P_1 = P_2

P = (m_1)*(v_1)+(m_2)(v_2)

P_1 = (3250kg)(8.1m/s)+(m_2)(0m/s)

P_1 = 26,325 kg*m/s

P_2 = (3250kg)(4.5m/s)+(m_2)(4.5m/s)

P_2 = 14,625kg*m/s+(4.5m/s)m_2

26,325 kg*m/s = 14,625kg*m/s+(4.5m/s)m_2

11,700kg*m/s = (4.5m/s)m_2

m_2 = 2600kg

Answer:

One standard kilogram is define as the mass of platinium iridium cylinder having equal diameter and height kept at the particular condition of international bureo of weigh and measure sevre near paris.

Explanation:

The principle of an electric motor is based on the current carrying conductor which produces magnetic field around it. A current carrying conductor is placed perpendicular to the magnetic field so that it experiences a force.

The largest electric motors are used for ship propulsion, pipeline compression and pumped-storage applications with ratings reaching 100 megawatts. Electric motors are found in industrial fans, blowers and pumps, machine tools, household appliances, power tools and disk drives.

Reverberation, in psychoacoustics and acoustics, is a persistence of sound after the sound is produced

Explanation:

I think it is right hope its helps

Answer:

D. A single sound undergoes several reflections due to multiple reflecting surfaces.

Explanation:

Sometimes, the source of a sound is surrounded by multiple reflecting surfaces. The waves traveling from the source strike these different surfaces, causing multiple reflections. For example, a single clap of thunder reflects on several clouds and the earth's surface, causing you to hear a rolling rumble instead of a single sound.

Answer:

2 m/s^2

Explanation:

a = ∆v / ∆t

a = (50 - 30) / 10

a = 20 / 10

a = 2 m/s^2

acceleration = u-v/t

50-30/10

=2m/s

Answer:

F = 9/5 C + 32     conversion from C to F

F = 9/5 * F/5 + 32

25 F = 9 F + 800

16 F = 800

F = 50  

Check:

C = 5/9 ( F - 32)  = 5/9 (50 - 32) = 10  as requested

Q = c m change in temp

Q = 1 cal/gm-deg C * 500 gm * 45 deg C = 22,500 calories

50 Fahrenheit heat required.

Heat transfer is a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy  between physical systems.

F = 9/5 C + 32     conversion from C to F

F = 9/5 * F/5 + 32

25 F = 9 F + 800

16 F = 800

F = 50

The answer is 50 Fahrenheit.

Learn more about heat transfer

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Answer:V₁=300ml

T₁=27°C

V₂=?

T₂= -3°C

as we know

V₁T₁=V₂T₂

By putting values in formula

300ml×27°C=V₂×(-3°C)

300ml×27°C/-3°C=V₂

8100ml/-3=V₂

-2700ml=V₂

or V₂=  -2700ml

Answer:

a)  [tex]n=5.98*10^{26}/m^3[/tex]

b)  [tex]i=2010000A/m^2[/tex]

c)  [tex]I_w=14.207A[/tex]

d)  [tex]V_e=1.92*10^{-4}m/s[/tex]

Explanation:

From the question we are told that:

Diameter [tex]d=3mm=>3*10^{-3}[/tex]

Conductivity [tex]\sigma= 6.7 10^7 (0.M),[/tex]

Electron mobility [tex]\phi= 0.0064 m2 /V sec[/tex]

Electric field [tex]E= 30 mV/m[/tex]

a)

Generally the equation for Charge Density is mathematically given by

[tex]\phi=\frac{\sigma}{n e}[/tex]

Therefore

[tex]n=\frac{\sigma}{\phi e}[/tex]

[tex]n=\frac{6.7 10^7}{1.6*10^{-19} *0.0064}[/tex]

[tex]n=5.98*10^{26}/m^3[/tex]

b)

Generally the equation for current density is mathematically given by

[tex]i=\sigma*E[/tex]

[tex]i= 30*10^{-3] *6.7 10^7[/tex]

[tex]i=2010000A/m^2[/tex]

c)

Generally the equation for current in wire is mathematically given by

[tex]I_w=iA[/tex]

[tex]I_w=i*\pi r^2[/tex]

[tex]I_w=(2010000)*\pi( 1.5*10^{-3})^2[/tex]

[tex]I_w=14.207A[/tex]

d)

Generally the equation for electron draft velocity. is mathematically given by

[tex]V_e=\phi E[/tex]

[tex]V_e=(0.0064)*(30*10^{-3})[/tex]

[tex]V_e=1.92*10^{-4}m/s[/tex]

Answer:

Snth = u + a/2 ( 2n - 1)

Explanation:

Do you need explanation based on graph, integration or other method?

