Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end of a rope is attached to the bus and the other is tied around Magnus's waist. If a force gauge placed halfway down the rope reads out a constant 2100 Newtons while Magnus pulls the bus a distance of 1.30 meters, how much work does the tension force do on Magnus

Answer:

The metalloids are located on the right side of the periodic table in a "step-like" arrangement.

All of the possible metalloids are:

boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), and polonium (Po)

No. of turns in the solenoid is an option (b) 144.

The self-inductance of a long solenoid depends only on its physical properties (such as the number of turns of wire per unit length and the volume), and not on the magnetic field or the current.

Self-inductance of solenoid = 39.1 μH

                                           = 39.1 × [tex]10^-^6[/tex] H

Length of the solenoid = 4.0 cm

Cross-sectional area = 0.60 cm²

Expression for the self-inductance of a coil ;

L = µ₀N²A / [tex]l[/tex]

where,

L = Self- Inductance

N = No. of turns.

A = Cross-sectional area

[tex]l=[/tex] Length of the solenoid

L =( 4π × [tex]10^-^7[/tex] × N² × 0.60 ) / 4.0

39.1× 4.0 / 4π × [tex]10^-^7[/tex] × 0.60 = N²

N² = 2.07 × [tex]10^6[/tex]

N = 144

Therefore, the no. of turns of the solenoid is 144.

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The velocity of a 55.0-kg person hitting the ground, is mathematically given as

vt=39.5983m/s

Generally, the equation for is  mathematically given as

mass of squirrel,

[tex]m=550 \mathrm{~g}\\\\Surface area, $A=945 \mathrm{~cm}^{2}=88 \times 10^{-3}$\\\\Height, $h-4 \mathrm{~m}$\\[/tex]

Terminal velocity is given by:

[tex]$v_{i}=\sqrt{\frac{2 m g}{\rho A C}}$[/tex]

where \rho is the density of fluid that is falling and it is given by

[tex]$\rho=\frac{m}{V}$[/tex]

since, volume =area * height

[tex]^{\rho=} \frac{0.55 \mathrm{Kg}}{0.0945 \mathrm{~m}^{2} \times 4.0 \mathrm{~m}}\\\\$\rho=0.1455 \mathrm{Kg} / \mathrm{m}^{3}$[/tex]

A is the surface area of squirrels.

C is the drag coefficient.

The surface area facing the fluid is given by:

[tex]A_{f}=\frac{0.0945 \mathrm{~m}^{2}}{2} \\\\\\ A_{f}=0.04725 \mathrm{~m}^{2}[/tex]

so, terminal velocity is :

[tex]$v_{t}=\sqrt{\frac{2 \times 0.55 \mathrm{Kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2}}{0.1455 \mathrm{Kg} / \mathrm{m}^{3} \times 0.04725 \mathrm{~m}^{2} \times 1}}$[/tex]

Vt=39.5983

In conclusion, the terminal velocity of the squirrel is 39.5983m/s

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Using the statistical definition of entropy, The entropy of a system W = 4 is 1.91×10²³.

Entropy is typically referred to as a measurement of a system's randomness or disorder. In 1850, a German physicist named Rudolf Clausius first proposed this idea. Entropy is a thermodynamic property used to characterize a system's behavior in terms of temperature, pressure, entropy, and heat capacity. This thermodynamic explanation took the systems' equilibrium condition into account.

Entropy can be calculated using a mathematical expression.

Entropy = Total change of heat / thermodynamic temperature

S= KblnW

S is the statistical entropy

The value of Boltzmann's constant is 1.38×10⁻²³.

S=  1.38×10⁻²³ ln(4)

S = 1.91×10⁻²³

Using the statistical definition of entropy, The entropy of a system W = 4 is 1.91×10²³.

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Answer: C.

1.91 × 10⁻²³ joules/kelvin

Explanation: edmentum

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

                           [tex]N_x=86.62N[/tex]

                           [tex]N_y=F_V=mg-Tsin59\\[/tex]

                           [tex]F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N[/tex]

                    [tex]N_y= (40*9.8)-(169.8*sin59)=246.4N[/tex]

                 [tex]N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N[/tex]

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

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The hi.nge will be subjected to a force of 261.12N from the beam.

We must understand the tension in order to choose the solution.

                       [tex]N_y=F_v=mg-Tsin59[/tex]

                         [tex]F_H=N_x=Tcos59\\T=\frac{F_H}{cos59} =169.84N[/tex]

                  [tex]N_y=(40*9.8)-(169.84*sin59)=246.4N[/tex]

                        [tex]N=\sqrt{N_x^2+N_y^2} =261.12N[/tex]

Thus, we can infer that the force the beam applies to the height is 261.12N in size.

