a rod of negligible mass is pivoted at a point that is off-center, so that length l1 is different from length l2. the figures show two cases in which masses are suspended from the ends of the rod. in each case the unknown mass m is balanced so that the rod remains horizontal.

The combined momentum of the passengers is 5000 kgm/s.

The combined momentum of the passengers is calculated as follows;

P = mv1 + mv2

where;

P = m(v1 + v2)

P = 5000(-1 + 2)

P = 5000 kgm/s

Thus, the combined momentum of the passengers is 5000 kgm/s.

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Answer:

b

Explanation:

Answer:

[tex]250000 \frac{n}{m {}^{2} } [/tex]

Explanation:

pressure is the force applied perpendicular to the surface area of an object per unit area.

for the question,

Force (F) = 5000N

[tex]Area = 200cm {}^{2} = 0.02m {}^{2} \\ Pressure = \frac{force}{area} \\ Pressure = \frac{5000}{0.02} \\ Pressure = 250000 \frac{n}{m {}^{2} } [/tex]

The acceleration lasted for 0.007s.

To find the answer, we need to know about the Newton's equation of motion.

Newton's equations are

V= final velocity, U= initial velocity, a= acceleration, t= time and S = distance

60²-0= 2a×0.2= 0.4a

Thus, we can conclude that the acceleration of the fireshell lasted for 0.007s.

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The power required to force the current of 4.13 A to flow through the conductor is 1927.43 watts

This is defined as the rate in which energy is consumed. Electrical power is expressed mathematically as:

Power (P) = square current (I²)× resistancet (R)

P = I²R

P = I²R

P = 4.13² × 113

P = 1927.43 watts

Thus, the power required is 1927.43 watts

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Supplemental oxygen shall be used in that flight time in excess of 30 minutes at those altitudes.

Most plane cabins are pressurized to 8,000 feet above sea level, an altitude that lowers the amount of oxygen inside the blood by approximately 4 percentage points, researchers say.

The higher the altitude, the less oxygen there is in the air and the lower the overall air pressure is. If flights were not pressurized, passengers could be prone to various physiological illnesses. Because of this, federal policies require that all commercial flights over 8,000 feet be pressurized.

The flight crew must use supplemental oxygen for the entire duration of flight operations above a cabin pressure altitude of 14,000 feet MSL.

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Answer:

2.45s

Explanation:

is explanation needed too?

The average acceleration in the first 10 seconds of the journey is 2 m/s²; option B.

Acceleration is the rate of change of velocity with time.

From the graph;

Acceleration in first  seconds = 20 - 0/5 = 4 m/s²

Acceleration in second 5 seconds = 0

Average acceleration = 4 + 0/2

Average acceleration = 2 m/s²

In conclusion, average acceleration is the average of the final and initial acceleration.

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A. The acceleration during the slide is 6.86 m/s²

B. The time taken to slide until he stops is 1.2 s

F = μN

F = 0.7 × 798.7

F = 559.09 N

a = F / m

a = 559.09 / 81.5

a = 6.86 m/s²

a = (v – u) / t

t = (v – u) / a

t = (0 – 8.23) / -6.86

t = 1.2 s

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Taking into account the definition of calorimetry,  the final temperature of the silver and water is 11.84 °C.

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

In this case, you know:

Replacing in the expression to calculate heat exchanges:

For silver: Qsilver= 240 [tex]\frac{J}{kgC}[/tex]× 0.5 kg× (Final temperature of silver - 250 C)

For water: Qwater= 4180 [tex]\frac{J}{kgC}[/tex] × 1 kg× (Final temperature of water - 5 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the silver gives up will be equal to the heat that the water receives. Therefore:

- Qsilver = + Qwater

And the final temperature of the silver is equal to the temperature of the water (Final temperature of silver= Final temperature of water= Final temperature). Then:

- 240 [tex]\frac{J}{kgC}[/tex]× 0.5 kg× (Final temperature - 250 C)= 4180 [tex]\frac{J}{kgC}[/tex] × 1 kg× (Final temperature - 5 C)

Solving:

- 120 [tex]\frac{J}{C}[/tex]× (Final temperature - 250 C)= 4180 [tex]\frac{J}{C}[/tex]× (Final temperature - 5 C)

120 [tex]\frac{J}{C}[/tex]× (250 C - Final temperature) = 4180 [tex]\frac{J}{C}[/tex]× (Final temperature - 5 C)

120 [tex]\frac{J}{C}[/tex]×250 C - 120 [tex]\frac{J}{C}[/tex]×Final temperature = 4180 [tex]\frac{J}{C}[/tex]×Final temperature - 4180 [tex]\frac{J}{C}[/tex]×5 C

30,000 J - 120 [tex]\frac{J}{C}[/tex]×Final temperature = 4180 [tex]\frac{J}{C}[/tex]×Final temperature - 20,900 J

50,900 J= 4300 [tex]\frac{J}{C}[/tex]×Final temperature

50,900 J÷ 4300 [tex]\frac{J}{C}[/tex]= Final temperature

11.84 °C= Final temperature

Finally,  the final temperature of the silver and water is 11.84 °C.

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Answer:

easy

Explanation:

it when the sunlight hits a rain droplets ,some of the light is reflected .The electromagnetic spectrum is made of light with many different wavelengths and each is reflected at a different angle . Thus spectrum is seperated producing a rainbow

The velocity of the photo electron is 6.11 × 105 ms

Given :

Supplied energy, Es = 5.3 x 10-19 J

Minimum energy of the electron to escape from the metal, E. = 3.6 x 10-19 J

To Find :

Velocity of photoelectron

Solution : The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So

5.3 × 10^-19 J = 3.6 × 10^-19 J + K

K = 5.3 x 10^-19 - 3.6 x 10^-19

K = 1.7 × 10^-19 J

The formula of kinetic energy is given by:

K = 1/2 mv^2

v = √2K/m

= √2 x 1.7 x 10^-19

√ 9.1 x 10^-31

v = 6.11 x 10^5 m/s

So, the velocity of the photo electron is 6.11 x 10^5 m/s

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From the calculation, the mass of the dummies head is 1647.44 Kg

We know that the momentum after collision is equal to the momentum before collision.

Mass of the headrest = 2005.6 kg

Initial velocity of the head rest = 4.524 m/s

Final velocity of the head rest =  4.487 m/s

Mass of the dummy = m

Initial velocity of the dummy = -1.005 m/s

Final velocity of the dummy = 9.965 m/s

Then;

(m *  -1.005) + (2005.6 * 4.524 ) = (2005.6 *   4.487) + (m * 9.965)

-1.005m + 9073.33 = 8999.13 + 9.965m

9073.33 - 8999.13  = 9.965m + 1.005m

18072.46 = 10.97m

m = 18072.46/ 10.97

m = 1647.44 Kg

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The acceleration of the body is found to be -4.05 m/s^2

We define the acceleration as the rate of change of the velocity with time. Hence, we can write that;

a = (μcosθ - sinθ)g

a = acceleration

μ = coefficient of kinetic friction

θ = angle of inclination

g = acceleration due to gravity

Thus;

a = (0.10cos 30 - sin 30) 9.8

a = -4.05 m/s^2

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B. The heat loss by the heat engine is 8,330 J.

The heat loss by the heat engine is calculated as follows;

H = E - W

where;

H = 10,750 J - 2,420 J

H = 8,330 J

Thus, the heat loss by the heat engine is 8,330 J.

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Answer:

St. Elmo's Fire - well known to old time sailors

A. The height above the ground reached by the mass is 2.3 m.

P.E = mgh

where;

h = P.E/mg

h = (342)/(15.2 x 9.8)

h = 2.3 m

Thus, the height above the ground reached by the mass is 2.3 m.

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The vertical components of velocity is 10.35 m/s and the horizontal component of velocity is 38.6 m/s

We know that velocity is a vector quantity, a vector often can be resolved into its components. The vertical components is V sinθ while the horizontal component is vcosθ.

Hence;

Vertical component = 40 m/s sin 15 degrees = 10.35 m/s

Horizontal component = 40 cos 15 degrees = 38.6 m/s

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Answer:

Gravity

Explanation:

The more massive the object is, the higher the gravity

Hence, a massive star, like our sun, has enough gravity to pull on planets and keeping them in orbit

The magnitude of acceleration of a block m₂ is 0.05 m/s² and the tension in the cord is 0.01 N.

Given:

mass of block 1, m₁ = 40 gm = 40×10⁻³ kg

mass of block 2, m₂ = 20 gm = 20×10⁻³ kg

Applied force, F = 0.03 N

Calculation:

Consider the free-body diagram of the system as shown below. Using Newton's second law of motion we get:

F = ma

where F is the applied force

            m is the total mass of the system

            a is the acceleration of block 2 (as it is pulled by horizontal force)

From the above equation we get:

0.03 N = (m₁+m₂) a

⇒ a = (0.03 N) / (m₁+m₂)

⇒ a = (0.03 N) / (40×10⁻³ kg + 20×10⁻³ kg)

⇒ a = 0.5 m/s²

Now, from the free-body diagram of block 2 as shown in figure 3, we get:

Balancing the forces along the horizontal:

∑Fₓ = 0

∴ T = m₂ a

where T is tension in the string

           a is the acceleration of block 2

Applying values in the above equation we get:

T = (20×10⁻³ kg) × (0.5 m/s²)

  = 0.01 N

Therefore, the acceleration of block 2 due to the applied horizontal force is 0.5 m/s² and the tension in the cord is 0.01 N.

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Answer:75 percent

Explanation:so in order tro fin d thge efficiency i used the forumla ,efficency=useful output energy/input energyx100%,in order to use this formula i needed the output,which i found by subtracting the input energy with wasted energy,that gave me the output,and after founding the output,i put that into the formula,

output energy=input energy - wasted energy

output energy=6000j-1500j

output energy=4500

put that into the formula

efficiency =output energy/input enrgy x100%

efficiency=4500/6000 multiplied by 100%

efficiency=0.75x100%

efficiency=75%

The Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x [tex]10^{24}[/tex] Kg

Linear speed is the product of angular speed and the maximum displacement of the particle. That is,

V = Wr

Where

Given that the earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.

a) To determine the Earth’s average orbital speed, we will make use of the below formula to calculate angular speed

W = 2[tex]\pi[/tex]/T

W = (2 x 3.143) / (365.26 x 24)

W = 6.283 / 876624

W = 7.2 x [tex]10^{-4}[/tex] Rad/hr

The Earth’s average orbital speed V = Wr

V = 7.2 x  [tex]10^{-4}[/tex] x 149.6 x [tex]10^{6}[/tex]

V = 107225.5 kilometers per hours.

b) Based on the information given in this question, to calculate the approximate mass of the Sun, we will use Kepler's 3rd law

M = (4[tex]\pi ^{2}[/tex][tex]r^{3}[/tex]) / G[tex]T^{2}[/tex]

M = (4 x 9.8696 x 3.35 x [tex]10^{24}[/tex]) / (6.67 x [tex]10^{-11}[/tex] x 7.68 x [tex]10^{11}[/tex])

M = 1.32 x [tex]10^{26}[/tex] / 51.226

M = 2.58 x [tex]10^{24}[/tex] Kg

Therefore, the Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x [tex]10^{24}[/tex] Kg

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Answer:

Kinematics is the study of the motion of mechanical points, bodies and systems without consideration of their associated physical properties and the forces acting on them. The study is often referred to as the geometry of motion, and it models these motions mathematically using algebra

Answer:

the branch of mechanics concerned with the motion of objects without reference to the forces which cause the motion is kinematics

Halfway between the like poles of two magnets, because the field lines bend away and do not enter this area.

How does a magnetic field diagram show where the field is strongest?

Where is magnetic field the strongest and weakest on a magnet?

Which two locations on the magnet would have the greatest attractive forces?

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If the density of water does not vary and the vents range in depth from about 1500 m to 3200 m below the surface, then the gauge pressure at a 2452-m deep vent is 224.268 atm.

Calculation:

Step-1:

It is given that the vents range in depth from about 1500 m to 3200 m below the surface. If we are assuming that the density of water does not vary. Then it is required to calculate the gauge pressure at a 2452-m deep vent.

The gauge pressure at a particular depth of ocean water is calculated as:

[tex]$$P=\rho g h$$[/tex]

Here [tex]\rho[/tex] is the density of water, P is the required pressure, h is the depth of water, and g is the gravitational acceleration.

Step-2:

Now we are substituting the values to calculate the pressure at the depth of 2452-m.

[tex]$$\\\begin{aligned}\\P&=\rho gh\\&=1030 (\text{ kg/m}^3)\times 9.8 (\text{ m/s}^2)\times 2452 \text{ m}\\&=24.75\times 10^6 \text{ Pa}\times\frac{1 \text{ atm}}{10.1325 \times10^4 \text{ Pa}}\\&=224.268 \text{ atm}\\\end{aligned}\\$$[/tex]

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The diameter of the largest orbit just before the protons exit the cyclotron is 39 cm.

The number of orbits completed by the proton during this 1.0 ms is 14000 revolutions.

The kinetic energy for the protons can be computed by using the formula:

mv² = 2 K.E

v = sqrt( 2*KE / M)

the kinetic energy of the medical isotopes = 6.5 MeV

substituting the values,

v = sqrt( 2* 6.5* 1.6* 10^-13 / 1.67* 10^-27 )

v = 3.53 × 10⁷ m/s

The radius of the orbit can be estimated by using the formula:

mv² / R = qvB

r = q*v / mv²

r = ( 1.67* 10^-27 * 3.53 × 10⁷ ) / ( 1.9* 1.6* 10^-19 )

r = 0.19415 m

Since diameter (D) = 2r,

D= 2(0.19415  m)

D= 0.39 m

D≅ 39 cm

The time period to complete a revolution around the spiral trajectory is:

T = 2πr / v

T = 2*3.14* 0.1941 / 3.53*10^7

T = 0.7 × 10⁻⁷ s

Finally, the number of orbits that the proton does to complete the revolution in 1 ms is:

n = t / T

n = 10^-3 / (0.7*10^-7)

n = 14285.71

n ≅ 14000 revolutions

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Your question is incomplete, but most probably full question was:

A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotron is 1.9 T.

(a) What is the diameter of the largest orbit, just before the protons exit the cyclotron? Express your answer with the appropriate units. d = 57 cm Previous

(b) A proton exits the cyclotron 1.0 ms after starting its spiral trajectory in the center of the cyclotron. How many orbits does the proton complete during this 1.0 ms?

Answer:

It is because of the blood pressure balances with our atmospheric pressure

The work is done by the 50 N force is 200 J.

Work done = Fd

where;

Work done = 50 N x 4 m = 200 J

Thus, the work is done by the 50 N force is 200 J.

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