Answer:

each resistor draws 1/3 of an amp or 0.33333 amps

Explanation:

V = I * R

V = 10 volts

R = 30 ohms

10 = I * 30       Divide by 30

10/30 = I

I = 0.33333

Answer:

Explanation:

Answer:

It was created during the French Revolution in 1799 and has enabled for the international exchange of scientific and technical information. Calculating with SI units is also a lot easier than using the English system.

Answer:

E) The indoor pollution experts' research shows that the amounts of lead that come into contact with the air people breathe, even in the office buildings that used the greatest amount of paint, are too low to affect workers

Explanation:

Answer:

friction reduces the efficiency of machines, thus we must reduce the friction force that is acting upon it.

Answer:

Because it causes a lot of wear and tear in machine parts that move against each other. It erodes the surfaces and destroys their symmetries

Explanation:

Answer:

A) How the size of a tree affect the bird species that prefer to live in it

Explanation:

I took the quiz

Answer:

1) The kinetic energy of the object after it drops for 2.0 seconds is approximately 384.9 Joules

2) The potential energy of the object after it drops for 2.0 seconds is approximately 204 J

Explanation:

1) The given mass of the object, m = 2.0 kg

The height from which the object is dropped, h = 30 m

The kinetic energy of the object after it drops for 2.0 seconds = Required

Kinetic energy, K.E. = (1/2)·m·v²

Where;

v = The velocity of the object

The kinematic equation for finding the velocity of the object is presented as follows;

v = u + g·t

Where;

u = The initial velocity of the object = 0

g = The acceleration due to gravity of the object ≈ 9.81 m/s²

t = The time of motion of the object = 2.0 seconds

∴ The velocity after 2 seconds, v ≈ 0 + 9.81 m/s² × 2 s = 19.62 m/s

The kinetic energy, K.E. after 2 seconds as the object drops is given as follows;

[tex]K.E._{(after \ two \ seconds)}[/tex] = (1/2) × 2.0 kg × (19.62 m/s)² = 384.9444 J ≈ 384.9 J

2) The total energy, M.E. of the object at the top, h = 30 m, u = 0, is given as follows;

The total mechanical energy, M.E. = P.E. + K.E.

M.E. = m·g·h + (1/2)·m·u²

∴ M.E. = 2.0 kg × 9.81 m/s² × 30 m + 0 = 588.6 J

M.E. = 588.6 J

Given that the total mechanical energy, M.E., is constant, we have;

At 2.0 seconds, M.E. = 588.6 J , K.E. ≈ 384.9 J, P.E. = M.E. - K.E.

∴ P.E. = 588.9 J - 384.9 J ≈ 204 J

The potential energy after it drops 2.0 seconds, P.E. ≈ 204 J

Answer:

[tex] \sf \: given \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bf \: mass \: \: \: m \: = 7.00 \times {10}^{ - 4} \: kg \\ \\ \bf \: E=mc^2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = > E=7.00 \times {10}^{ - 4} \times ({3 \times {10}^{8} })^{2} \\ \\ = > \green{ \boxed{ E = 6.3 \times {10}^{13} \: J}}[/tex]

Answer:

8392

Explanation:

d=s/t

Answer:

Explanation:

A plane mirror is the kind you look into when you look into a "regular" mirror. The image you see is right-side-up. These images are virtual. Real images are always upside down and are made by mirrors that are "parabolic" in shape. Virtual images are always right-side-up.

All three cars experience the same impulse.

Impulse is equal to change in momentum.

Each car starts with the same amount of momentum and ends up with zero, so the magnitudes of all three changes are equal.

