A 46.0-kg girl is standing on a 157-kg plank. Both originally at rest on a frozen lake that constitutes a frictionless, flat surface. The girl begins to walk along the plank at a constant velocity of 1.48î m/s relative to the plank.

(a) What is her velocity relative to the surface of ice?
(b) What is the velocity of the plank relative to the surface of ice?

Answers

Answer 1

To solve this problem we will apply the linear motion kinematic equations. Just as we will also find the relative speed of the body through the conservation of momentum. Our data is given as

[tex]M = 157kg[/tex]

[tex]m = 46kg[/tex]

[tex]v_1 = 1.48m/s[/tex]

PART A)

From the conservation of momentum,

[tex]\text{Momentum of Plank+girl}+\text{Mometum of girl} = 0[/tex]

[tex](M+m)v_2+mv_1 = 0[/tex]

[tex](M+m)v_2 = -m_v1[/tex]

[tex]v_2 = \frac{-Mv_1}{M+m}[/tex]

[tex]v_2 = \frac{-(46)(1.48)}{(157+46)}[/tex]

[tex]v_2 = -0.3353m/s[/tex]

Since the ice surface is frozen lake and girl is moving on it so the relative velocity will get added. Therefore the velocity of the girl relative to the ice surface is as,

[tex]v_1+v_2 = 1.48+(-0.33353)[/tex]

[tex]v_1+v_2 = 1.14647m/s[/tex]

The velocity of the girl relative to the ice surface is 1.14647m/s

PART B) The velocity of the plank plus girl is [tex]v_2 = -0.3353m/s[/tex]

Since the ice surface is frozen lake and plank is moving with girl on it so the relative velocity will get added. Therefore the velocity of the plank relative to the ice surface is as:

[tex]v_2 = -0.3353m/s[/tex]

Answer 2

"The correct answers are: (a) The girl's velocity relative to the surface of ice is 1.48m/s. (b) The velocity of the plank relative to the surface of ice is 0 m/s.

 (a) Since the girl is walking on the plank with a constant velocity of 1.48 m/s relative to the plank, and the plank itself is at rest on the frictionless ice surface, there are no external forces acting on the girl-plank system in the horizontal direction.

According to the principle of inertia, an object at rest will stay at rest, and an object in motion will stay in motion with a constant velocity unless acted upon by an external force. Therefore, the girl's velocity relative to the surface of ice is the same as her velocity relative to the plank, which is 1.48 m/s.

(b) The plank is initially at rest on the frictionless ice surface, and since there are no external forces acting on it in the horizontal direction, it will remain at rest relative to the ice surface. This means that the velocity of the plank relative to the surface of ice is 0 m/s. The girl walking on the plank does not affect the plank's velocity because her motion is internal to the girl-plank system, and there is no friction to transfer her momentum to the plank.

In summary, the girl's motion is relative to the plank, and since the plank remains stationary on the ice, her velocity relative to the ice is the same as her walking velocity on the plank. The plank itself does not move because of the lack of friction and external forces, thus its velocity relative to the ice remains 0 m/s."


Related Questions

A Capacitor is a circuit component that stores energy and can be charged when current flows through it. A current of 3A flows through a capacitor that has an initial charge of 2μC (micro Coulombs). After two microseconds, how much is the magnitude of the net electric charge (in μC) of the capacitor?

Answers

Answer:

[tex]8\mu C[/tex]

Explanation:

t = Time taken = [tex]2\mu s[/tex]

i = Current = 3 A

q(0) = Initial charge = [tex]2\mu C[/tex]

Charge is given by

[tex]q=\int_0^t idt+q(0)\\\Rightarrow q=\int_0^{2\mu s} 3dt+2\mu C\\\Rightarrow q=3(2\mu s-0)+2\mu C\\\Rightarrow q=6\mu C+2\mu C\\\Rightarrow q=8\mu C[/tex]

The magnitude of the net electric charge of the capacitor is [tex]8\mu C[/tex]

A fullback with a mass of 100kg and a velocity of 3.5 m/s duewest collides head-on with a defensive back with a mass of 80 kgand a velocity of 6 m/s due east.
a.)What is the initial momentum of each player?b.)What is the total momentum of the system before thecollision?c.)If they stick together and external forces can be ignored,what direction will they be traveling immediately after theycollide?

Answers

Answer

given,

mass of full back, M = 100 Kg

velocity of full back  in east = 3.5 m/s

mass of the defensive back,m = 80 Kg

velocity of defensive back, due east = 6 m/s

a) initial momentum of each player

  for full back

   P₁ = M v  =  100 x 3.5 = 350 kg.m/s

 for defensive back

   P₂= m v = 80 x 6 = 480 kg.m/s

b) total momentum before collision

     P = P₁ + P₂

taking west direction as positive

     P = 350  + (- 480 )

     P = -130 kg.m/s

c) speed of the them when they stick together.

    using conservation of momentum

   initial momentum = final momentum

    -130 = (M + m ) V

    180 V = -130

    V = -0.722 m/s

velocity after the collision is equal to 0.722 m/s in direction of east.

a) P₁=350 kg.m/s and P₂= 480 kg.m/s

b) Total momentum before collision is -130 kg.m/s.

c) 0.722 m/s in direction of east

Given,

Mass of full back, M = 100 Kg

Velocity of full back  in east = 3.5 m/s

Mass of the defensive back, m = 80 Kg

Velocity of defensive back, due east = 6 m/s

Momentum

It is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction.

To find:

a) initial momentum of each player

For full back  

P₁ = M*v  =  100 * 3.5 = 350 kg.m/s

For defensive back

P₂= m*v = 80*6 = 480 kg.m/s

b) total momentum before collision

P = P₁ + P₂

taking west direction as positive

P = 350  + (- 480 )

P = -130 kg.m/s

c) speed of the them when they stick together.

Using conservation of momentum

initial momentum = final momentum

-130 = (M + m ) V

180 V = -130

V = -0.722 m/s

Velocity after the collision is equal to 0.722 m/s in direction of east.

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A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplitude 1.80 mm travels along the wire.

(a) Calculate the average power carried by the wave.
(b) What happens to the average power if the wave amplitude is halved?