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Answer:

Object A's momentum is larger

Explanation:

as the formula for momentum goes:

P = M * V

where P is momentum, M is mass, V is velocity.

so where Va (Object A's velocity) is larger than Vb (Object B's velocity) we get:

( i ) Va > Vb

as the masses of both objects are equal, we mark:

( ii) Ma = Mb = M

we multiply both sides of ( i ) by M to get:

( iii ) Va × M > Vb × M

and we finally get:

( iv ) Pa > Pb

Answer: Object A's momentum of larger than compared to that of object B.

Explanation:

The speed of the ball at the given distance and time of motion is 4 m/s.

The speed of an object is the rate of change of distance traveled by the object with time.

Speed is a scalar quantity because it has only magnitude and no direction. It is measured in meters per second.

For a ball that travelled a distance of 20 m and displacement of 10 m in 5 seconds​, the speed of the ball at the given distance and time of motion is calculated as follows;

Speed of the ball = distance traveled by the ball / time of motion

speed of the ball = 20 m / 5 s

speed of the ball = 4 m/s

Thus, the speed of the ball at the given distance and time of motion is 4 m/s.

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The number of half-lives that would go by in 44 h is 7 half-lives

This is the time taken for half a substance to decay

The number of half-lives that will elaspe after 44 h can be obtained as illustrated below:

n = t / t½

n = 44 / 6.01

n = 7

Thus, 7 half-lives will elaspe after 44 h

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(a) The minimum force F he must exert to get the block moving is 38.9 N.

(b) The acceleration of the block is 0.79 m/s².

The minimum force F he must exert to get the block moving is calculated as follows;

Fcosθ = μ(s)Fₙ

Fcosθ = μ(s)mg

where;

F = [0.1(36)(9.8)] / [(cos(25)]

F = 38.9 N

F(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32

a = F(net)/m

a = 28.32/36

a = 0.79 m/s²

Thus, the minimum force F he must exert to get the block moving is 38.9 N.

The acceleration of the block is 0.79 m/s².

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The minimum  necessary height of the IV bag above the position of the needle is 0.37 m.

we know that,

P = ρgh

where,

P = 35.7mmHg

  = 4759.609 Pa

g = 9.8[tex]\frac{m}{s^{2} }[/tex]

ρ = 1.308 kg / m^3

now, substituting all the values, we get,

4759.609 = 1.308 × 9.8 × h

h = 0.37 m

The minimum  necessary height of the IV bag above the position of the needle = 0.37 m.

what is an IV bag ?

A reagent, also called as an analytical reagent, is a substance or compound that is added to a system in chemistry to bring about a chemical reaction or examine to see if one happens. Even though the terms "reagent" and "reactant" are frequently used synonymously, "reactant" refers to a substance that is consumed during a chemical reaction.

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The direction of magnetic field is south-east and the magnitude is

[tex]23.66[/tex] × [tex]10^{3} N/C[/tex].

Here, the magnitude of CD and BC will be cancelled, as they both are in the opposite direction and equal to each other.
the magnitude, towards the diagonal AC will result in CP, which is the direction of the electric field.

magnitude of electric field can be defined as :- The force per charge on the test charge is a straight forward way to define the size of the electric field.

To find the magnitude of the electric field use the formula

[tex]E = kq/ r^{2}[/tex]

inserting the values,

[tex]E = 9. 10^{9}[/tex] × [tex]4.05[/tex] × [tex]10^{-6} / 1.1 \sqrt{2}[/tex]

[tex]E= 36.45[/tex] × [tex]10^{3}[/tex] / [tex]1.54[/tex]

[tex]E = 23.66[/tex] × [tex]10^{3}[/tex] N/C

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The symbol for a variable resistor is a line with a zig zag pattern with a diagonal line drawn across it.

What is a variable resistor ?

A resistor whose electric resistance value may be altered is known to as a variable resistor. A variable resistor, which works usually by sliding a contact (wiper) over a variable resistor, is basically an electro-mechanical transducer.

The flow of electrical current is restricted by a resistor. The resistivity of a fixed resistor is steady. By adjusting a slider's position, the resistance of this resistor can be altered. Some volume controls and dimmer switches employ variable resistors.

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Answer: The force acting on the proton of charge ​1.5 x10-19 C moving with velocity 1.5 x10-19 C under the influence of a magnetic field of 0.7 T will be 3.15×10^-13 N.

Explanation: To find the answer we need to know more about the Lorentz magnetic force.

                                  F= q (v× B)

                               [tex]F= qvBsin\alpha[/tex]  where, [tex]\alpha[/tex] is the angle between v and B.

                                [tex]B=0.7 T\\q=1.5*10^{-19}C\\v= 3*10^{6}m/s.\\\alpha =90 degree.\\[/tex] because, from the question it is clear that the proton is moving along x axis and the magnetic field is along the y axis.

                      [tex]F=qvBsin\alpha =1.5*10^{-19}C*3*10^6 m/s*0.7T*sin (90)\\F=3.15*10^{-13}N[/tex]

Thus, we can conclude that the Lorentz force acting on the proton will be 3.15×10^-13 N.

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In response to a magnetic field of 0.7 T, a proton with a charge of 1.5 x 10-19 C travelling at a speed of 1.5 x 10-19 C will experience a force of 3.15 x 10^-13 N.

We need to learn more about the Lorentz magnetic force in order to locate the solution.

                                      F= Q (v× B)

                                   [tex]F=QvBsin\alpha[/tex]

[tex]\alpha[/tex] the angle between v and B is, where.

                           [tex]F=3.15*10^{-13}N[/tex]

Thus, we can infer that the proton will be subject to a 3.15 10^-13 N Lorentz force.

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A round cactus with many spines is the adaptation of a cactus that protects it from predators.

A cactus, unlike other plants, has unique adaptations in its roots, leaves, and stems that allow it to flourish in hot and dry surroundings.

The one adaptation that protects the cactus from predators is spines.

A cactus does not have any parts that resemble leaves if you could look at one closely.

Instead, the leaves are transformed into spines, which protrude from the plant's tiny bumps known as areoles.

Herbivores that live in the desert may be enticed to eat the cactus. The spines prevent these predators by modifying leaves into spines.

Other than protection, Spines perform many functions like

1) Since evaporation is a problem in a desert since water is scarce, the spines prevent excessive evaporation.

2)  The spines also impede airflow and prevent evaporation by trapping air.

3) Collecting dew from the early-morning fog is another crucial job that the spines do.

The gathered dew turned into liquid water and ran down to the earth below. The plant then absorbs this water.

Hence, the adaptation of a cactus that protects it from predators is round cactus with many spines.

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Answer:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

Explanation:

Answer:

90.3N

Explanation:

⊥mg = (0.170 m)(1.20 kg) 9.81 m/s

τ ball = r⊥Wball = (0.340 m)(1.42 N) = − 0.483 N ⋅m

F − 2.001− 0.483 N ⋅m = 0

F = 2.484 N ⋅m

0.0275 m = 90.3 N

The net torque acting on the forearm and hand is 90.3N

Torque is a measure of the pressure that can motivate an object to rotate about an axis. simply as pressure is what causes an object to accelerate in linear kinematics, torque is what reasons an item to collect angular acceleration. Torque is a vector amount.

⇒mg = (0.170 m)(1.20 kg) 9.81 m/s

⇒torque = r⊥weight of the ball

⇒ (0.340 m)(1.42 N) = − 0.483 N ⋅m

⇒F = − 2.001− 0.483 N ⋅m = 0

⇒F = 2.484 N ⋅m

⇒0.0275 m = 90.3 N

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These given conditions satisfy the second law of thermodynamics.

As the process is isobaric

So there will be a straight line of P= 200kPa in P-v and P-T planes

P1 = P2 = 100kPa

For perfect ideal gas, v-T plane:

[tex]v = (\frac{R}{P}) T[/tex]

[tex]v_{1} = (\frac{R}{P_{1} }) T_{1}[/tex] = 287 × 500/200000 = 0.717 m³/kg

[tex]v_{2} = (\frac{R}{P_{2} }) T_{2}[/tex] = 287 × 600/200000 = 0.861m³/kg

As it is the calorically perfect gas

de = [tex]c_{v}[/tex]dT

Integration on both sides

e2 - e1 = [tex]c_{v}[/tex](T2 - T1)

           = ( 716.5J/kg/K) (600-500)

           = 71650 J/kg

also,

Tds = de + Pdv

Tds = [tex]c_{v}[/tex]dT +Pdv

For ideal gas

V = RT/P        

dv = Rdt/P - RTdp/P²

Tds = [tex]c_{v}[/tex]dT + Rdt - RTdp/P

ds = ([tex]c_{v}[/tex] + R)dT/T - RdP/P

ds = ([tex]c_{v} + c_{p} -c_{v}[/tex])dT/T - RdP/P

ds = [tex]c_{p}[/tex]dT/T - RdP/P

Integration on both sides

s2 - s1 = [tex]c_{p}[/tex]ln (T2/T1) - R ln (P2/P1)

Since P is constant

s₂ - s₁ = [tex]c_{p}[/tex] ln (T2/T1)

           = 1003.5 ln (600/500)

           = 1003.5 × 0.182

           = 182.95 J/kg/K

w = Pdv

[tex]w_{12}[/tex] = P(v₂ - v₁)

     = 2,00,000 ( 0.861 - 0.717)

     = 28,800 J/kg

de = δq -δw

δq = de + δw

q₁₂ = (e₂ - e₁) +  w₁₂

    =  71,650 + 28,800 = 1,00,450 J/kg

Now in this process, the gas is heated from 500 K to 600 K. We would expect at a minimum that the surroundings were at 600 K.

Let’s check for second law satisfaction.

s₂ - s₁ ≥ q₁₂ / Tₓ

182.95 ≥ 1,00,450 / 600 K

182.95 J/kg/K ≥ 167.41 J/kg/K

Hence this condition satisfies the second law of thermodynamics

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Answer:

more durability and easier to pound a hammer with the heavy weight.

Explanation:

One should use an iron hammer instead of the wooden hammer because the force developed with the help of the iron hammer is far greater than the wooden hammer

Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.

The mathematical expression for Newton's second law is as follows

F = m*a

where F represents the force applied

m is the mass of the object

a is the acceleration of the object

As we know that force is a product of mass and acceleration, let us assume that both iron and wooden hammers have the same dimension , considering the acceleration applied is the same while hammering ,

As for the same dimension the mass of iron hammer would be greater as result force applied through the iron hammer would be grater

Since the iron hammer's force is so much stronger than the wooden hammer's force, one should use an iron hammer instead of a wooden hammer.

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The elastic potential energy of the block-spring system is 0.906 J.

Velocity of the block , v = 0.83 m/s.

Elastic potential energy is the energy stored in a stretched or compressed elastic material.

The elastic potential energy of the block-spring system = 725 * 0.05²/2 The elastic potential energy of the block-spring system = 0.906 J

The elastic potential energy of the spring is converted to kinetic energy of the block.

1/2 mv² = 0.906 J

where v is velocity

v = √(0.906 * 2)/2.6

v = 0.83 m/s.

In conclusion, elastic potential energy is present in compressed or stretched elastic materials.

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The work done by gas is 0.753 J and change in internal energy is 4.247J

So we are given that mass is 2kg , radius 1 cm and the amount of heat is 5 cm

The piston raised by 2.4cm

As we know that Work done is PΔV

Where ΔV is change in volume

Therefore ΔV =  πr^2 h = π x (.01)^2 x .024 =7.53×10^(-6)m^3

Here pressure is 10^5 pa

So W = [tex]10^5\times7.53\times10^-(6)[/tex]

Therefore W = 0.753 J

Now coming to change in internal energy

Change in Internal Energy = Heat Added - Energy lost in work

∴ 5J - 0.753 J = 4.247J

Hence the change in internal energy is 4.247 J

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1.06 is the maximum refractive index that the liquid may have for the light to be totally reflected.

Only when a light source passes from a denser to a rarer medium can it completely reflect.

When the angle of incidence surpasses a specific critical value, specular reflection occurs in the more highly refractive of the two mediums at their interface, and this reflection is known as total reflection.

sin [tex]i_{c}[/tex] = μ[tex]_{r}[/tex] / μ[tex]_{d[/tex]

From the diagram

Angle of incidence = 60°

sin60°  ≥ sin[tex]i_{c}[/tex] = μ[tex]_{r}[/tex]/μ[tex]_{d}[/tex]

μ[tex]_{r}[/tex] ≤ μ[tex]_{d}[/tex] sin60°

μ[tex]_{r}[/tex] ≤ √1.5 × √3/2

   = 1.06

Hence, the maximum index that the liquid may have for the light to be totally reflected is 1.06

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The Answer is Option C. Friction.

Explanation:

The friction force acts in the forward direction on the rear wheel and it acts in the backward direction on the front wheel. The magnitude of friction force on the rear wheel can be more, equal or less than that on the front wheel.

10,378.82 m/s is the second cosmic speed.

69,801 km is the radius of the synchronous orbit of a satellite.

Given

Mass of planet = 4.74 × [tex]10^{24}[/tex] kg

Radius of planet = 5870 km = 5870000m

First Cosmic speed = 7.34 km/sec

1) Second cosmic speed i.e. the minimum speed required for a satellite to break free permanently from the planet is also known as the escape velocity of a satellite.