Answer:

F = 32.28 N

Explanation:

For this exercise we must use the rotational equilibrium relation

          Σ τ = 0

In the initial configuration it is in equilibrium, for which all the torque and forces are compensated. By the time the payment lands on the bar, we assume that the counter-clockwise turns are positive.

          W_bird  L / 2 - F_left 0.595 - F_right 0.595 = 0

we assume that the magnitude of the forces applied by the hands is the same

          F_left = F_right = F

          W_bird L / 2 - 2 F 0.595 = 0

          F = [tex]\frac{m_{bird} \ g L} { 4 \ 0.595}[/tex]

we calculate

         F = 0.560 9.8 14.0 /2.38

         F = 32.28 N

Answer:

13.759 % of the initial mechanical energy is lost as thermal energy.

Explanation:

By the First Law of Thermodynamics we know that increase in internal energy of the object ([tex]U[/tex]), in joules, is equal to the lost amount of the change in gravitational potential energy ([tex]U_{g}[/tex]), in joules:

[tex]\frac{x}{100} \cdot \Delta U_{g} = \Delta U[/tex] (1)

Where [tex]x[/tex] is the percentage of the energy loss, no unit.

By definition of the gravitational potential energy and internal energy, we expand this equation:

[tex]\frac{x\cdot m \cdot g \cdot h}{100} = m\cdot c\cdot \Delta T[/tex] (1b)

Where:

[tex]m[/tex] - Mass of the object, in kilograms.

[tex]g[/tex] - Gravitational acceleration, in meters per square second.

[tex]h[/tex] - Initial height of the object above the lunar ground, in meters.

[tex]c[/tex] - Specific heat of aluminium, in joules per degree Celsius-kilogram.

[tex]\Delta T[/tex] - Temperature increase due to collision, in degree Celsius.

If we know that [tex]m = 0.95\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]h = 75\,m[/tex], [tex]c = 920\,\frac{J}{kg\cdot ^{\circ}C}[/tex] and [tex]\Delta T = 0.11\,^{\circ}C[/tex], then the percentage of energy loss due to collision is:

[tex]x = \frac{100\cdot c\cdot \Delta T}{g\cdot h}[/tex]

[tex]x = \frac{100\cdot \left(920\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (0.11\,^{\circ}C)}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (75\,m)}[/tex]

[tex]x = 13.759\,\%[/tex]

13.759 % of the initial mechanical energy is lost as thermal energy.

Answer:

i. The pressure of due to the water, P, is given according to the following equation;

P = ρ·g·h

Where;

ρ = The density of the water (a constant) = 997 kg/m³

g = The acceleration due to gravity = 9.81 m/s²

h = The height of the water (minimum h = h₁, maximum h = h₂)

The pressure is directly proportional to the water height, and we have;

The pressure, P, will be maximum when the water height, h, is maximum or h = h₂, which is the level DC

ii. The thrust = The force acting on the body = Pressure × Area

The maximum areas exposed to the water are on side AB and DC

However, the pressure at level DC, which is the location of the maximum pressure, is larger than the pressure at level AB, therefore, the maximum thrust will be at the level DC

Explanation:

Answer:

78.34

Explanation:

1.3/30=78.3m

Answer:

Explanation:

We are basically needing to solve for the time in the equation d = rt, where d is the distance around Mars (aka the circumference), r is the velocity, and t is time. We need to find the circumference and the velocity. We will begin with the velocity.

Because the gravitational attraction between Phobos and Mars provides the centripetal acceleration necessary to keep Phobos in its (sort of) circular path, the equation we use for this is:

[tex]F_g=F_c[/tex] which says that Force supplied by gravity is equal to the centripetal force. Expanding that:

[tex]\frac{Gm_{Phobos}m_{Mars}}{r^2}=\frac{m_{Phobos}v^2}{r}[/tex]

When we move that around mathematically to solve for the velocity value, what we end up with is:

[tex]v=\sqrt{\frac{Gm_{Mars}}{r}[/tex] and filling in:

[tex]v=\sqrt{\frac{(6.67*10^{-11})(6.42*10^{23})}{9.38*10^6} }[/tex] and we get that

v = 2100 m/s

Now for the circumference:

C = 2πr and

C = 2(3.1415)(9.38 × 10⁶) so

C = 5.9 × 10⁷

Putting that all together in the C = vT equation:

5.9 × 10⁷ = 2100T so

T = 2.8 × 10⁴ sec or 7.8 hours

Answer:

Option A.

