Multiple Concept Example 9 deals with the concepts that are important in this problem. A grasshopper makes four jumps. The displacement vectors are (1) 40.0 cm, due west; (2) 26.0 cm, 32.0 ° south of west; (3) 19.0 cm, 50.0 ° south of east; and (4) 18.0 cm, 60.0 ° north of east. Find (a) the magnitude and (b) direction of the resultant displacement. Express the direction as a positive angle with respect to due west.

Answer: ok, so total time is Tt = 6.65 seconds = time falling + time of sound traveling.

the time of the rock falling is given by:  [tex](g/2)*t₀^{2} -h = 0[/tex] (1)

where h is the heigth of the hole, and t₀ is the time it took to reach the bottom.

the time of the sound traveling is given by 343*t₁ - h = 0 (2)

so t₁ + t₀ = 6.65s = total time

then you know that t₁ = 6.65s -  t₀

so now you have two variables, t₀ and h, and we want to know the value of h, so we want to write t₀ as a function of h.

in the second equation we have now: 343*(6.65 -  t₀) = h

so  t₀ = (-h/343 + 6.65)s

replacing this on the first equation you have:

 [tex](g/2)*(-h/343 + 6.65)^{2} -h = 0[/tex]

now you want to take h to the right side so

w

so if we replace g by 9.8 meters over seconds square we get

h = [tex]\frac{-1.19 -+ \sqrt[2]{1.56} }{2*0.00041}[/tex]

where you will take the positive root

h = 61 meters

Answer:

19 degrees is the answer

Explanation:

Answer:

For diatomic oxygen:V=539.06 m/s

For carbon dia oxide:V=459.71 m/s

For dia atomic hydrogen:V=2156.25 m/s

Explanation:

As we know that

Root mean square velocity V

[tex]V=\sqrt{\dfrac{3RT}{M}}[/tex]

Where

R is the gas constant

[tex]R=8.31\ \frac{kg.m^2}{s^2.mol.K}[/tex]

T is the temperature (K).

M is the molecular weight.

For diatomic oxygen:

M=32 g/mol

T=273+100 = 373 K

[tex]R=8.31\ \frac{kg.m^2}{s^2.mol.K}[/tex]

[tex]V=\sqrt{\dfrac{3RT}{M}}[/tex]

[tex]V=\sqrt{\dfrac{3\times 8.31\times 373}{32\times 10^{-3}}}[/tex]

V=539.06 m/s

For carbon dia oxide

M=44 g/mol

T=273+100 = 373 K

[tex]R=8.31\ \frac{kg.m^2}{s^2.mol.K}[/tex]

[tex]V=\sqrt{\dfrac{3RT}{M}}[/tex]

[tex]V=\sqrt{\dfrac{3\times 8.31\times 373}{44\times 10^{-3}}}[/tex]

V=459.71 m/s

For dia atomic hydrogen:

M= 2 g/mol

T=273+100 = 373 K

[tex]R=8.31\ \frac{kg.m^2}{s^2.mol.K}[/tex]

[tex]V=\sqrt{\dfrac{3RT}{M}}[/tex]

[tex]V=\sqrt{\dfrac{3\times 8.31\times 373}{2\times 10^{-3}}}[/tex]

V=2156.25 m/s

Answer:

(a). The time is 26.67 sec.

(b). The distance traveled during this period is 1066.9 m.

Explanation:

Given that,

Speed = 80 m/s

Acceleration = 3 m/s

Initial velocity = 0

(a). We need to calculate the time

Using equation of motion

[tex]v = u+at[/tex]

[tex]t = \dfrac{v-u}{a}[/tex]

Put the value into the formula

[tex]t = \dfrac{80-0}{3}[/tex]

[tex]t =26.67\ sec[/tex]

The time is 26.67 sec.

(b). We need to calculate the distance traveled during this period

Using equation of motion

[tex]s = ut+\dfrac{1}{2}at^2[/tex]

[tex]s = 0+\dfrac{1}{2}\times3\times(26.67)^2[/tex]

[tex]s =1066.9\ m[/tex]

The distance traveled during this period is 1066.9 m.

Hence, This is the required solution.