Answers

The new power is: New P_avg = 2.52 W / 4 ≈ 0.63 W

Average Power Carried by a Wave

To solve this problem, we need the following information:

Mass of piano wire: 2.95 g = 0.00295 kg

Length of wire: 79.0 cm = 0.79 m

Tension: 29.0 N

Frequency: 105 Hz

Amplitude: 1.80 mm = 0.00180 m

First, calculate the linear mass density (μ) of the wire:

μ = mass / length = 0.00295 kg / 0.79 m ≈ 0.00373 kg/m

Next, find the wave speed using the formula for the speed of a wave on a string:

v = [tex]\sqrt{Tension / \mu}[/tex] =[tex]\sqrt{29.0 N / 0.00373 kg/m}[/tex]≈ 88.19 m/s

Now, we calculate the average power (P_avg) carried by the wave using the formula:

P_avg = 0.5 x μ x v x ω² x A²

Where:

ω = 2πf (angular frequency)

ω = 2 x π x 105 ≈ 659.73 rad/s

Therefore,

P_avg = 0.5 x 0.00373 kg/m x 88.19 m/s x (659.73 rad/s)² x (0.00180 m)² ≈ 2.52 W

Average Power if Amplitude is Halved

If the amplitude (A) is halved, the new amplitude is:

New A = 0.00180 m / 2 = 0.00090 m

Since power is proportional to the square of the amplitude (A²), halving the amplitude reduces the power by a factor of 4.

Thus, the new power is:

New P_avg = 2.52 W / 4 ≈ 0.63 W

The car's motion can be divided into three different stages: its motion before the driver realizes he's late, its motion after the driver hits the gas (but before he sees the police car), and its motion after the driver sees the police car. Which of the following simplifying assumptions is it reasonable to make in this problem? a. During each of the three different stages of its motion, the car is moving with constant acceleration. b. During each of the three different stages of its motion, the car is moving with constant velocity. c.The highway is straight (i.e., there are no curves). d. The highway is level (i.e., there are no hills or valleys).Enter all the correct answers in alphabetical order without commas. For example, if statements C and D are correct, enter CD.

Answers

Final answer:

The assumptions we can reasonably make in this scenario include the highway being straight and level, corresponding to options C and D. Assuming constant acceleration or velocity for each stage of the car's motion, options A and B, is not necessarily accurate.

Explanation:

In analyzing the motion of a car in different stages, we can make reasonable assumptions to simplify the problem. For each stage, assuming the car is moving with constant acceleration (option a) or velocity (option b) is not necessarily accurate because acceleration and velocity may change due to various factors like interaction with driver, road conditions, or appearance of a police car. The assumptions more likely to hold are that the highway is straight (option c), meaning there are no curves, and the highway is level (option d), indicating no hills or valleys. Thus, the correct answers would be C and D.

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(a) What is the escape speed on a spherical asteroid whose radius is 545 km and whose gravitational acceleration at the surface is 2.9 m/s2?

Answers

Answer:

1777.92 m/s

Explanation:

R = Radius of asteroid = 545 km

M = Mass of planet

g = Acceleration due to gravity = 2.9 m/s²

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Acceleration due to gravity is given by

[tex]g=\dfrac{GM}{R^2}\\\Rightarrow M=\dfrac{gR^2}{G}[/tex]

The expression of escape velocity is given by

[tex]v=\sqrt{\dfrac{2GM}{R}}\\\Rightarrow v=\sqrt{\dfrac{2G}{R}\dfrac{gR^2}{G}}\\\Rightarrow v=\sqrt{2gR}\\\Rightarrow v=\sqrt{2\times 2.9\times 545000}\\\Rightarrow v=1777.92\ m/s[/tex]

The escape speed is 1777.92 m/s

Answer:

1.78 km/s

Explanation:

radius, R = 545 km = 545000 m

acceleration due to gravity, g = 2.9 m/s²

The formula for the escape velocity is given by

[tex]v=\sqrt{2gR}[/tex]

[tex]v=\sqrt{2\times 545000\times 2.9}[/tex]

v = 1777.92 m/s

v = 1.78 km/s

Thus, the escape velocity on the surface of asteroid is 1.78 km/s.

Two parallel square metal plates that are 1.5 cm and 22 cm on each side carry equal but opposite charges uniformly spread out over their facing surfaces. How many excess electrons are on the negative surface if the electric field between the plates has a magnitude of 18,000 N/C? (k = 1/4 pi epsilon_0 = 9.0 times 10^9 N m^2/C^2, e = 1.6 times 10^-19 C)

Answers

Answer:

The number of excess electrons are on the negative surface is [tex]4.80\times10^{10}\ electrons[/tex]

Explanation:

Given that,

Distance =1.5 cm

Side = 22 cm

Electric field = 18000 N/C

We need to calculate the capacitance in the metal plates

Using formula of capacitance

[tex]C=\dfrac{\epsilon_{0}A}{d}[/tex]

Put the value into the formula

[tex]C=\dfrac{8.85\times10^{-12}\times(22\times10^{-2})^2}{1.5\times10^{-2}}[/tex]

[tex]C=0.285\times10^{-10}\ F[/tex]

We need to calculate the potential

Using formula of potential

[tex]V=Ed[/tex]

Put the value into the formula

[tex]V=18000\times1.5\times10^{-2}\ V[/tex]

[tex]V=270\ V[/tex]

We need to calculate the charge

Using formula of charge

[tex]Q=CV[/tex]

Put the value into the formula

[tex]Q=0.285\times10^{-10}\times270[/tex]

[tex]Q=76.95\times10^{-10}\ C[/tex]

Here, the charge on both the positive and negative  plates

[tex]Q=+76.95\times10^{-10}\ C[/tex]

[tex]Q=-76.95\times10^{-10}\ C[/tex]

We need to calculate the number of excess electrons are on the negative surface

Using formula of number of electrons

[tex]n=\dfrac{q}{e}[/tex]

Put the value into the formula

[tex]n=\dfrac{76.95\times10^{-10}}{1.6\times10^{-19}}[/tex]

[tex]n=4.80\times10^{10}\ electrons[/tex]

Hence, The number of excess electrons are on the negative surface is [tex]4.80\times10^{10}\ electrons[/tex]

what is the distance on a map in centimeters between two features if they are 7.6 km apart on the ground and the map has a scale of 1. 125000

Answers

Answer:

6.08 cm

Explanation:

We are given that

Ratio =1:125000

Let x be the distance on a map between two features .

The distance between two features on ground=y=7.6 km

According to question

[tex]\frac{x}{y}=\frac{1}{125000}[/tex]

Substitute the values then we get

[tex]\frac{x}{7.6}=\frac{1}{125000}[/tex]

[tex]x=\frac{7.6}{125000}=0.0000608 km[/tex]

We know that

[tex]1km=100000 cm[/tex]

0.0000608 km=[tex]0.0000608\times 100000=6.08cm[/tex]

Hence, the distance between two features on the map=6.08 cm

Final answer:

The distance on the map between the two features is 6.08 centimeters.