Explanation:

We define the half time T as the time such that an initial quantity A reduces to its half.

So we can model the quantity as a function of time like:

P(t) = A*e^(-k*t)

Then for the half time, T, we will have:

P(T) = A/2 = A*e^(-k*T)

solving for k, we get:

A/2 = A*e^(-k*T)

1/2 = e^(-k*T)

ln(1/2) = ln( e^(-k*T)) = -k*T

-ln(1/2)/T = k

Here we know that the half time is T = 26.5s

if we input that in the above equation, we get:

-ln(1/2)/26.5s = k = 0.0262 s^-1

Then the correct option is A

Answer:

The atomic number of phosphorus is 15.

Explanation:

It’s found after Si(Silicon) and before S(sulphur)

Answer:

The orbital velocity an aircraft orbiting around a heavenly body is found as follows;

At the orbital velocity, [tex]F_G[/tex] = [tex]F_C[/tex]

Where;

[tex]F_G[/tex] = The gravitational force = [tex]\dfrac{G \cdot M \cdot m}{R_E^2}[/tex]

[tex]F_C[/tex] = The centripetal force = [tex]\dfrac{m \cdot v_0^2}{R_E}[/tex]

Therefore

[tex]v_0 = \sqrt{\dfrac{G \cdot M}{R_E} }[/tex]

Therefore, at the orbital velocity of the spacecraft, the centripetal force attracting the person away from the central region heavenly body is equal to the gravitational force pulling the person towards the center of the heavenly body (which was felt as her or his weight), and the person feels weightless while inside the orbiting spacecraft

Explanation:

Answer:

Q. 1. Newton's Law of gravitation states that all bodies in the universe exerts a force of attraction on all other bodies in the universe with a proportional force to both the product of the masses of the bodies and inversely proportional to the square of the distance between their centers

Mathematically, we have;

[tex]F = G \times \dfrac{m_1 \times m_2}{R^2}[/tex]

Where;

m₁, and m₂ are the masses of the bodies

R = The distance between their centers

G = The gravitational constant = 6.6743 × 10⁻¹¹ N·m²/kg²

The gravitational constant, G, is the Newton's law of gravitation's constant of proportionality between the force of attraction that exist two bodies and the product of their masses divided by the square of the distance between their centers

Q. 2. Newton's law of gravitation in vector form is presented as follows;

[tex]\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{12}[/tex]

The above equation gives the gravitational force of attraction of body 1 on body 2, with the negative sign and unit vector indicating that the force of of gravity is towards body 1

The force of gravity of body 2 on 1 is presented as follows;

[tex]\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{12}^2} \cdot \hat R_{21}[/tex]

The gravitational force of attraction of body 2 on body 1 is therefore, equal in magnitude and opposite in direction of the gravitational force of body 1 on body 2 (towards body 2)

[tex]-\underset{F_{12}}{\rightarrow} = G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{12} = G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot -(\hat R_{21}) = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{21}[/tex]

[tex]-\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{21} = \underset{F_{21}}{\rightarrow}[/tex]

[tex]-\underset{F_{12}}{\rightarrow} = \underset{F_{21}}{\rightarrow}[/tex]

Explanation:

Answer:

d. Is indicated when a magnetometer measures intervals of magnetism that are weaker than expected

Explanation:

Magnetic field intensity is measured with a magnetometer on the surface of the Earth. Areas in which the magnetic field strength is lower or more than average are known as areas with magnetic anomalies, which may be due to  the presence of rocks that have a different magnetic characteristics

Where the magnetic anomaly is negative, it is indicative of a magnetic field strength reading that is lower than average magnetic field which is generally obtainable

Therefore, the correct option is indicated when a magnetometer measures intervals of magnetism that are weaker than expected