Find the coefficient of kinetic friction between a 3.80-kg block and the horizontal surface on which it rests if an 87.0-N/m spring must be stretched by 6.50 cm to pull it with constant speed. Assume that the spring pulls in the horizontal direction.

Answer:

When a teen driver drives with a lot of his peers as passengers they may lead to distraction which may later end up in accident as the driver was distracted

Overconfidence, lack of focus, and phone while driving are the factors  contribute to accidents when a teen driver controls other teens as passengers,

When a teen driver drives with a lot of his peers as passengers they may direct to distraction which may later end up in casualty as the driver was distracted.

Several studies have indicated that passengers substantially increase the chance of crashes for young, novice drivers. This improved risk may result from distractions that young passengers complete for drivers.Teens driving with a teen or young adult passengers existence of teen or young adult passengers raises the crash risk of unsupervised teen drivers. This risk grows with each additional teen or a young adult passenger.

Crash risk is two- to six times more significant for those who utilize a cellphone while driving resembled for drivers who are not distracted. Using a phone delays reaction time increases lane deviations, and forces drivers to look away from the road for extended times.

Overconfidence, lack of focus, and phone while driving are the factors  contribute to accidents when a teen driver controls other teens as passengers,

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Answer: 66.67 m, 44.44 s

Explanation:

Given

Velocity of flow is [tex]u=1.5\ m/s[/tex]

Mary can row with speed [tex]v=3.6\ m/s[/tex]

Width of the river [tex]y=160\ m[/tex]

Flow will drift the Mary towards east, while Mary boat will cause it to travel in North direction

time taken to cross river

[tex]\Rightarrow t=\dfrac{160}{3.6}\\\\\Rightarrow t=\dfrac{400}{9}\ s[/tex]

Flow will drift Mary by

[tex]\Rightarrow x=ut\\\\\Rightarrow x=1.5\times \dfrac{400}{9}\\\\\Rightarrow x=66.67\ m[/tex]

Velocity w.r.t shore is

[tex]\Rightarrow v_{net}=\sqrt{3.6^2+1.5^2}\\\Rightarrow v_{net}=\sqrt{15.21}\\\Rightarrow v_{net}=3.9\ m/s[/tex]

Answer:

: No, a clinical thermometer cannot be used to measure the temperature of boiling water because it has a small range and might break due to extreme heat. ... The temperature is around 100 degrees Celsius.

Explanation:

First, we convert the energy from eV to Joules:

[tex]2.90\:\text{eV}×\left(\dfrac{1.6×10{-19}\:J}{1\:\text{eV}} \right)[/tex]

[tex]= 4.64×10^{-19}\:\text{eV}[/tex]

We know from definition that

[tex]E=h\nu = \dfrac{hc}{\lambda}[/tex]

so the wavelength of the photon is

[tex]\lambda = \dfrac{hc}{E} = 4.28×10^8\:\text{m}[/tex]

Answer:

Reverse Osmosis

Explanation:

Reverse osmosis is a type of filtration that involves passing a solvent through a semipermeable membrane in the opposite direction that natural osmosis does. Separation is always enforced through the use of pressure in this process. Ions, fine dust particles, molecules, and larger particles are typically removed from solvents using this method. The technique is particularly popular in the treatment and purification of water.

Answer:

filtration is used to separate

Answer:

E_total = - 50 N / A

Explanation:

The electric field is a vector magnitude whereby

          E_total = Eₐ + E_b

where the bold letters indicate vectors, in this case the charges of the two objects A and B are the same and they are on the same line

         E_total = - E_a - E_b

the electric field for a point charge is

        E_a = [tex]k \ \frac{q_a}{r_a^2 }[/tex]

        qₐ= Eₐ rₐ² / k

indicates that Eₐ = 40.0 N / C

        qₐ = 40.0 0.250²/9 10⁹

        qₐ = 2.777 10⁻¹⁰ C

indicates that the charge of the two points is the same

        qₐ = q_b

      E_total = - k qₐ / rₐ² - k qₐ / (2 rₐ)²

      E_total = [tex]-k \ \frac{q_a}{r_a^2} \ ( 1 + \frac{1}{4} )[/tex]

we calculate

       E_total = - 40.0 (5/4)

       E_total = - 50 N / A

Answer:

The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.

Explanation:

By Newton's Laws of Motion and definition of maximum friction force, we derive the following two formulas for the static and kinetic coefficients of friction:

[tex]\mu_{s} = \frac{f_{s}}{m\cdot g}[/tex] (1)

[tex]\mu_{k} = \frac{f_{k}}{m\cdot g}[/tex] (2)

Where:

[tex]\mu_{s}[/tex] - Static coefficient of friction, no unit.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.