Answer: 0.123 N/C

Explanation: In order to solve this question we have to use the  electric  Force on a particle produced by an electric field which is given by:

F=q*E

so E=F/q= -2,58* 10^-6/-2.1*10^-5= 0.123 N/C

Answer:

t=444.4s

Explanation:

m=1.5*10^7 kg

F=7.5*10^5 N

v=80km/h*(1h/3600s)*(1000m/1km)=22.22m/s

Second Newton's Law:

F=ma

a=F/m=7.5*10^5/(1.5*10^7)=0.05m/s^2

Kinematics equation:

vf=vo+at=at      

vo: initial velocity equal zero

t=vf/a=22.22/0.05=444.4s

To calculate the time it takes to increase the speed of a train from rest to 80 km/h, you can use Newton's second law of motion.

To calculate the time it takes to increase the speed of the train from rest to 80 km/h, we can use Newton's second law of motion. First, we need to calculate the force required to accelerate the train. Given the mass of the train, 1.5 x 10^7 kg, and the acceleration, we can use the formula: force = mass x acceleration

force = (1.5 x 10^7 kg) x (80 km/h to m/s conversion)

Next, we can use the formula: force = mass x acceleration to find the time it takes to accelerate the train:

time = force / (7.5 x 10^15 N)

Plugging in the values, we can calculate the time it takes to increase the speed of the train from rest to 80 km/h.

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Answer:

t=2.97h

t= 10692s

t= 178.2 min

Explanation:

We propose the following ratio:

[tex]\frac{t}{26.22mi} =\frac{1h}{8.83mi}[/tex]

[tex]t=\frac{1h*26.22mi}{8.883mi}[/tex]

t=2.97h

Equivalences

1h=3600s

1h=60 min

Calculation of t in minutes (min) and seconds(s):

t=2.97h*3600s/h= 10692s

t=2.97h*60min/h= 178.2 min

Answer:

The charge on sphere A and C is [tex]\frac{3q}{8} C[/tex]

Solution:

As per the question:

Charge on sphere A = q C

Now,

When sphere A with 'q' charge is brought in contact with B then conduction takes place and equal charges are distributed on both the spheres as:

[tex]\frac{q + q}{2} = \frac{q}{2}[/tex]

Charge on each A and B is [tex]\frac{q}{2}[/tex]

Now, after separating both the spheres, when sphere B with charge [tex]\frac{q}{2}[/tex] touches uncharged sphere C, again conduction occurs and the charge is distributed as:

[tex]\frac{\frac{q}{4} + \frac{q}{4}}{2} = \frac{q}{4}[/tex]

Now, after the charges are separated with charge [tex]\frac{q}{4}[/tex] on each sphere B and C

Now,

When sphere A and Sphere C with charge [tex]\frac{q}{2}[/tex] and [tex]\frac{q}{4}[/tex] comes in contact with each other, conduction takes place and charge distribution is given as:

[tex]\frac{\frac{q}{2} + \frac{q}{4}}{2} = \frac{3q}{8}[/tex]

Answer:

E(final)/E(initial)=2

Explanation:

Applying the law of gauss to two parallel plates with  charge density equal σ:

[tex]E=\sigma/\epsilon_{o}=Q/(L^{2}*\epsilon_{o})\\[/tex]

So, if the charge is doubled the Electric field is doubled too

E(final)/E(initial)=2

Answer:

The wavelength of the wave is 20 m.

Explanation:

Given that,

Amplitude = 10 cm

Radial frequency [tex]\omega = 20\pi\ rad/s[/tex]

Bulk modulus = 40 MPa

Density = 1000 kg/m³

We need to calculate the velocity of the wave in the medium

Using formula of velocity

[tex]v=\sqrt{\dfrac{k}{\rho}}[/tex]

Put the value into the formula

[tex]v=\sqrt{\dfrac{40\times10^{6}}{10^3}}[/tex]

[tex]v=200\ m/s[/tex]

We need to calculate the wavelength

Using formula of wavelength

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

[tex]\lambda=\dfrac{v\times2\pi}{\omega}[/tex]

Put the value into the formula

[tex]\lambda=\dfrac{200\times2\pi}{20\pi}[/tex]

[tex]\lambda=20\ m[/tex]

Hence, The wavelength of the wave is 20 m.

Answer:

[tex]f'=2176.256983Hz[/tex]

Explanation:

The relationship between observed frequency f' and the emitted frequency f is given by the doppler effect equation. In this case the observer and the source are moving in opposite direction away from each other, so:

[tex]f'=\frac{c-v_0}{c+v_s} f[/tex]

Where:

[tex]c=Speed-of-the-sound-waves=343m/s[/tex]

[tex]v_0=Velocity -of-observer=25m/s[/tex]

[tex]v_s=Velocity -of-source=15m/s[/tex]

[tex]f=Emitted -frequency=2450Hz[/tex]

[tex]f'=Observed-frequency[/tex]

Evaluating the data in the equation:

[tex]f'=\frac{343-25}{343+15}*2450=2176.256983Hz[/tex]

Answer:

1.5cm

Explanation:

The place where the bead will come to rest is the place where the force between the left charge and the bead is the same, but in opposite direction, as he force between the right charge and the bead.

[tex]F_{left} = F_{right}\\K\frac{q_{left}*q_b}{r_{left}^2} =K\frac{q_{right}*q_b}{r_{right}^2}[/tex]

Also:

[tex]r_{left}+r_{right}=0.045m\\r_{right} = 0.045m-r_{left}[/tex]

[tex]q_{left}= q\\q_{right}=4q[/tex]

So, we replace and simplify. Notice that the charges and the Coulomb constant will cancel:

[tex]K\frac{q_{left}*q_b}{r_{left}^2} =K\frac{q_{right}*q_b}{r_{right}^2}\\\frac{q*q}{r_{left}^2} = \frac{4q*q}{(0.045m-r_{left})^2}\\(0.045m-r_{left})^2 =4r_{left}^2\\0.002025m - 2*0.045r_{left} + r_{left}^2 = 4r_{left}^2\\3r_{left}^2 + 0.09r_{left} - 0.002025m = 0[/tex]

[tex]r_{left} = \frac{-(0.09)+-\sqrt{(0.09)^2-4*(3)(-0.002025)}}{2(3)}\\ r_{left} = 0.015m| -0.045m[/tex]

But the only solution that would place the bead on the wire is 0.015m or 1.5cm, so this is our answer.

The effective spring stiffness of the interatomic force within a magnesium wire, when subject to stretching by a 104 kg mass, is approximately 120451.067 N/m. This value is calculated using Hooke's Law, which relates the force applied to the extension of the spring (wire) and the spring's stiffness.

To find the effective spring stiffness of the interatomic force in a magnesium wire, we can use Hooke's Law, which states that the force F needed to extend or compress a spring by some distance x is proportional to that distance. This is mathematically represented as F = kx, where k is the spring stiffness (or force constant), and x is the extension or compression of the spring (in this case, the wire).

First, we calculate the force applied by the mass hung from the magnesium wire. The force due to gravity is F = mg, where m is the mass (104 kg), and g is the acceleration due to gravity (approximately 9.8 m/s2).

F = 104 kg * 9.8 m/s2 = 1019.2 N.

The elongation of the wire (x) is given as 8.46 mm, which is 8.46 * 10-3 m.

Using Hooke's Law, we can solve for the spring stiffness, k, by rearranging the equation: k = F / x.

k = 1019.2 N / (8.46 * 10-3 m) = 120451.067 N/m (approximately).

This calculation gives us the effective spring stiffness of the interatomic force within the magnesium wire when a 104 kg mass is hung from it, causing it to stretch by 8.46 mm.

Answer:

a)

velocity unit = [tex]\frac{inches}{days}[/tex]

acceleration unit = [tex]\frac{inches}{days^{2} }[/tex]

b) No

Explanation:

a)

Velocity is a vector expression of the displacement respect to time of something, with magnitude and a defined direction. Then, to know the magnitude of the vector (v) we have to divide the distance (Δx) by the time (Δt) it takes to travel that distance:

v = Δx / Δt

Besides, the unit of velocity is a compound unit between distance units and time units. If we choose inches as our basic unit of distance and days as our basic unit of time, then the unit of velocity would be:

[tex]\frac{inches}{days}[/tex]

On the other hand, acceleration is a vector defined as the rate at which an object changes its velocity. The math expression divides the change o velocity (Δv) by the time (Δt) it takes to make this change to obtain the acceleration magnitude (a):

a = Δv / Δt

So, the unit of acceleration is a compound unit between velocity units and time units. Again, if we choose inches as our basic unit of distance and days as our basic unit of time, then the unit of velocity in this system would be [tex]\frac{inches}{days}[/tex] , as shown. Finally, the unit of acceleration would be:

[tex]\frac{inches}{days^{2} }[/tex]

In resume:

velocity unit = [tex]\frac{inches}{days}[/tex]

acceleration unit = [tex]\frac{inches}{days^{2} }[/tex]

b)

To answer whether this units´ system is a good choice or not for measuring the acceleration of an automobile, let´s think about its normal values.

A car normally takes between 4 to 8 seconds to change its velocity from 0 to 100 km/h, then normal acceleration values for the media of 6 seconds (or its equivalent in hours 0.0016 h) are:

a = Δv / Δt = [tex]\frac{100\frac{km}{h}-0\frac{km}{h} }{0.0016 h}[/tex] = 62,500 [tex]\frac{km}{h^{2} }[/tex]

If we want to use the units system of this exercise, then we have to use the equivalences between inches and kilometers (1 km = 39,370.1 inches) and the other between days and hours (24 h = 1 day). Then,

a = 62,500 [tex]\frac{km}{h^{2} }[/tex] * [tex]\frac{39370.1 inches}{1 km}[/tex] * [tex](\frac{24 h}{1 days}) ^{2}[/tex]

a = 62500 [tex]\frac{km}{h^{2} }[/tex] * [tex]\frac{39,370.1 inches}{1 km}[/tex] * [tex]\frac{576 h^{2} }{days^{2} }[/tex]

a = 1.42 ˣ10¹² [tex]\frac{inches}{days^{2} }[/tex]

If we choose this units´ system to express the acceleration of an automobile, it results in a very high number that introduces a difficulty just to quantify the acceleration. Then, this system is not a good choice for that purpose.

Answer:

F = 1.58*10^{11} N

Explanation:

given data:

length of steel beam = 27 m

cross sectional area of rail = 35 cm

[tex]\Delta T = 39[/tex] Degree celcius

change in length of steel beam is given as

[tex]\Delta L = L_O \alpha \Delta T[/tex]

            [tex]= 20*1.1*10^{-5}*39[/tex]

           [tex] =8.58*10^{-3}[/tex] m

Young's modulus is

[tex]Y = \frac{FL}{A\Delta L}[/tex]

[tex]F = \frac{ YA\Delta L}{L}[/tex]

[tex]= \frac{2.0*10^{11}*25*10^{-4}8.58*10^{-3}}{27}[/tex]

F = 1.58*10^{11} N

Final answer:

The force exerted by the steel beam when it is heated to 39°C is approximately 9,191 N.

Explanation:

To calculate the force that the beam exerts when it is heated, we can use the formula for thermal expansion:

ΔL = αLΔT

where ΔL is the change in length, α is the coefficient of linear expansion, L is the original length, and ΔT is the change in temperature.

In this case, the original length of the beam is 27m, the coefficient of linear expansion for steel is 1.1 × 10^−5/C°, and the change in temperature is 39°C - 0°C = 39°C.

Plugging in these values, we can calculate the change in length:

ΔL = (1.1 × 10^−5/C°)(27m)(39°C) = 0.011m

Next, we can calculate the change in cross-sectional area due to the expansion:

ΔA = αAΔT

where ΔA is the change in cross-sectional area, α is the coefficient of linear expansion, A is the original cross-sectional area, and ΔT is the change in temperature.

In this case, the original cross-sectional area of the beam is 35cm², the coefficient of linear expansion for steel is 1.1 × 10^−5/C°, and the change in temperature is 39°C - 0°C = 39°C.

Plugging in these values, we can calculate the change in cross-sectional area:

ΔA = (1.1 × 10^−5/C°)(35cm²)(39°C) = 0.015cm²

Next, we can use Hooke's Law to calculate the stress:

σ = F/A

where σ is the stress, F is the force, and A is the cross-sectional area.

In this case, the change in force is equal to the change in stress times the change in area:

ΔF = σΔA

Since the cross-sectional area changes and the force is equal to the change in force plus the initial force:

F = ΔF + σA

Substituting the values we know, we can calculate the force:

F = (σΔA) + (YAΔL)

where Y is the Young's modulus for steel.

Plugging in these values, we can calculate the force:

F = (σΔA) + (Y(0.011m)(35cm²))

Finally, we can calculate the force:

F = (σΔA) + (Y(0.011m)(35cm²)) = (σ(0.015cm²)) + (2.0 × 10^11 N/m²(0.011m)(35cm²))

After performing the calculations, we find that the force exerted by the beam when it is heated to 39°C is approximately 9,191 N.

Answer:

1.5 ohm, 9 ohm

Explanation:

Total resistance, R = 10.5 ohm

Let the resistance of one piece is r and then the resistance of another piece is 10.5 - r.

According to question, the resistance of one piece is equal to the 6 times the resistance of another piece.

so, 6 r = 10.5 - r

7 r = 10.5

r = 1.5 ohm

Thus, the resistance of one piece is 1.5 ohm and resistance of another piece is 10.5 - 1.5 = 9 ohm.

Answer:

Astronaut in spacecraft while orbiting earth experience weightlessness because there is no gravity of earth or moon is acting on the body of an astronaut.                      

while on earth, we experience weight because the gravity of earth is acting on our body which is pulling us downward.

Both spacecraft and the astronauts both are in a free-fall condition.

The mass of an ice cube that would bring coffee to 65 degrees Celsius from 87 degrees Celsius is determined via the principles of heat transfer. This involves considering the ice warming, phase change and water warming steps and setting this total heat to be equal to the heat lost by the coffee which results in calculating the mass of the ice.

To find out the mass of an ice cube that would cool down the coffee to 65 degrees Celsius from 87 degrees Celsius, we need to use the principles of heat transfer, specific heat capacities, and phase changes. Knowing that ice comes from a -15.0 degrees Celsius freezer it has to first warm up from -15C to 0C (icing warming), then melt at 0C (phase change), and the resulted water to be warmed from 0C to 65C (water warming). The sum of the heat absorbed during these three steps will be equal to the heat lost by the coffee.

During icing warming and water warming we use q=m*c*ΔT where m is the mass we don't know, c is the specific heat capacity (2.09 J/g°C for ice and 4.18 J/g°C for water) and ΔT is the change in temperature. During phase change we use q=m*L where m is the mass we are looking for and L is the heat of fusion of water 334 J/g.

Setting the sum of three heats equal to the heat lost by the coffee q=m*c*ΔT where mass is known (300g because density of coffee is similar to water's), c is again 4.18 J/g°C and ΔT is the change in temperature of the coffee, we can find out the mass of the ice cube needed.

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To cool the coffee from 87.0 degrees Celsius to 65 degrees Celsius, we can use the equation Q = mcΔT to calculate the mass of the ice cube needed.

To cool the coffee from 87.0 degrees Celsius to 65 degrees Celsius, we need to calculate the amount of heat transferred from the coffee to the ice cube. We can use the equation:

Q = mcΔT

Where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature. We know the temperature change and the specific heat capacity of water. We can find the mass of the ice cube by rearranging the equation and plugging in the values:

m = Q / (cΔT)

Finally, we can calculate the mass of the ice cube and provide an answer to the question.

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Answer:

3.2 m/s²

Explanation:

Initial velocity of the red blood cell = 0 m/s = u

Initial velocity of the red blood cell = 0.8 m/s = v

Time taken by the red blood cell to reach its final velocity = 250 ms = t

Equation of motion

[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{0.8-0}{250\times 10^{-3}}\\\Rightarrow a=3.2\ m/s^2[/tex]

Average acceleration of a red blood cell as it leaves the heart is 3.2 m/s²

The average acceleration of a red blood cell as it leaves the heart, based on given velocity and time, is calculated to be 3.2 m/s².

To calculate the average acceleration of a red blood cell as it leaves the heart, we can use the formula for acceleration: a = Δv/Δt. Δv is the change in velocity, and Δt is the change in time. Here, the initial velocity of the blood cell is assumed to be zero, and the final velocity is 0.80 m/s, as given in the question. The change in time is the duration of the contraction of the left ventricle, which is 250 ms, or 0.25 sec. So we have:

a = (0.80 m/s - 0 m/s) / 0.25 s = 3.2 m/s²

Therefore, the average acceleration of a red blood cell as it leaves the heart is 3.2 meters per second squared (m/s²).

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Answer:Decreases

Explanation:

Given

Volume is held constant that is it is a isochoric process.

We know that

PV=nRT

as n,V& R are constant therefore only variables are

P & T

so [tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]

[tex]\frac{P_1}{P_2}=\frac{T_1}{T_2}[/tex]

As [tex]T_1[/tex] is decreasing therefore Pressure must also decrease so that ratio remains constant.

Answer:

(C) zero (there is no net horizontal component of the E-field)

Explanation:

If we subdivide the bar into small pieces, each piece (dx) contains a charge (dq), the electric field of each piece is equivalent to the field of a punctual electric charge, and has a direction as shown in the attached figure. For each piece (dx) in the negative axis there is another symmetric piece (dx) in the positive axis, and as we see in the figure for symmetry the sum of their electric fields gives a resultant in the Y axis (because its components in X are cancelled by symmetry).

Then the resultant of the electric field will be only in Y.

(C) zero (there is no net horizontal component of the E-field)

The x-component of the electric field at point (0,a) due to a uniform continuous line charge on the x-axis is zero, due to the symmetrical distribution of charge and corresponding cancellation of horizontal electric field components.

The student is asking about the direction of the electric field at a point on the positive y-axis due to a uniform continuous line charge distributed along the x-axis. To find the direction of the x-component of the electric field at the point (0,a), we can consider the symmetry of the charge distribution. For any small element of charge on the positive side of the x-axis, there is an identical element of charge on the negative side at the same distance from the origin. The electric fields produced by these two elements at point (0,a) on the y-axis will have the same magnitude but opposite x-components. These x-components will cancel each other out, resulting in a net x-component of the electric field being zero. Therefore, the correct answer to the student's question is (C) zero (there is no net horizontal component of the electric field).

Answer:

Time of flight  A is greatest

Explanation:

Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.

So

H = u₁² sin²θ₁ /2g

H = u₂² sin²θ₂ /2g

H = u₃² sin²θ₃ /2g

On the basis of these equation we can write

u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃

For maximum range we can write

D = u₁² sin2θ₁ /g

1.5 D = u₂² sin2θ₂ / g

2 D =u₃² sin2θ₃ / g

1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁

1.5 = u₂ cosθ₂ /u₁ cosθ₁      ( since , u₁ sinθ₁ =u₂ sinθ₂ )

u₂ cosθ₂ >u₁ cosθ₁

u₂ sinθ₂ < u₁ sinθ₁

2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g

Time of flight B < Time of flight  A

Similarly we can prove

Time of flight C < Time of flight B

Hence Time of flight  A is greatest .

Answer:

The net force on the block is 19.40 N.

Explanation:

Given that,

Coefficient of static friction = 0.4500

Coefficient of kinetic friction = 0.1500

Mass of block = 4.400 kg

We need to calculate the net force required to make the block slide

[tex]F=\mu_{s} mg[/tex]

Put the value into the formula

[tex]F=0.4500\times4.400\times9.8[/tex]

[tex]F=19.40\ N[/tex]

Hence, The net force on the block is 19.40 N.

Final answer:

The net force on the block the instant it starts to slide is 19.548 N.

Explanation:

The net force on the block the instant it starts to slide can be found by multiplying the coefficient of static friction (μmk) by the normal force (N).

Given that the coefficient of static friction is 0.4500 and the mass of the block is 4.400 kg, we can calculate the normal force using the formula N = mg, where g is the acceleration due to gravity (9.8 m/s²).

Therefore, the net force on the block the instant it starts to slide is N × μmk = 4.400 kg × 9.8 m/s⁲ × 0.4500 = 19.548 N.

Answer:

The camera lands in t=2.91s with a velocity:

[tex]v=-30.45m/s[/tex]

Explanation:

The initial velocity of the camera is the same as the hot-air ballon:

[tex]v_{o}=-1.9m/s[/tex]

[tex]y_{o}=47m[/tex]

Kinematics equation:

[tex]v(t)=v_{o}-g*t[/tex]

[tex]y(t)=y_{o}+v_{o}t-1/2*g*t^{2}[/tex]

when the camera lands, y=0:

[tex]0=47-1.9t-4.91*t^{2}[/tex]

We solve this equation to find t:

t1=-3.29s       This solution have not sense in our physical point of view

t2=2.91s    

So, the camera lands in t=2.91s

We replace this value in v(t):

[tex]v=v_{o}-g*t=-1.9-9.81*2.91=-30.45m/s[/tex]

Answer:

1.6 g

Explanation:

When they change temperatures their diameters will change following these equations:

D'(t1) = D(t0) * (1 + a(Cu) * (t1 - t0(Cu)))

d'(t1) = d(t0) * (1 + a(Al) * (t1 - t0(Al)))

The sphere goes inside the ring when they are at thermal equilibrium (at the same temperature, which is t1). Their diameters will be the same.

D'(t1) = d'(t1)

D(t0) * (1 + a(Cu) * (t1 - t0(Cu))) = d(t0) * (1 + a(Al) * (t1 - t0(Al)))

D(t0) + D(t0) * a(Cu) * t1 - D(t0) * a(Cu) * t0(Cu) = d(t0) + d(t0) * a(Al) * t1 - d(t0) * a(Al) * t0(Al)

D(t0) * a(Cu) * t1 - d(t0) * a(Al) * t1 = D(t0) * a(Cu) * t0(Cu) - d(t0) * a(Al) * t0(Al) - D(t0) + d(t0)

(D(t0) * a(Cu) - d(t0) * a(Al)) * t1 = D(t0) * a(Cu) * t0(Cu) - d(t0) * a(Al) * t0(Al) - D(t0) + d(t0)

t1 = (D(t0) * a(Cu) * t0(Cu) - d(t0) * a(Al) * t0(Al) - D(t0) + d(t0)) / (D(t0) * a(Cu) - d(t0) * a(Al))

t1 = (3.71382 * 17*10^-6 * 0) - 3.72069 * 23*10^-6 * 83 - 3.71382 + 3.72069) / (3.71382 * 17*10^-6 - 3.72069 * 23*10^-6)

t1 = 10.37 C

To reach this temperature they both exchange heat.

Q(Al) + Q(Cu) = 0

The copper will gain heat (positive) and the aluminum will lose heat (negative)

Q(Al) = -Q(Cu)

mAl * CpAl * (t1 - t0Al) = - mCu * CpCu* (t1 - t0Cu)

mAl = -(mCu * CpCu* (t1 - t0Cu)) / (CpAl * (t1 - t0Al))

mAl = -(0.026 * 386 * (10.37 - 0)) / (900 * (10.37 - 83))

mAl = 1.6*10^-3 kg = 1.6 g

Answer:

Explanation:

The ball is going down with velocity. It must have momentum . It is stopped by

catcher so that its momentum becomes zero . There is change in momentum . So force is applied on the ball  by the gloves  .

The rate of change of momentum gives the magnitude of force. This force must be in upward direction to stop the ball. So force is in positive direction .

Let us measure the force applied on the ball .

Final velocity after the fall by 66 m

V = √ 2gh

= √ 2x9.8 x 66

35.97 m /s

Momentum = m v

0.145 x 35.97

= 5.2156 kgms⁻¹

Change in momentum

= 5.2156 - 0

= 5.2156

Rate of change of momentum

=  Change of momentum / time = 5.2156 / .0118

= 442 N

if R be the force exerted by gloves to stop the ball

R - mg represents the net force which stops the ball so

R - mg = 442

R = 442 + mg

= 442 + .145 x 908

443.43 N

According to Newton's second law of motion rate of change of momentum is equal to the applied force. The force exerted by the glove of the player on the ball will be 497.4 N.

According to Newton's second law of motion rate of change of momentum is equal to the applied force.

Momentum is given by the product of mass and velocity. it is denoted by P.it leads to the impulsive force.

P = mv

ΔP = mΔv

[tex] \rm{ F = \frac{\delP}{\delt} } [/tex]

[tex]V = \sqrt{2gh} [/tex]

[tex]\rm{V = \sqrt{2\times9.81\times60} [/tex]

v = 34.31 m/sec

ΔV = v -u

ΔV = 34.31-0

ΔV = 34.31 m /sec.

Δt = 0.0118 sec

[tex]\rm{\frac{\delv}{\delt} } [/tex]= 34.31

F = [tex]m\rm{\frac{\delv}{\delt} }[/tex]

F = [tex] 0.145\times3430 [/tex]

F = 497.4 N

Thus the force exerted by the glove of the player on the ball will be 497.4 N.

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Answer:

(a) The constants required describing the rod's density are B=2.6 and C=1.325.

(b) The mass of the road can be found using [tex]A\int_0^{12}\left(B+Cx)dx[/tex]

Explanation:

(a) Since the density variation is linear and the coordinate x begins at the low-density end of the rod, we have a density given by

[tex]2.6\frac{g}{cm^3}+\frac{18.5\frac{g}{cm^3}-2.6\frac{g}{cm^3}}{12 cm}x = 2.6\frac{g}{cm^3}+1.325x\frac{g}{cm^2}[/tex]

recalling that the coordinate x is measured in centimeters.

(b) The mass of the rod can be found by having into account the density, which is x-dependent, and the volume differential for the rod:

[tex]m=\int\rho dv=\int\left(B+Cx\right)Adx=5\int_0^{12}\left(2.6+1.325x\right)dx=126.6[/tex],

hence, the mass of the rod is 126.6 g.

The distance light travels in 1.0 µs is approximately 3.00 × 10^2 m.

The speed of light is approximately 3.00 × 108 m/s. To find out how far light travels in 1.0 µs, we need to multiply the speed of light by the time taken.

Distance = Speed × Time
Distance = (3.00 × 108 m/s) × (1.0 × 10-6 s)

Calculating this value gives us:
Distance = 3.00 × 102 m

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To find how far light travels in 1.0μs, multiply the speed of light (2.998×108 m/s) by the time interval (1.0×10-6 s), yielding an expression for distance with units in meters.

To calculate the distance light travels in 1.0μs (microsecond), you can use the formula for distance where distance traveled is speed multiplied by time. The speed of light (c) in a vacuum is a fundamental constant, and we will use the value 2.998×108 m/s for our calculations.

The time given is 1.0μs, which needs to be converted into seconds to match the units of the speed of light. Remember that 1μs is equal to 1×10-6 seconds. Therefore, the distance light travels in 1.0μs is a matter of multiplying the speed of light by this time.

The math expression without doing any calculation will be:

Distance = c × time

Distance = (2.998×108 m/s) × (1.0×10-6 s)

The units for the final answer will be in meters (m).

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Answer:

Charge, q = 1.402 C

Explanation:

Given that,

Current from lightning bolt, [tex]I=7.01\times 10^4\ A[/tex]

Time, t = [tex]t=20\ \mu s=2\times 10^{-5}\ s[/tex]

Let q is the charge transferred in this event. We know that the total charge divided by time is called current. Mathematically, it is given by :

[tex]I=\dfrac{q}{t}[/tex]

[tex]q=I\times t[/tex]

[tex]q=7.01\times 10^4\times 2\times 10^{-5}[/tex]

q = 1.402 C

So, 1.402 coulomb of charge is transferred in this event. Hence, this is the required solution.

Answer:

(a) electrons being added.

(b) 7.5 × 10¹⁰

Explanation:

(a)

The plastic rod acquires a charge of -12 nC which means it contains more negative charge.

Thu, the electrons have been added to the rod. Removing protons are very difficult as they are in the nucleus and are bind by nuclear forces. Thus, by rubbing, electrons are transferred and thus the rod become negatively charged which means that the electrons being added.

(b)

The total charge = -12 nC

Also,

1 nC = 10⁻⁹ C

So,

Charge = -12 × 10⁻⁹ C

Charge on electron = -1.6 × 10⁻¹⁹ C

So, number of electrons transferred = -12 × 10⁻⁹ C / -1.6 × 10⁻¹⁹ C = 7.5 × 10¹⁰

a) Electrons have been added to the plastic rod.

b) 7.5 x 10¹⁰ electrons have been added to the plastic rod.

a) When a plastic rod is rubbed with a cloth, electrons are transferred from the cloth to the rod. This gives the rod a negative charge, because it now has an excess of electrons.

b) The charge on an electron is -1.6 x 10⁻¹⁹ C. The rod has a charge of -12 nC, which is -12 x 10⁻⁹ C. So, 7.5 x 10¹⁰ electrons have been added to the rod.

Charge: Charge is a property of matter that can be positive, negative, or neutral. Objects with the same charge repel each other, while objects with opposite charges attract each other.

Electrons: Electrons are negatively charged particles that are found in all atoms. They are much smaller than protons and neutrons, the other two types of particles found in atoms.

Protons: Protons are positively charged particles that are found in the nucleus of atoms. They are about the same size as neutrons.

Rubbing: When two objects are rubbed together, electrons can be transferred from one object to the other. This is because the electrons in the two objects are not always evenly distributed.

When a plastic rod is rubbed with a cloth, electrons are transferred from the cloth to the rod. This is because the cloth has a stronger attraction for electrons than the rod does. The rod now has an excess of electrons, which gives it a negative charge.

The number of electrons that are transferred can be calculated by dividing the rod's charge by the charge on an electron. In this case, the rod has a charge of -12 nC, and the charge on an electron is -1.6 x 10⁻¹⁹ C. So, 7.5 x 10¹⁰ electrons have been added to the rod.

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