Explanation:

To find the distance on a map in centimeters between two features if they are 7.6 km apart on the ground and the map has a scale of 1:125000, you can use the scale factor formula: Distance on Map = Distance on Ground / Scale

Plugging in the values: Distance on Map = 7.6 km / 125000 = 0.0000608 km

Since 1 km = 100000 centimeters, multiply by 100000 to convert km to cm:

Distance on Map = 0.0000608 km × 100000 cm/km = 6.08 cm

Therefore, the distance on the map between the two features is 6.08 centimeters.

An electron falls through a distance d in a uniform electric field of magnitude E. Thereafter, the direction of the field is reversed (keeping its magnitude the same) and now a proton falls through the same distance. Compare, using quantitative reasoning, the time of fall in each case. Contrast this situation with that of objects falling freely under gravity.

Answers

Answer:

Please see below as the answer is self-explanatory

Explanation:

Assuming that the electric field is pointing upward this will produce a downward force on the electron. Neglecting the effect of gravity, according to Newton's 2nd Law, the force on the electron due to the field, produces an acceleration, that can be found solving the following equation:

F = me*a = qe*E ⇒ a = qe*E / me

If the electric field is uniform, the acceleration that produces is constant, so, we can use the kinematic equation that relates displacement and acceleration with time:

x = v₀*t + 1/2*a*t² = v₀*t +1/2*(qe*E/me)*t²

Now, for a proton falling, if  the direction of the field  is reversed (pointing downward) it will accelerate the proton downward.

Using the  same reasoning as above, we get the value  of the  acceleration as follows:

F = mp*a = qp*E ⇒ a = qp*E / mp

The equation for displacement is just the following:

x = v₀*t + 1/2*a*t² = v₀*t +1/2*(qp*E/mp)*t²

We know that qe = qp = 1.6*10⁻¹⁹ coul, but mp = 1,836 me, so, for the same displacement, the time must be much less for the electron, that has an acceleration 1,836 times higher.

When both objects fall freely the same distance under  the sole influence of gravity, if the initial velocity is the same, the time must be the  same  also, as the fall time doesn't depend on the mass of the object.

Answer:

Explanation:

We know that the electric force can be expressed as: F=qE. According to Newton's Second Law of Motion, force can also be expressed as: F=ma. Therefore: a=F/m. We can substitute the electric force expression for "F" in this equation. We get: a=qE/m. We can see from this equation that acceleration is inversely proportional to mass and directly proportional to the electric field and charge. Since the electric field is being reversed and since the charges on the proton and electron differ only by the + or - sign respectively, the numerator of this fraction will remain constant in this scenario. The only variables that are effectively changing are the mass and the resultant acceleration. From the inverse relationship of acceleration and mass, we can say that the proton - having a significantly larger mass than the electron - should experience a smaller acceleration, and should thus take longer to fall distance "d." The electron, on the other hand, should experience a greater acceleration due to its significantly smaller mass, and should fall through distance "d" in a shorter amount of time.

Under the influence of gravity (on the surface of the Earth, for example), objects released from the same height should fall freely with the same acceleration at any given time, regardless of mass. It makes sense, however, that subatomic particles interacting with the electric field are hardly affected by gravity, given how weak gravitational forces are on the microscopic scale.

As additional resistors are connected in series to a constant voltage source, how is the power supplied by the source affected?

(A) The power supplied by the sources remains constant.
(B) The power supplied by the source increases.
(C) The effect on the power supplied by the source cannot be determined without knowing the voltage of the source.
(D) The power supplied by the source decreases.

Answers

Final answer:

As additional resistors are connected in series to a constant voltage source, the power supplied by the source decreases.

Explanation:

As additional resistors are connected in series to a constant voltage source, the power supplied by the source decreases.

This can be explained by Ohm's Law, which states that power is equal to the voltage squared divided by the resistance. When resistors are connected in series, the total resistance increases, which leads to a decrease in the power supplied by the source.

For example, if you connect two resistors in series to a constant voltage source, the total resistance would be the sum of the resistances of the two resistors. As a result, the power supplied by the source would decrease.

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Final answer:

Adding more resistors in series to a constant voltage source results in an increase in total resistance, which leads to a decrease in current. Because power is proportional to the square of the current, the power supplied by the voltage source decreases.

Explanation:

When additional resistors are connected in series to a constant voltage source, the power supplied by the source is affected in a specific way. Given that power (P) is calculated by P = I^2R for a given resistance (R) and current (I), and by P = V^2/R for a given voltage (V) and resistance, we can understand the impact of adding resistors in series.

In a series circuit, the current remains constant throughout the resistors, but the total resistance (sum of all individual resistances) increases as more resistors are added. Since the voltage is constant, an increase in the total resistance will result in a decrease in current according to Ohm's law (I = V/R). Consequently, if the current decreases, the power supplied by the source also decreases (P = I^2R), because the power is proportional to the square of the current flowing through the circuit. Therefore, the correct answer is (D) The power supplied by the source decreases.

Maximum current problem. If the current on your power supply exceeds 500 mA it can damage the supply. Suppose the supply is set for 37 V. What is the smallest resistance you can measure?

Answers

To solve this problem we will apply Ohm's law. The law establishes that the potential difference V that we apply between the ends of a given conductor is proportional to the intensity of the current I flowing through the said conductor. Ohm completed the law by introducing the notion of electrical resistance R. Mathematically it can be described as

[tex]V = IR \rightarrow R = \frac{V}{I}[/tex]

Our values are

[tex]I = 500mA = 0.5A[/tex]

[tex]V = 37V[/tex]

Replacing,

[tex]R = \frac{V}{I}[/tex]

[tex]R = \frac{37}{0.5}[/tex]

[tex]R = 74 \Omega[/tex]

Therefore the smallest resistance you can measure is [tex]74 \Omega[/tex]

(I) The springs of a 1700-kg car compress 5.0 mm when its 66-kg driver gets into the driver's seat If the car goes over abump, what will be the frequency of oscillations? Ignore damping.

Answers

Answer:

[tex]f=7 \ Hz[/tex]

Explanation:

Simple Harmonic Motion

The spring-mass system is a typical case of a simple harmonic motion, since the distance traveled by the mass describes an oscillatory behaviour. The natural angular frequency of a spring-mass system is computed by

[tex]{\displaystyle \omega ={\sqrt {\frac {k}{m}}}}[/tex]

And the frequency is

[tex]{\displaystyle f=\frac {w}{2\pi}[/tex]

Thus

[tex]{\displaystyle f =\frac {1}{2\pi}{\sqrt {\frac {k}{m}}}}[/tex]

The total mass of the car and the driver is

[tex]m=1700+66=1766\ kg[/tex]

They both weigh

[tex]W=m.g=1766\ kg*9.8\ m/s^2[/tex]

[tex]W=17306.8\ N[/tex]

We need to know the constant of the spring. It can be found by using the formula of the Hook's law:

[tex]F=k.x[/tex]

We know the spring stretches 5 mm (0.005 m) when holding the total weight of the car and the driver. Solving for k

[tex]\displaystyle k=\frac{F}{x}[/tex]

[tex]\displaystyle k=\frac{17306.8}{0.005}[/tex]

[tex]k=3,461,360\ N/m[/tex]

Thus, the frequency of oscillations is

[tex]{\displaystyle f =\frac {1}{2\pi}{\sqrt {\frac {3,461,360}{1,766}}}}[/tex]

[tex]\boxed{f=7 \ Hz}[/tex]

Final answer:

The frequency of oscillation of a car going over a bump can be calculated using principles of simple harmonic motion. By calculating the spring constant from the car's weight and the amount the springs are compressed, we can find the angular frequency and then the normal frequency of the oscillation.

Explanation:

The question is about understanding the oscillation concepts of a car once it goes over a bump. It can be treated as a simple harmonic oscillator. To calculate the frequency, we need to first find the spring constant (k) based on Hooke's Law which states that the force F on the spring exerted by the car's weight equals k multiplied by the distance the spring is compressed (x), or F = kx.

In this case, the force F is the weight of the car plus the driver: (1700kg + 66kg) * 9.81m/s² gravitational acceleration. The spring compression x is 5.0mm, or 0.005m. Solving F = kx for k gives us the spring constant.

The angular frequency w of oscillation is equal to √(k/m), where m is the total mass: 1700 kg (car mass) + 66 kg (driver mass), and k is the spring constant we found earlier. The frequency (f) can then be found by dividing w by 2π.

This frequency measures the number of oscillations the car makes per second after going over a bump.

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An 8900-pF capacitor holds plus and minus charges of 1.85×10−7 C . Part A What is the voltage across the capacitor?

Answers

Final answer:

The voltage across the capacitor is 20.79 V.

Explanation:

Capacitance is a measure of how much charge can be stored in a capacitor per unit potential difference. The voltage across the capacitor can be calculated using the formula V = Q/C, where V is the voltage, Q is the charge, and C is the capacitance. In this case, the charge is 1.85×10-7 C and the capacitance is 8900 pF. Converting the capacitance to farads, we get C = 8900 × 10-12 F. Plugging in these values, we have V = 1.85×10-7 C / (8900 × 10-12 F), which simplifies to V = 20.79 V.

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During a car collision, the knee, thighbone, and hip can sustain a force no greater than 4000 N/ Forces that exceed this amount could cause dislocations or fractures. Assume that in a collision a knee stops when it hits the car's dashboard. Also assume that the mass of the body parts stopped by the knee is about 20% of the total body mass. a) What is the minimum stopping time interval in needed to avoid injury to the knee if the person is initially traveling at 15 m/s (34 mi/h)? b) what is the minimum stopping distance?

Answers

Complete Question:

During a car collision, the knee, thighbone, and hip can sustain a force no greater than 4000 N/ Forces that exceed this amount could cause dislocations or fractures. Assume that in a collision a knee stops when it hits the car's dashboard. Also assume that the mass of the body parts stopped by the knee is about 20% of the total body mass, which is 62 kg.

a) What is the minimum stopping time interval in needed to avoid injury to the knee if the person is initially traveling at 15 m/s (34 mi/h)? b) what is the minimum stopping distance?

Answer:

a) t= 46.5 msec. b) 0.35 m

Explanation:

Applying Newton´s 2nd Law to the mass supported by the knee (20% of the total mass), we can get the maximum acceleration allowable in order to avoid an injury, as follows:

a = F/m = 4000 N / 0.2*62 kg = 322.6 m/s²  

Applying the definition of acceleration, and taking into account that the knee finally come to an stop, we have:

a = vf - v₀ / Δt = -15 m/s / Δt

Solving for Δt :

Δt = -15 m/s / -322.6 m/s² = 0.0465 sec = 46.5 msec.

b) Assuming the acceleration remains constant during this time interval, we can find the distance needed to come to an ⇒stop, applying any of the kinematic equations, as this one:

vf² - v₀² = 2*a*Δx

⇒Δx = (vf²-v₀²) / 2*a

⇒Δx = -(15 m/s)² / 2*(-322.6 m/s²) = 0.35 m

This question involves the concepts of the equations of motion and Newton's Second Law of Motion.

a) The minimum stopping time interval needed to avoid knee injury is "0.05 s".

b) The minimum stopping distance is "0.37 m".

a)

First, we will use Newton's Second Law of Motion to find out the acceleration:

[tex]F=ma\\\\a=\frac{F}{m}[/tex]

where,

a = acceleration = ?

m = mass supported by knee = 20% of total mass = (0.2)(65 kg) = 13 kg

Assuming the average mass of a person to be 65 kg.

Force = - 4000 N (reaction force)

Therefore,

[tex]a=\frac{-4000\ N}{13\ kg}\\\\a=-307.7\ m/s^2[/tex] (negative sign shows decelration)

Now, for the minimum time interval, we will use the first equation of motion:

[tex]v_f=v_i+at\\\\t=\frac{v_f-v_i}{a}[/tex]

where,

t = time interval = ?

vf = final speed = 0 m/s

vi = initial speed = 15 m/s

Therefore,

[tex]t=\frac{0\ m/s-15\ m/s}{-307.7\ m/s^2}[/tex]

t = 0.05 s = 50 ms

b)

Now, we will use the second equation of motion to find out the stopping distance:

[tex]s=v_it+\frac{1}{2}at^2\\s=(15\ m/s)(0.05\ s)+\frac{1}{2}(-307.7\ m/s^2)(0.05\ s)^2\\[/tex]

s = 0.75 m - 0.38 m

s = 0.37 m

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The attached picture shows the equations of motion.

A 90 kg man lying on a surface of negligible friction shoves a 69 g stone away from himself, giving it a speed of 4.0 m/s.
What speed does the man acquire as a result?