[tex]f_{s}[/tex] - Static friction force, in newtons.

[tex]f_{k}[/tex] - Kinetic friction force, in newtons.

[tex]m[/tex] - Mass, in kilograms.

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

If we know that [tex]f_{s} = 83\,N[/tex], [tex]f_{k} = 55\,N[/tex], [tex]m = 31\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the coefficients of friction are, respectively:

[tex]\mu_{s} = \frac{83\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]\mu_{s} = 0.273[/tex]

[tex]\mu_{k} = \frac{55\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]\mu_{k} = 0.181[/tex]

The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.

Answer:

If the mass of one of the objects is tripled, then the force of gravity between them is tripled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces

Answer:

If the mass of B is m and the temperature change is the same, the mass of B will be 2m.

Explanation:

Q = mcT

T = mc/Q

M = 4Q/2cT........... (1)

T = Q/mc

Plug this in equation 1.

M = 4Q/(2c × Q/mc)  = 4Q ÷ 2Q/m  = 4Q × m/2Q = 2m

Answer:

A = 26.875 rad/s²

Explanation:

Given the following data;

Initial angular speed, Uw = 150 rads/s.

Final angular speed, Vw = 580 rads/s.

Time = 16 seconds.

To calculate the angular acceleration;

From kinematics equation;

At = Vw - Uw

Where;

Substituting into the formula, we have;

A*16 = 580 - 150

16A = 430

A = 430/16

A = 26.875 rad/s²

Answer:

0.857 cm

Explanation:

We are given that:

The focal length for a convex lens to be (f) = 1.5cm

The object distance (u) = - 2.0 cm

We are to determine the image distance (v) = ??? cm

By applying the lens formula:

[tex]\dfrac{1}{f} = \dfrac{1}{u}+\dfrac{1}{v}[/tex]

By rearrangement and making (v) the subject of the above formula:

[tex]v = \dfrac{uf}{u-f}[/tex]

replacing the given values:

[tex]v = \dfrac{(-2.0)(1.5)}{(-2.0 -1.5)}[/tex]

[tex]v = \dfrac{-3.0}{(-3.5)}[/tex]

v = 0.857 cm

Answer:

Power = Work / Time

P = 400 kg * 9.8 m/s * 4.5 m / 3600 sec = 4.9 J/s = 4.9 Watts

Also, 4.9 Watts / (746 Watts / Horsepower) = .0066 Hp

Answer:

4 kg of force

Explanation:

Force = (mass x distance to fulcrum) / length of fulcrum to end

Subsitute values

F = (10 x 20)/50

F =4

Answer:

the correct is C

Explanation:

The concept of a frame of reference is of crucial importance in physics, because it is the system from which measurements are made. Therefore, the relationships between the different reference frames must be clear so that the measurements made can be compared correctly.

In this case, the first observed sees the movement of the ball, suppose it moves a distance r, the second observed is next to me, separated by a distance x, therefore a frame of reference located in the movement of the ball. ball r '.

Consequently, the measurement carried out is related by

             r = r’ + x

where the bold letters indicate wind blowers.

With these explanations we review the different answers, the correct one is C

Answer:

Pressure is more in the open container than the closed one.

Explanation:

The pressure due to the fluid at a depth is given by

Pressure = depth x density of fluid x gravity

So, when the container is open, the atmospheric pressure is also add  up but when  the container is closed only the pressure due to the fluid is there.

So, when the container is open, the pressure is atmospheric pressure + pressure due to the fluid.

hen the container is closed only the pressure due to the fluid is there.

Answer:

Explanation:

For starters begin with a warning not to touch the brake drums. All of the KE is transferred to the brake drums. The result is a large rise in temperature. Heat. If you press hard on the brakes, rubber is left on the road and there is heat involved in that too.

Answer:

KInetic energy reduces.

Explanation:

Application of breaks reduces velocity. Reduction of velocity constitutes velocity reduction.

For more help contact +1 (304) 223-3136

Answer:

B = 0.118 T

Explanation:

From Biot-Savart Law:

[tex]B = \frac{\mu_o I}{2\pi r}[/tex]

where,

B = strength of magnetic field = ?