It can be calculated by

v = √2GM/r where,

v= Escape velocity of the satellite

G = Gravitational constant

M = Mass of planet

r = Radius of planet

v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]

v = 10,378.82 m/s

2) Speed of the satellite at the given period

v = 2πr/T where,

T= Time period of rotation = 16.6 × 3600 seconds

r = v×T/2π

r = (7,338.93 x 16.6 x 3600 s) / (2π)

r = 69,801 km

Hence

The Second Cosmic Speed i.e. the minimum speed required for a satellite to break free permanently from the planet is 10,378.82 m/s.

And the radius of the synchronous orbit of a satellite is 69,801 km.

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Based on the calculations, the sound intensity level is equal to 1.0 × 10⁻⁵ W/m².

Mathematically, sound intensity level can be calculated by using this formula:

[tex]\beta = 10 log(\frac{I}{I_{ref}} )[/tex]

Where:

I is the intensity of the sound.

Note: The reference value of sound intensity is equal to 1.0 × 10⁻¹² W/m².

Rewriting the formula, we have:

β/10 = logI - logIo

Substituting the parameters into the formula, we have;

60/10 = logI - log(1.0 × 10⁻¹²)

6 = logI + 12

logI = 6 - 12

logI = -6

I = 1.0 × 10⁻⁵ W/m².

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Answer:

I think it's force

Explanation:

Mark as brainliest if it is right

Answer: A

Explanation:

Force can cause a stationary object to start moving or a moving object to change its speed or direction or both

(a) The tension the musician must stretch it is 147.82 N.

(b) The percent increase in tension is needed to increase the frequency is 26%.

v = √T/μ

where;

v = Fλ

in fundamental mode, v = F(2L)

v = 2FL

v = 2 x 65.4 x 0.6 = 78.48 m/s

v = √T/μ

v² = T/μ

T = μv²

T = 0.024 x (78.48)²

T = 147.82 N

v = 2FL = 2 x 73.4 x 0.6 = 88.08 m/s

T = μv²

T = (0.02)(88.08)²

T = 186.19 N

= (186.19 - 147.82)/(147.82)

= 0.26

= 0.26 x 100%

= 26 %

Thus, the tension the musician must stretch it is 147.82 N.

The percent increase in tension is needed to increase the frequency is 26%.

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The ratio of the speed of the lighter bullet to the speed of the heavier will be four times the heavier bullet.

Both can do same amount of work.

Kinetic energy of a body is the energy due to the motion of the body.

Kinetic energy = mv²/2

where m is mass and v is velocity of the object.

Since both objects have the same kinetic energy but one bullet has twice the mass of the other, the ratio of the speed of the lighter bullet to the speed of the heavier will be four times the heavier bullet.

The kinetic energy of both bullets is the same. Hence, they can do equal amount of work.

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The correct response for each of the condition given in the questions are,

To find the answer, we have to know about the Archimedes principle.

                               [tex]F-W=0\\F=W\\W=mg, where.\\m=density*volume=d*V\\V=\frac{4}{3} \pi r^3[/tex]

                            [tex]d_wVg = d_0(\frac{4}{3}\pi r^3 ) g \\d_wV = d_0(\frac{4}{3}\pi r^3 )[/tex]           (1)

Case 1:

Case 2:

Case 3:

Thus, we can conclude that,

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The minimum necessary height of the IV bag above the position of the needle, if the density of the solution is 1.308kg/L, and the pressure in the vein is 35.7 mmHg is, 2.78m.

To find the answer, we need to know more about the pressure exerted by a liquid column.

                                                [tex]P=[/tex] ρgh

                                           [tex]density=1.308 kg/L\\\\P=35.7 mmHg.\\[/tex]

                                       [tex]h=\frac{P}{density*g} =\frac{35.7}{1.308*9.8}\\\\ h=2.78 m[/tex]

Thus, we can conclude that, the height of the Iv bag is 2.78 m above the position of needle.

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If the solution has a density of 1.308 kg/L and the vein pressure is 35.7 mmHg, the minimum height of the IV bag above the needle position is 2.78 m.

We need to learn more about the pressure that a liquid column exerts in order to determine the solution.

                            [tex]P=[/tex] ρgh

                        [tex]Density=1.308kg/L\\P=35.7mmHg[/tex]

                           [tex]h=\frac{P}{density*g}\\\\ h=2.78m[/tex]

As a result, we may say that the I.V. bag is 2.78 m above the ground.

Learn more about the pressure exerted by a liquid column here:

https://brainly.com/question/28106909

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