Answers

Answer:

v = 3.06 x 10⁻³ m/s

Explanation:

given,

mass of man,M = 90 kg

mass of stone , m= 69 g = 0.069 Kg

speed of the stone, u = 4 m/s

speed of the man, v = ?

using conservation of momentum

initial velocity of the man and stone is equal to zero

(M + m)V = M v + m u

(M + m) x 0 = 90 x v + 0.069 x 4

90 v = 0.276

 v = 0.00306 m/s

v = 3.06 x 10⁻³ m/s

speed of the man on the frictionless  surface is equal to 3.06 x 10⁻³ m/s

Final answer:

The man's velocity after shoving the stone is -0.003067 m/s, indicating movement in the direction opposite to the stone, as per the conservation of momentum.

Explanation:

This problem is a classic example of the conservation of momentum, which is a fundamental concept in physics. According to the law of conservation of momentum, if no external forces are acting on a system, the total momentum of the system remains constant. In this scenario, the man and the stone together constitute an isolated system because the surface is frictionless and there are no external forces acting on the system.

The total momentum before the man shoves the stone is zero since both are initially at rest. When the man shoves the stone, the stone acquires momentum in one direction, and the man must acquire momentum in the opposite direction to conserve the total momentum of the system. The magnitude of the momentum gained by the stone and the man will be equal because their initial total momentum was zero.

The man's velocity (v) can be calculated using the formula:

Momentum of stone = mass of stone (m_stone) * velocity of stone (v_stone)

Momentum of man = mass of man (m_man) * velocity of man (v)


Since total momentum is conserved, Momentum of stone = -Momentum of man.

We can then solve for the man's velocity:

(m_stone * v_stone) = -(m_man * v)


To find v, we rearrange the equation:

v = - (m_stone * v_stone) / m_man

Plugging in the given values:

v = - (0.069 kg * 4.0 m/s) / 90 kg

v = - (0.276 kg·m/s) / 90 kg
v = -0.003067 m/s

The negative sign indicates that the man's velocity is in the opposite direction of the stone's velocity.

Two carts are colliding on an airtrack (neglect friction). The first cart has a mass of m1=40 g and an initial velocity of v1=2 m/s. The second cart has a mass of m2=47 g and an initial velocity of v2=-5 m/s. Two experiments are conducted. In the first experiment, the first cart has a final velocity of v1'=-1.11 m/s.
What is the velocity of the second cart?

For the second experiment, the bumpers of the carts are modified, but the carts are started with the same initial velocities as before. Now the first cart has a final velocity of v1'=-3.8942 m/s, and the second cart a final velocity of v2'=0.0163 m/s.
How much (if any) energy was lost in the collision?

In good approximation, what kind of collision was the second experiment?

Answers

Answer:

v₂ = -2.35m/s

This is an Inelastic collision

Explanation:

Law of conservation of momentum

This states that for a collision occurring between two object in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

Given from the question

m₁=40 g

initial velocity of m₁: v₁=2 m/s

m₂=47 g

initial velocity of m₂: v₂ = -5 m/s

final velocity of m₁: v'₁ = -1.11 m/s.

final velocity of  m₂ = ?

second experiment final velocity of m₁ : v₁' = -3.8942 m/s

second experiment final velocity of m₂ : v₂' = 0.0163 m/s

Considering the first experiment

Apply the knowledge of conservation of momentum

[tex] m₁v₁ + m₂v₂ =m₁v₁ + m₂v₂[/tex]

[tex](0.040)(2) + (0.047)(-5) = (0.040)(-1.11) + (0.040)v₂[/tex]

[tex]v₂ = -2.35m/s[/tex]

Considering the second experiment

initial kinetic energy [tex]KEₓ  = (1/2)m₁v₁² + (1/2)m₂v₂²[/tex]

     [tex]KEₓ = (1/2)(0.040)(2)² + (1/2)(0.047)(-5)²[/tex]

     [tex]KEₓ = 0.6675 J[/tex]

Final kinetic energy[tex] KEₙ = (1/2)m₁v₁² + (1/2)m₂v₂² [/tex]

[tex]KEₙ = (1/2) (0.040) (-3.8942)² + (1/2) (0.047) (0.0163)² [/tex]

[tex]KEₙ = 0.3033 J  [/tex]

Loss in kinetic energy  [tex]ΔKE = KEₓ - KEₙ [/tex]

ΔKE = 0.6675 - 0.3033

ΔKE = 0.3642 J

This collision is a perfectly inelastic collision because the maximum  kinetic energy is lost this means that the kinetic energy before the collision is not the same as the kinetic energy after the collision.Though momentum is conserved kinetic energy is not conserved.

A magnetic field of magnitude 1.30x10-3 T is measured a distance of 0.03 m from a long straight wire. What is the current through the wire?

Answers

Final answer:

The current through the wire is 1.95 A.

Explanation:

To find the current through the wire, we can use Ampere's law. Ampere's law states that the magnetic field around a long straight wire is directly proportional to the current through the wire and inversely proportional to the distance from the wire.

So, we can use the equation B = μ0 * I / (2π * r), where B is the magnetic field, μ0 is the magnetic constant, I is the current, and r is the distance from the wire.

Plugging in the given values, we have 1.30x10-3 T = (4πx10-7 T*m/A) * I / (2π * 0.03 m). Solving for I, we get I = 1.30x10-3 * (2*0.03)/(4x10-7) = 1.95 A.

A helicopter is flying horizontally with a speed of 339 m/s over a hill that slopes upward with a 2% grade. What is the component of the helicopter's velocity perpendicular to the sloping surface of the hill?

Answers

We will use the trigonometric ratios to know the helicopter's deceleration speed. Later applying the concept of speed as a vector component of the value found we will find the vertical speed.

The 2% grade indicates that for every 100 meters traveled in the x direction, there is an ascent / descent of 2 meters. Therefore we will have the relationship

[tex]\rightarrow \frac{y}{x} = \frac{2}{100}[/tex]

Now you would have the value o[tex]v_y = 6.78m/s[/tex]f the angle tangent would be

[tex]tan \theta = \frac{y}{x}[/tex]

[tex]tan \theta = \frac{2}{100}{[/tex]

[tex]tan\theta = 0.02[/tex]

From this relationship we could conclude that the vertical speed would be

[tex]v_y = v*tan\theta[/tex]

[tex]v_y = 339*0.02[/tex]

[tex]v_y = 6.78m/s[/tex]

Therefore the component of the helicopter's velocity perpendicular to the sloping surface of the hill is 6.78m/s

Final answer:

The component of the helicopter's velocity perpendicular to the sloping surface of the hill is approximately 6.78 m/s.

Explanation:

In this scenario, we can analyze the helicopter's velocity vector into two components: the parallel component along the sloping surface of the hill and the perpendicular component. The component of the velocity perpendicular to the sloping surface is equal to the total velocity multiplied by the grade percentage as a decimal.

Given that the helicopter's speed is 339 m/s and the hill has a 2% grade, we can calculate the perpendicular component as follows:

Perpendicular component = Total velocity × Grade percentage
Perpendicular component = 339 m/s × 0.02
Perpendicular component ≈ 6.78 m/s

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The position of a particle is given by the function x=(4t3−6t2+12)m, where t is in s.
A.) at what time does the particle reach its minimum velocity
B.) what is (vx)min
C.) at what time is the acceleration zero

Answers

Answer

given,

x = 4 t³ - 6 t² + 12

velocity, [tex]v = \dfrac{dx}{dt}[/tex]

[tex]\dfrac{dx}{dt}=\dfrac{d}{dt}(4t^3-6t^2+12)[/tex]

[tex]v =12t^2-12t[/tex]

For minimum velocity calculation we have differentiate it and put it equal to zero.

[tex]\dfrac{dv}{dt} =\dfrac{d}{dt}12t^2-12t[/tex]

[tex]\dfrac{dv}{dt} =24t-12[/tex].........(1)

putting it equal to zero

24 t - 12 =0

t = 0.5 s

At t = 0.5 s velocity will be minimum.

b) minimum velocity

    v = 12t² -12 t

    v = 12 x 0.5² -12 x 0.5

    v = -3 m/s

c) derivative of velocity w.r.t. time is acceleration

from equation 1

     a = 24 t - 12

time at which acceleration will be zero

     0 = 24 t - 12

     t = 0.5 s

At t = 0.5 s acceleration will be zero.

Part A. The particle reaches its minimum velocity at 0.5 seconds.

Part B. The minimum velocity of the particle is -3 m/s.

Part C. The acceleration of the particle will be zero at the time t = 0.5 seconds.

How do you calculate the minimum velocity and acceleration?

Given that the position of a particle is given by the function x.

[tex]f(x) = 4t^2 -6t^2 +12[/tex]

The function of the velocity of a particle can be obtained by the time function.

[tex]v= \dfrac {dx}{dt}[/tex]

[tex]v = \dfrac {d}{dt} ( 4t^3-6t^2 +12)[/tex]

[tex]v = 12t^2 -12 t[/tex]

The velocity function of the particle is [tex]v = 12t^2 - 12t[/tex].

Part A

The minimum velocity of the particle is obtained by the differentiation of velocity function with respect to the time and put it equal to zero.

[tex]\dfrac {dv}{dt} = \dfrac {d}{dt} (12t^2 - 12t) = 0[/tex]

[tex]\dfrac {dv}{dt} = 24 t-12 = 0[/tex]

[tex]t = 0.5\;\rm s[/tex]

Hence we can conclude that the particle reaches its minimum velocity at 0.5 seconds.

Part B

The velocity function is [tex]v = 12t^2 - 12t[/tex]. Substituting the value of t = 0.5 s to calculate the minimum velocity.

[tex]v = 12(0.5)^2 - 12(0.5)[/tex]

[tex]v = 3 - 6[/tex]

[tex]v = -3 \;\rm m/s[/tex]

The minimum velocity of the particle is -3 m/s.

Part C

The acceleration is defined as the change in the velocity with respect to time. Hence,

[tex]a = \dfrac {dv}{dt}[/tex]

[tex]a = 24 t-12[/tex]

Substituting the value of a = 0, we get the time.

[tex]0 = 24t - 12[/tex]

[tex]t = 0.5 \;\rm s[/tex]

The acceleration of the particle will be zero at the time t = 0.5 seconds.

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Why isn't Coulomb's law valid for dielectric objects, even if they are spherically symmetrical?

Answers

Answer:

Explanation:

The "traditional" form of Coulomb's law, explicitly the force between two point charges. To establish a similar relationship, you can use the integral form for a continuous charge distribution and calculate the field strength at a given point.

In the case of moving charges, we are in presence of a current, which generates magnetic effects that in turn exert force on moving charges, therefore, no longer can consider only the electrostatic force.

g what are the bands of electromagneitc radiation and how are they related to energy frequency and wavelength

Answers

Answer:

The details of bands is given in explanation.

Explanation:

The electromagnetic waves are differentiated into different bands based upon their wavelengths and frequencies. The names of different bands are as follows:

1. Radio Waves

2. Micro Waves

3. Infra-red

4. Visible light

5. Ultra Violet

6. X-rays

7. Gamma Rays

The frequency of every region or rays increases from 1 through 7. The energy of rays also increase from 1 through 7. Since, the wave length is inversely related to energy and frequency, thus the wavelength of rays decrease from 1 through 7.

A detailed information of the bands is provided in the picture attached.

12 of 15 Constants The predominant frequency of a certain fire truck's siren is 1500 Hz when at rest.

Answers

Answer:

The detect frequency is 1622.72 Hz.

Explanation:

Given that,

Frequency = 1500 Hz

Suppose you move with a speed of 27.0 m/s toward the fire engine. what frequency do you detect ?

We need to calculate the frequency

Using formula of frequency

[tex]f=(\dfrac{v+v_{0}}{v})f_{0}[/tex]

Where, v = speed of sound

v₀ = speed of source

f₀ = frequency of siren

Put the value into the formula

[tex]f=\dfrac{330+27.0}{330}\times1500[/tex]

[tex]f=1622.72\ Hz[/tex]

Hence, The detect frequency is 1622.72 Hz.

How many times a minute does a boat bob up and down on ocean waves that have a wavelength of 40.0 m and a propagation speed of 5.00 m/s?

Answers

Answer:

n = 7.5 times/minute

Explanation:

Given that,

Wavelength of the ocean wave, [tex]\lambda=40\ m[/tex]

The speed of the ocean wave, v = 5 m/s

To find,

Number of times a minute does a boat bob up and down on ocean waves.

Solution,

The relation between the speed of wave, wavelength and frequency is given by :

[tex]v=f\times \lambda[/tex]

[tex]f=\dfrac{v}{\lambda}[/tex]

[tex]f=\dfrac{5\ m/s}{40\ m}[/tex]

f = 0.125 Hz

The number of times per minute the bob moves up and down is given by :

[tex]n=f\times t[/tex]

[tex]n=0.125\times 60[/tex]

n = 7.5 times/minute

So, its will move up and down in 7.5 times/minute. Therefore, this is the required solution.

Taking into account the definition of wavelength, frecuency and propagation speed, the number of times per minute the bob moves up and down is 7.5 times per minute.

Wavelength

Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).

Frequency

Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).

Propagation speed

Finally, the propagation speed is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave disturbance propagates along its displacement.

The propagation speed relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation:

v = f× λ

Amount of times in a minute that a boat bob up and down on ocean waves

In this case, you know:

v= 5 [tex]\frac{m}{s^{2} }[/tex]f= ?λ= 40 m

Replacing in the definition of propagation speed:

5 [tex]\frac{m}{s^{2} }[/tex]= f× 40 m

Solving:

f= 5 [tex]\frac{m}{s^{2} }[/tex]÷ 40 m

f= 0.125 Hz= 0.125 [tex]\frac{1}{seconds}[/tex]

Then, a boat bob up and down on ocean waves 0.125 times in a second.

So, the number of times per minute the bob moves up and down is given by:

n= f× time

n= 0.125 Hz× 60 minutes in 1 second

n=7.5 times per minute

Finally, the number of times per minute the bob moves up and down is 7.5 times per minute.

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A roller coaster car crosses the top of a circular loop-the-loop at twice the critical speed. Part A What is the ratio of the normal force to the gravitational force? What is the ratio of the normal force to the gravitational force? n/FG=2n/FG=2 n/FG=3n/FG=3 n/FG=4n/FG=4 n/FG=5n/FG=5

Answers

Answer:

n/(FG) = 3.

Explanation:

At the top of the loop-the-loop, the normal force is directed downwards as well as the weight of the car. So, the total net force of the car is

[tex]F_{net} = N + mg[/tex]

By Newton's Second Law, this force is equal to the centripetal force, because the car is making circular motion in the loop.

[tex]F_{net} = ma = \frac{mv^2}{R}\\N + mg = \frac{mv^2}{R}[/tex]

The critical speed is the minimum speed at which the car does not fall. So, at the critical speed the normal force is zero.

[tex]0 + mg = \frac{mv_c^2}{R}\\v_c = \sqrt{gR}[/tex]

If the car is moving twice the critical speed, then

[tex]N + mg = \frac{m(2v_c)^2}{R} = \frac{m4gR}{R} = 4mg\\N = 3mg[/tex]

Finally, the ratio of the normal force to the gravitational force is

[tex]\frac{3mg}{mg} = 3[/tex]

While floating in space a 100.-kg robot throws a 0.800-kg wrench at 12.0 m/s toward its human partner working on the spaceship. How fast will the robot recoil?

Answers

Answer:

   v = - 0.096 m/s

Explanation:

given,

mass of the robot, M = 100 Kg

mass of wrench, m = 0.8 Kg

speed of the wrench,v' = 12 m/s

recoil of the robot,v = ?

initial speed of the robot and the wrench is equal to zero

using conservation of motion

(M + m) V = M v + m v'

(M + m) x 0 = 100 x v + 0.8 x 12

100 v = -9.6

   v = - 0.096 m/s

negative sign shows that velocity of robot is in opposite direction of wrench.

hence, the recoil velocity of the robot is equal to 0.096 m/s

An electric current of 202.0 mA Transports 56.0 C of charge. Calculate the time this took.

Answers

Answer:

277 s

Explanation:

Given data

Electric current (I): 202.0 mA = 202.0 × 10⁻³ A = 0.2020 A = 0.2020 C/sTransported charge (C): 56.0 CoulombsElapsed time (t): to be determined

We can find the elapsed time using the following expression.

I = C/t

t = C/I

t = 56.0 C/(0.2020 C/s)

t = 277 s

It took 277 seconds.

A sailor drops a wrench from the top of a sailboat's vertical mast while the boat is moving rapidly and steadily straight forward. Where will the wrench hit the deck?

(A) ahead of the base of the mast
(B) at the base of the mast
(C)behind the base of the mast
(D)on the windward side of the base of the mast
(E)None of the above choices

Answers

Answer:

B

Explanation:

The sailor, the boat and the wrench are all moving at he same constant rate, so the wrench will appear to fall straight down. This due to that fact there is no relative motion among them and all are at rest w.r.t to one another. Hence the correct answer would be B.

The correct option is (B). The wrench will fall straight down relative to the moving boat and will land directly at the base of the mast.

To determine where the wrench will hit the deck, let's analyze the motion of the wrench.

1. Horizontal Motion:

When the wrench is dropped from the top of the mast, it has the same horizontal velocity as the sailboat. This is because, in the absence of air resistance and assuming no external horizontal forces act on the wrench, it retains the horizontal component of its velocity that it had while it was still in the sailor's hand. Therefore, as the wrench falls, it will continue to move forward with the same horizontal velocity as the boat.

2. Vertical Motion*: The wrench will accelerate downwards due to gravity.

Since the wrench maintains its horizontal velocity and is only influenced vertically by gravity, it will fall straight down from the perspective of someone moving with the boat. From the perspective of an observer on the boat, the wrench will fall directly downwards.

Starting fom rest, a car accelerates at a constant rate, reaching 88 km/h in 12 s. (a) What is its acceleration? (b) How far does it go in this time?

Answers

Answer:

(A)  [tex]a=2.0.37m/sec^2[/tex]

(B) s = 146.664 m

Explanation:

We have given car starts from the rest so initial velocity u = 0 m /sec

Final velocity v = 88 km/hr

We know that 1 km = 1000 m

And 1 hour = 3600 sec

So [tex]88km/hr=88\times \frac{1000}{3600}=24.444m/sec[/tex]

Time is given t = 12 sec

(A) From first equation of motion v = u+at

So [tex]24.444=0+a\times 12[/tex]

[tex]a=2.0.37m/sec^2[/tex]

So acceleration of the car will be [tex]a=2.0.37m/sec^2[/tex]

(b) From third equation of motion [tex]v^2=u^2+2as[/tex]

So [tex]24.444^2=0^2+2\times 2.037\times s[/tex]

s = 146.664 m

Distance traveled by the car in this interval will be 146.664 m

A particle's position along the x-axis is described by. x(t)= At+Bt^2where t is In seconds: x is in meters: and the constants A and B are given below.Randomized Variables A= -3.5 m/s B= 3.9 m/s^2 a. What is the velocity, in meters per second. of the particle at the time t1= 3.0 s? b. What is the velocity, in meters per second: of the particle when it is at the origm (x=0) at time to> 0?

Answers

Answer

given,

position of particle

x(t)= A t + B t²

A = -3.5 m/s

B = 3.9 m/s²

t = 3 s

a)  x(t)= -3.5 t + 3.9 t²

   velocity of the particle is equal to the differentiation of position w.r.t. time.

[tex]\dfrac{dx}{dt}=\dfrac{d}{dt}(-3.5t + 3.9t^2)[/tex]

[tex]v= -3.5 + 7.8 t [/tex]------(1)

velocity of the particle at t = 3 s

  v = -3.5 + 7.8 x 3

 v = 19.9 m/s

b) velocity of the particle at origin

  time at which particle is at origin

  x(t)= -3.5 t + 3.9 t²

   0 = t (-3.5  + 3.9 t )

   t = 0, [tex]t=\dfrac{3.5}{3.9}[/tex]

   t = 0 , 0.897 s

speed of the particle at t = 0.897 s

from equation (1)

 v = -3.9 + 7.8 t

 v = -3.9 + 7.8 x 0.897

  v = 3.1 m/s

To solve the problem we should know about velocity.

Velocity

Velocity is the rate of change of its position with respect to time.

[tex]V = \dfrac{dy}{dt}[/tex]

Given to us

x(t)= At+Bt^2A= -3.5 m/s B= 3.9 m/s^2

Velocity of Particle

x(t)= At+Bt²

[tex]V(t) = \dfrac{dy}{dt} = \dfrac{d(At+Bt^2)}{dt} = A+2Bt[/tex]

A.)  the velocity, in meters per second. of the particle at the time t1= 3.0 s,

Velocity of particle(t = 3.0 s)

[tex]V(t) = A +2Bt[/tex]

Substituting the values,

[tex]V(t_1=3) = (-3.5) +2(3.9)(3.0)\\\\V(t_1=3) = 19.9\ m/s[/tex]

B.)  the velocity, in meters per second: of the particle when it is at the origin (x=0) at t ≥ 0

Displacement, x = 0

[tex]x(t)= At+Bt^2\\\\0 = At+Bt^2\\\\[/tex]

Taking t as common,

[tex]0 = t(A+Bt)\\\\[/tex]

[tex]0 = (A+Bt)\\\\[/tex]

Substituting the values and solving or t,

[tex]0 = A+ Bt\\0 = -3.5 + (3.9)t\\3.5=3.9t\\t= \dfrac{3.5}{3.9}\\\\t = 0.8974\ s[/tex]

Velocity of particle(t = 0.8974 s)

Substituting the value in the formula of velocity,

[tex]V(t) = A +2Bt[/tex]

Substituting the values,

[tex]V(t_1=0.8974) = (-3.5) +2(3.9)(0.8974)\\\\V(t_1=3) = 3.5 \ m/s[/tex]

Hence, the velocity of the particle is 3.5 m/s.

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A 0.105-g sample of X2 contains 8.92x1020 molecules. The compound MX3 consists of 54.47% X by mass. What are the identities of M and X, and what is the correct name for MX3?

Answers

Answer:

The element X is chlorine.

The element M is Ytterium.

Ytterium (III) chloride is the correct name for [tex]YCl_3[/tex]

Explanation:

Molecules of [tex]X_2, N = 8.92\times 10^{20} [/tex]

Moles of [tex]X_2, n= ?[/tex]

[tex]N=n\times N_A[/tex]

[tex]n=\frac{N}{N_A}=\frac{8.92\times 10^{20}}{6.022\times 10^{23} mol^{-1}}[/tex]

n = 0.001481 mol

Mass of [tex]X_2, m=0.105 g[/tex]

Molar mass of [tex]X_2=M[/tex]

[tex]m=M\times n[/tex]

[tex]M=\frac{m}{n}=\frac{0.105 g}{0.001481 mol}[/tex]

M = 70.87 g/mol

Atomic mass of X = [tex]\frac{70.87 g/mol}{2}=35.435 g/mol[/tex]

X is a chlorine atom.

The compound [tex]MX_3[/tex] consists of 54.47% X by mass.

M' = Molar mass of compound [tex]MX_3[/tex]

Percentage of X in compound = 54.47%

[tex]54.47\%=\frac{3\times 35.435 g/mol}{M'}\times 100[/tex]

M' = 195.16 g/mol

Atomic mass of M = a

[tex]195.16 g/mol=a+3\times 35.435 g/mol[/tex]

[tex]a = 195.16 g/mol-3\times 35.435 g/mol[/tex]

a = 88.86 g./mol

The element M is Ytterium.

Ytterium (III) chloride is the correct name for [tex]YCl_3[/tex]

The element X is Chlorine and the element M is Ytterium.

Ytterium Chloride is the correct name for [tex]MX_3[/tex] that is [tex]YCl_3[/tex]

Number of molecules of [tex]X_2={8.92*10^{20}[/tex]

Number of moles of X:

[tex]n=\frac{8.92*10^{20}}{6.023*10^{23}}[/tex]  

n = 0.001481 mol = 0.105g

Molar mass of [tex]X_2[/tex] molecule

[tex]M= 0.105/0.001481\\\\M=70.87g/mol[/tex]

Therefore molar mass of X atom = molar mass of [tex]X_2[/tex]/2

m = 35.435g/mol

X is a chlorine atom.

Given that the compound  consists of 54.47% X by mass.

Let M' = Molar mass of compound [tex]MX_3[/tex]

Percentage of X in compound = 54.47%

54.47 = {amount of X}/{mola mass of compound}*100

[tex]54.47 = \frac{3*35.435}{M_{'} }*100\\\\M^{'}= 195.16g/mol[/tex]

let Atomic mass of M = A

A = 195.16 - 3 × 35.435

A = 88.86 g/mol

The element M is Ytterium.

Therefore the compound [tex]MX_3[/tex] is Ytterium Chloride [tex]YCl_3[/tex]

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Suppose the B string on a guitar is 24" long and vibrates with a frequency of about 247 Hz. You place your finger on the 5th fret, which changes the length of the string to 18". Which note do you hear when you play the string?
a. D (about 294 Hz)
b. E (about 329 Hz)
c. G (about 392 Hz)

Answers

Answer:

b. E (about 329 Hz)

Explanation:

Given data:

Initial length of the string l1= 24 in

initial frequency f1= 247 Hz

changed length l2= 18 in

Then we have to find the changed frequency f2= ?

We already now that

frequency f ∝ 1/length of the string l

therefore,

[tex]\frac{f_1}{f_2} =\frac{l_1}{l_2}[/tex]

⇒[tex]{f_2}=\frac{l_1}{l_2}\times{f_1}[/tex]

⇒[tex]{f_2}=\frac{24}{18}\times{247}[/tex]

⇒[tex]{f_2}=329.33 Hz[/tex]

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