μ₀ = 4π x 10⁻⁷ Tm/A

I = current enclosed = 474.1 A

r = radius = 0.8 mm = 8 x 10⁻⁴ m

Therefore,

[tex]B = \frac{(4\pi\ x\ 10^{-7}\ Tm/A)(474.1\ A)}{2\pi(8\ x\ 10^{-4}\ m)}[/tex]

B = 0.118 T

Answer:

a. 282.84 cm/s b. 100 cm/s c. 400 cm/s d. 200 cm/s

Explanation:

The speed of the wave v = √(T/μ) where T = tension and μ = mass per unit length = m/l where m = mass of string and l = length of string.

So, v = √(T/μ)

v = √(T/m/l)

v = √(Tl/m)

a. The string's tension is doubled?

If the tension is doubled, T' = 2T the new speed is

v' = √(T'l/m)

v' = √(2Tl/m)

v' = √2(√Tl/m)

v' = √2v

v' = √2 × 200 cm/s

v' = 282.84 cm/s

b. The string's mass is quadrupled (but its length is unchanged)?

If the mass is quadrupled, m' = 4m the new speed is

v' = √(Tl/m')

v' = √(Tl/4m)

v' = (1/√4)(√Tl/m)

v' = v/2

v' = 200/2 cm/s

v' = 100 cm/s

c. The string's length is quadrupled (but its mass is unchanged)?

If the length is quadrupled, l' = 4l the new speed is

v' = √(Tl'/m)

v' = √(T(4l)/m)

v' = √4)(√Tl/m)

v' = 2v

v' = 200 × 2 cm/s

v' = 400 cm/s

d. The string's mass and length are both quadrupled?

If the length is quadrupled, l' = 4l and mass quadrupled, m' = 4m, the new speed is

v' = √(Tl'/m')

v' = √(T(4l)/4m)

v' = √(Tl/m)

v' = v

v' = 200 cm/s

Answer:

a) He found the same value of q/m for different cathode materials.

b)      y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex] ,  c)  v = [tex]\frac{E_o}{B_o}[/tex]

Explanation:

In Thomson's experiments he was able to measure the deflection of the light beam under the effect of the magnetic field and with these results find the e / m relationship, which in all cases is the same, therefore the most important conclusion is that the value e E / m is constant for all materials.

b) In the part of the plates the electrons are accelerated by the electric field,

              F = ma

             - e E = m a

              a = - (e/m)  E₀

the distance traveled is          

X axis

          x = v₀ t

the separation of the plates is x = d

          t = vo / d

Y axis

          y = v_{oy} t + ½ to t²

          y = ½ a t²

          y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex]

c) In this case there is a magnetic field B₀ and the electrons have no deflection

         F = - e E + e v x B

if there is no deviation F = 0

         e E = e v B

         v = [tex]\frac{E_o}{B_o}[/tex]

Answer:

increase

Explanation:

hope it will help you:)

Answer:

(d)

Explanation:

because dU = Q -W so ,that the option d(D) is correct

Answer:

D) the two spheres remain of equal size.

Explanation:

Answer:

The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm

Explanation:

Given;

wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m

maximum kinetic energy of the ejected electron, K.E = 0.92 eV

let the work function of the aluminum metal = Ф

Apply photoelectric equation:

E = K.E + Ф

Where;

Ф is the minimum energy needed to eject electron the aluminum metal

E is the energy of the incident light

The energy of the incident light is calculated as follows;

[tex]E = hf = h\frac{c}{\lambda} \\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\c \ is \ speed \ of \ light = 3 \times 10^{8} \ m/s\\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{248\times 10^{-9}} \\\\E = 8.02 \times 10^{-19} \ J[/tex]

The work function of the aluminum metal is calculated as;

Ф = E - K.E

Ф = 8.02 x 10⁻¹⁹  -  (0.92 x 1.602 x 10⁻¹⁹)

Ф =  8.02 x 10⁻¹⁹ J   -  1.474 x 10⁻¹⁹ J

Ф = 6.546 x 10⁻¹⁹ J

The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;

[tex]\phi = hf = \frac{hc}{\lambda_{max}} \\\\\lambda_{max} = \frac{hc}{\phi} \\\\\lambda_{max} = \frac{(6.626\times 10^{-34}) \times (3 \times 10^8) }{6.546 \times 10^{-19}} \\\\\lambda_{max} = 3.037 \times 10^{-7} m\\\\\lambda_{max} = 303.7 \ nm[/tex]

Answer:

20 seconds

Explanation:

We are given 2 givens in the first statement

v0=0 and a=5

And we are trying to find time needed to cover 1km or 1000m.

So we use

x-x0=v0t+1/2at²

Plug in givens

1000=0+2.5t²

solve for t

t²=400

t=20s

Answer:

non biodegradable

Explanation:

It is non biodegradable because plastic cannot dispose off easily ..

Answer:

17.07 kJ

Explanation:

The work done against gravity by the man W equals the potential energy change of the man and can of paint, ΔU

W = ΔU = mgΔy where m = mass of man and can of paint = 200 lb + 10 lb = 210 lb = 210 × 1 kg/2.205 lb, g = acceleration due to gravity = 9.8 m/s² and Δy = height of silo = 60 ft = 60 × 1m/3.28 ft

Since W = mgΔy, we substitute the values of the variables into the equation.

So,

W = mgΔy

W = 210 lb × 1 kg/2.205 lb × 9.8 m/s² × 60 ft × 1m/3.28 ft

W = 123480/7.2324 J

W = 17073.2 J

W = 17.0732 kJ

W ≅ 17.07 kJ

"down/up the plane" suggests an inclined plane, but no angle is given so I'll call it θ for the time being.

The free body diagram for the crate in either scenario is the same, except for the direction in which static friction is exerted on the crate. With the P = 100 N force holding up the crate, static friction points up the incline and keeps the crate from sliding downward. When P = 350 N, the crate is pushed upward, so static friction points down. (see attached FBDs)

Using Newton's second law, we set up the following equations.

• p = 100 N

∑ F (parallel) = f + p cos(θ) - mg sin(θ) = 0

∑ F (perpendicular) = n - p sin(θ) - mg cos(θ) = 0

• P = 350 N

∑ F (parallel) = P cos(θ) - F - mg sin(θ) = 0

∑ F (perpendicular) = N - P sin(θ) - mg cos(θ) = 0

(where n and N are the magnitudes of the normal force in the respective scenarios; ditto for f and F which denote static friction, so that f = µn and F = µN, with µ = coefficient of static friction)

Solve for n and N :

n = p sin(θ) + mg cos(θ)

N = P sin(θ) - mg cos(θ)

Substitute these into the corresponding equations containing µ, and solve for µ :

µ = (mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ))

µ = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))

Next, you would set these equal and solve for m :

(mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ)) = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))

...

Once you find m, you back-substitute and solve for µ, but as you might expect the result will be pretty complicated. If you take a simple angle like θ = 30°, you would end up with

m ≈ 36.5 kg

µ ≈ 0.256

The coefficient of static friction between the plane and the crate is μ = 0.256 and the mass of the crate is m=36.4 kg.

From the given,

The force that opposes the crate by sliding is P = 100N

In X-axis, the sum of forces is zero.

ΣF = 0

Pcosθ - mgsinθ-Ff = 0

Ff = Pcosθ - mgsinθ

In Y-axis

Psinθ - mgcosθ - N = 0

N = Psinθ-mgcosθ

Frictional force, Ff = μN, μ is the coefficient of friction

Ff = μN

Pcos30- mgsin30 + μ( Psin30+mgcos30) = 0

μ = mgsin30-Pcos30/Psin30+mgcos30 ------1

The block is sliding with the horizontal force, F = 350N

X-axis

P₂cosθ - mgsinθ-Ff = 0

Y-axis

P₂sinθ - mgcosθ - N = 0

N = P₂sinθ-mgcosθ

μ = P₂cos30-mgsin30/P₂sin30-mgcos30   -----2

Equate equations 1 and 2

mgsin30-Pcos30/Psin30+mgcos30 =P₂cos30-mgsin30/P₂sin30-mgcos30

4.905m-86.6/50+8.49 = 303.1-4.905m/175+8.49

41.7m² + 123m - 1.516×10⁴ = 0

-41.7m² +2330m -1.516×10⁴(4.905-86.6)(175+8.49) =(303.1-4.905)(50+8.49)

83.4m² - 2207m -3.03×10⁴ = 0

m= 36.4 kg

Hence, the mass of the crate is 36.4 Kg.

Substitute the value of m in equation 1,

μ = 4.905(36.4) - 86.6 / 50 + 8.49

μ  = 0.256

Thus, the coefficient of static friction is 0.256.

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Answer:

v₂ = 53.23 m/s

Explanation:

Given that,

The mass of a golf club, m₁ = 158 g = 0.158 kg

The initial speed of a golf club, u₁  =  48.2 m/s

The mass of a golf ball, m₂ = 46 g = 0.046 kg

It was at rest, u₂ = 0

Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s, v₁ = 32.7 m/s

We use the conservation of energy to find the speed of the golf ball just after impact as follows :

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.158(48.2)-0.158(32.7)}{0.046}\\\\=53.23\ m/s[/tex]

So, the speed of the golf ball just after the impact is equal to 53.23 m/s.

Answer:

Explanation: