If an object, initially at rest, accelerates at the rate of 25m/s2, what will the magnitude of the displacement be after 50s

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:

-1.0778×10⁻¹⁰ N/C

Explanation:

Applying,

E = kq/r²................ equation 1

Where E = elctric field, q = charge, r = distance, k = coulomb's law

From the question,

Given: q = -3.0×10 C, r = 5.0 m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values in equation 1

E = (-3.0×10)(8.98×10⁹)/5²

E = -1.0778×10⁻¹⁰ N/C

Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C

Answer:

Liquid volume is usually measured using either a graduated cylinder or a buret. As the name implies, a graduated cylinder is a cylindrical glass or plastic tube sealed at one end, with a calibrated scale etched (or marked) on the outside wall.

Answer:

The intensity of 4 lawn movers is 86 dB.

Explanation:

Intensity of one lawnmower = 80 dB

Let the intensity is I.

Use the formula of intensity

[tex]dB = 10 log\left ( \frac{I}{Io} \right )\\\\80=10log\left ( \frac{I}{Io} \right )\\\\10^8 = \frac{I}{10^{-12}}\\\\I = 10^{-4} W/m^2[/tex]

Now the intensity of 4 lawn movers is

[tex]dB = 10 log\left ( \frac{4I}{Io} \right )\\\\dB=10log\left ( \frac{4\times10^{-4}}{10^{-12}} \right )\\\\dB = 86 dB\\[/tex]

Answer:

17.8cm

Explanation:

1.3kg --> 4cm

1kg --> 3, 1/13cm

5.8kg --> 18.8cm

Answer:

Force is zero.

Explanation:

According to the Newton's second law, when an object is moving with an acceleration the force acting on the object is directly proportional to the rate of change of momentum of the object.

F = m a

if the object is moving with uniform velocity, the acceleration is zero, and thus, the force is also zero.  

Answer: Near the moon’s surface, a thrust over 11,250 N but under 13,500 N would make it travel at a constant vertical velocity.

Explanation: .Newton’s first law of motion states that an object in motion continues to move in a straight line at a constant velocity unless acted upon by an unbalanced force. In accordance with this law, the lunar lander moves in a downward direction toward the surface of the moon under the influence of force due to gravity. A thrust somewhere between 11,250 and 13,500 balances this gravitational force out.

Answer:

The answer is below

Explanation:

Specific heat capacity is an intensive property of a material. The specific heat of a material is the amount of energy required to raise the temperature of one unit mass m of material by one unit of temperature.

a) Temperature is inversely proportional to specific heat capacity. If the same amount of heat is applied to all three balls, the ball that will reach the highest temperature is the ball with the least specific heat capacity.

Hence lead will have the highest temperature since it has the least specific heat capacity.

b) The quantity of heat is directly proportional to the specific heat capacity. Hence if all balls experience the same temperature change, the ball that have the most energy will be that with the highest specific heat capacity.

Hence aluminum will have the most heat since it has the highest specific heat capacity.

Answer:

Minimum angular spread (in rad) = 547.45 x 10⁻⁶ rad

Explanation:

GIven;

Wavelength of manganese vapor laser beam = 534 nm = 534 x 10⁻⁹ m

Diameter =  1.19 mm = 1.19 x 10⁻³ m

Find:

Minimum angular spread (in rad)

Computation:

Minimum angular spread (in rad) = 1.22[Wavelength / Diameter]

Minimum angular spread (in rad) = 1.222[(534 x 10⁻⁹) / (1.19 x 10⁻³)]

Minimum angular spread (in rad) = 2[448.73 x 10⁻⁶]

Minimum angular spread (in rad) = 547.45 x 10⁻⁶ rad

Complete question:

A transverse wave on a rope is given by [tex]y \ (x, \ t) = (0.75 \ cm) \ cos \ \pi[(0.400 \ cm^{-1}) x + (250 \ s^{-1})t][/tex]. The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.

Answer:

The tension on the rope is 1.95 N

Explanation:

The general equation of a progressive wave is given as;

[tex]y \ (x,t) = A \ cos(kx \ + \omega t)[/tex]

Compare the given equation with the general equation of wave, the following parameters will be deduced.

A = 0.75 cm

k = 0.400π cm⁻¹

ω = 250π s⁻¹

The frequency of the wave is calculated as;

ω = 2πf

2πf = 250π

2f = 250

f = 250/2

f = 125 Hz

The wavelength of the wave is calculated as;

[tex]\lambda = \frac{2\pi}{k} \\\\\lambda = \frac{2\pi }{0.4 \pi} = 5 \ cm = 0.05 \ m[/tex]

The velocity of the wave is calculated as;

v = fλ

v = 125 x 0.05

v = 6.25 m/s

The tension on the rope is calculated as;

[tex]v = \sqrt{\frac{T}{\mu}} \\\\where;\\\\T \ is \ the \ tension \ of \ the \ rope\\\\\mu \ is \ the \ mass \ per \ unit \ length = 0.05 \ kg/m\\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (6.25)^2\times (0.05)\\\\T = 1.95 \ N[/tex]

Therefore, the tension on the rope is 1.95 N

Answer:

E = m c^2 = 2.5 * (3 * 10E8)^2 = 2.25 * 10E17 Joules

Answer:

The answer is D. 2.25 × 1017 J

Explanation:

got it right on edge 2021

Answer:

x ’= 368.61 m,  y ’= 258.11 m

Explanation:

To solve this problem we must find the projections of the point on the new vectors of the rotated system  θ = 35º

            x’= R cos 35

            y’= R sin 35

The modulus vector can be found using the Pythagorean theorem

            R² = x² + y²

            R = 450 m

we calculate

            x ’= 450 cos 35

            x ’= 368.61 m

            y ’= 450 sin 35

            y ’= 258.11 m

Answer:

7.5 kg

Explanation:

We are given that

[tex]m_1=10 kg[/tex]

Length of plank, l=3 m

Distance of fulcrum from one end of the plank=1 m

[tex]m_2=20 kg[/tex]

We have to find the mass must be on the other end if the plank remains balanced.

Let m be the mass must be on the other end if the plank remains balanced.

In balance condition

[tex]20\times 1=10\times (1.5-1)+m\times (1.5+0.5)[/tex]

[tex]20=10(0.5)+2m[/tex]

[tex]20=5+2m[/tex]

[tex]2m=20-5=15[/tex]

[tex]\implies m=\frac{15}{2}[/tex]

[tex]m=7.5 kg[/tex]

Hence, mass 7.5 kg   must be on the other end if the plank remains balanced.

Answer:

The mass at the other end is 7.5 kg.

Explanation:

Let the mass is m.

Take the moments about the fulcrum.

20 x 1 = 10 x 0.5 + m x 2

20 = 5 + 2 m

2 m = 15

m = 7.5 kg

Answer:

105.62°F

Explanation:

When the body temperature having fever is measured to be 40.9 on a scale then it must be a Celsius scale thermometer because 37°C is the normal temperature of a healthy human. In case of fever the given temperature is measured on a standard Celsius scale.

The relation between Fahrenheit and Celsius scale is:

[tex]\frac{C}{5}=\frac{F-32}{9}[/tex]

[tex]F=\frac{9C}{5} +32[/tex]

[tex]F=105.62^{o}F[/tex]

It is a high fever and an immediate medical help must be taken.

Answer:

the answer to the question is known as D

Answer:

Step down transformers are used in power adaptors and rectifiers to efficiently decrease the voltage. They are also used in electronic SMPS.

Explanation:

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Answer:

Explanation:

a)

The force on electron acts opposite to the velocity , and direction of force on electron is always opposite to direction of electric field .

Hence direction of electric field must be in the same  in which electrons travels.

Hence option iii is correct.

b )

deceleration a = force / mass

= qE / m

= 1.6 x 10⁻¹⁶ x 5.6 x 10⁵ / 9.1 x 10⁻³¹

= .98 x 10²⁰ m /s²

v² = u² - 2 a s

0 = (8.06 x 10⁶ )² - 2 x .98 x 10²⁰ s

s = 64.96 x 10¹² / 1.96 x 10²⁰

= 33.14 x 10⁻⁸ m

c ) time required

= 8.06 x 10⁶ / .98 x 10²⁰

= 8.22 x 10⁻¹² s .

d ) Its speed will be same as that in the beginning ie 8.06 x 10⁶ m/s .

Answer:

(a) Option (i)

(b) 6.6 x 10^-4 m  

(c) 8.2 x 10^-11 s

Explanation:

initial velocity, u = 8 .06 x 10^6 m/s

Electric field, E = 5.6 x 10^5 N/C

(a) The direction of field is opposite.

Option (i).

(b) Let the distance is s.  

Use third equation of motion

[tex]v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 \times \frac{qE}{m}\times s\\\\8.06\times 10^6\times 8.06\times 10^6 = \frac {1.6\times 10^{-19}\times 5.6\times 10^5}{9.1\times 10^{-31}} s\\\\s = 6.6\times 10^{-4} m[/tex]

(c) Let the time is t.

Use first equation of motion.

[tex]v = u + a t \\\\0 = u - \times \frac{qE}{m}\times t\\\\8.06\times 10^6 = \frac {1.6\times 10^{-19}\times 5.6\times 10^5}{9.1\times 10^{-31}} t\\\\t = 8.2\times 10^{-11} s[/tex]

Answer:

1) [tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex], 2) [tex]T_{B} \approx 1.137\cdot T_{A}[/tex], where [tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex].

Explanation:

1) Pendulum A is a simple pendulum, whose period ([tex]T_{A}[/tex]) is determined by the following formula:

[tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex] (1)

Where:

[tex]l[/tex] - Length of the massless bar.

[tex]g[/tex] - Gravitational acceleration.

2) Pendulum B is a physical pendulum, whose period ([tex]T_{B}[/tex]) is determined by the following formula:

[tex]T_{B} = 2\pi \cdot \sqrt{\frac{I_{O}}{m\cdot g\cdot l} }[/tex] (2)

Where:

[tex]m[/tex] - Total mass of the pendulum.

[tex]g[/tex] - Gravitational acceleration.

[tex]l[/tex] - Length of the uniform bar.

[tex]I_{O}[/tex] - Moment of inertia of the pendulum with respect to its suspension axis.

The moment of inertia can be found by applying the formulae of the moment of inertia for a particle and the uniform bar and Steiner's Theorem:

[tex]I_{O} = \frac{1}{2} \cdot m\cdot l^{2}+\frac{1}{24}\cdot m\cdot l^{2} + \frac{3}{4}\cdot m\cdot l^{2}[/tex]

[tex]I_{O} = \frac{31}{24}\cdot m\cdot l^{2}[/tex] (3)

By applying (3) in (2) we get the following expression:

[tex]T_{B} = 2\pi \cdot \sqrt{\frac{\frac{31}{24}\cdot m \cdot l^{2} }{m\cdot g \cdot l} }[/tex]

[tex]T_{B} = 2\pi \cdot \sqrt{\frac{31\cdot l}{24\cdot g} }[/tex]

[tex]T_{B} = \sqrt{\frac{31}{24} } \cdot \left(2\pi \cdot \sqrt{\frac{l}{g} }\right)[/tex]

[tex]T_{B} \approx 1.137\cdot T_{A}[/tex]

1. The period of pendulum A for small oscillations is  

[tex]T_A=2\pi\sqrt{\dfrac{L}{g}}[/tex]

2. The period of pendulum B for small oscillations.

[tex]T_B=1.137.T_A[/tex]

Simple harmonic motion is the periodic motion or back and forth motion of any object with respect to its equilibrium or mean position. The restoring force is always acting on the object which try to bring it to the equilibrium.

1) Pendulum A is a simple pendulum, whose period () is determined by the following formula:

[tex]T_A=2\pi\sqrt{\dfrac{L}{g}}[/tex]

Where:

l - Length of the massless bar.

g - Gravitational acceleration.

2) Pendulum B is a physical pendulum, whose period () is determined by the following formula:

[tex]T_A=2\pi\sqrt{\dfrac{I_o}{mgl}}[/tex] .............................2

Where:

m - Total mass of the pendulum.

g - Gravitational acceleration.

l - Length of the uniform bar.

Io- Moment of inertia of the pendulum with respect to its suspension axis.

The moment of inertia can be found by applying the formulae of the moment of inertia for a particle and the uniform bar and Steiner's Theorem:

[tex]I_o=\dfrac{1}{2}ml^2+\dfrac{1}{24}ml^2+\dfrac{3}{4}ml^2[/tex]

[tex]I_o=\dfrac{31}{24}ml^2[/tex]..................................3

By applying (3) in (2) we get the following expression:

[tex]T_B=2\pi\sqrt{\dfrac{\frac{31}{24}ml^2}{mgl}[/tex]

[tex]T_B=2\pi\sqrt{\dfrac{31l}{24g}}[/tex]

[tex]T_B=\sqrt{\dfrac{31}{24}}. (2\pi\sqrt{\dfrac{l}{g}})[/tex]

[tex]TB=1.137.T_A[/tex]

Thus to know more about Simple harmomnic motion follow

https://brainly.com/question/17315536

Answer:

I am pretty sure it is C

Explanation:

It can be found all over the universe

Answer:

KE = 2800 J

Explanation:

Usually a velocity is expressed as m/s. Then the energy units are joules.

[tex]\frac{108 km}{hr} * \frac{1000m}{1 km} * \frac{1 hour}{3600 seconds} =\frac{108*1000 m}{3600sec}[/tex]

v = 30 m / sec

KE = 1/2 * 4 * (30)^2

KE =2800 kg m^2/sec^2

KE = 2800 Joules

Answer:

D

The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.

Explanation:

Since block moves with constant speed

So, frictional force

f = FCosq

Work done by friction

W = - fd

W = - fd Cos q

The answer may be greater or less than - fdSinq. It depends on the value of q which is less than, or equal to 45°.

Answer:

because minerals can be gotten from the bottom

Explanation:

it's self explanatory

Answer:

Average Velocity = 3.63 m/s

Explanation:

First, we will calculate the total displacement of the quarterback, taking forward direction as positive:

Total Displacement = 15 m - 3 m + 24 m = 36 m

Now, we will calculate the total time taken for this displacement:

Total Time = 3 s + 1.71 s + 5.2 s = 9.91 s

Therefore, the average velocity will be:

[tex]Average\ Velocity = \frac{Total\ Displacement}{Total\ Time}\\\\Average\ Velocity = \frac{36\ m}{9.91\ s}[/tex]

Average Velocity = 3.63 m/s

Answer:

30J

Explanation:

Given data

The total quantity of heat recieved= 50J

Quantity of heat used to do work= 20J

Hence the net change is

ΔU= Total Heat - Net work

ΔU= 50-20

ΔU= 30J

Hence the change in the internal energy is 30J

Explanation:

Given:

L = 0.02 H

C = [tex]2\:\mu \text{F}[/tex]

f = 200 Hz

The general form of the impedance Z is given by

[tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]

Since this is a purely inductive/capacitive circuit, R = 0 so Z reduces to

[tex]Z = \sqrt{(X_L - X_C)^2} = \sqrt{\left(\omega L - \dfrac{1}{\omega C} \right)^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{\left(2 \pi L - \dfrac{1}{2 \pi f C} \right)^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{\left[2 \pi (200\:\text{Hz})(0.02\:\text{H}) - \dfrac{1}{2 \pi (200\:\text{Hz})(2×10^{-6}\:\text{F})} \right]^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{(25.13\:\text{ohms} - 397.89\:\text{ohms})^2}[/tex]

[tex]\:\:\:\:\:\:\:=372.66\:\text{ohms}[/tex]

Answer:

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Explanation:

haha

Answer:

in t seconds, Car A sweep out t radian { i.e θ = t radian }

Explanation:

Given the data in the question;

4 toy racecars are racing along a circular race track.

They all start at 3 o'clock position and moved CCW

Car A is constantly 2 feet from the center of the race track and moves at a constant speed

so maximum distance from the center = 2 ft

The angle Car A sweeps out increases at a constant rate of 1 radian per second.

Rate of change of angle = dθ/dt = 1

Now,

since dθ/dt = 1

Hence θ = t + C

where C is the constant of integration

so at t = 0, θ = 0, the value of C will be 0.

Hence, θ = t radian

Therefore, in t seconds, Car A sweep out t radian { i.e θ = t radian }

Answer:

150 N

Explanation:

Given that,

We are asked to calculate force required.

[tex]\longrightarrow[/tex] F = ma

[tex]\longrightarrow[/tex] F = (50 × 3) N

[tex]\longrightarrow[/tex] F = 150 N

Answer:

Explanation:

When the path difference is equal to wave length or its integral multiple, constructive interference occurs . If it is odd multiple of half wave length , then destructive interference occurs.

For constructive interference , path diff = n λ

For destructive interference path diff = ( 2n+ 1 ) λ /2

where λ is wave length of wave , n is an integer.

a )

path diff = 10 cm which is half the wavelength , so maximum destructive interference will occur.

b )

path diff = 15 cm which is neither  half the wavelength nor full wavelength , so in between is the right option.

c )

path diff = 20 cm which is equal to  the wavelength , so maximum constructive  interference will occur.

d)

path diff = 30 cm which is 3 times half the wavelength , so maximum destructive interference will occur.

e)

path diff = 35 cm which is neither integral multiple of half the wavelength , nor integral multiple of wavelength so in between is th eright answer.

f )

path diff = 40 cm which is 2 times the wavelength , so maximum constructive  interference will occur

Answer:

Explanation:

This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is

[tex][m_1v_1+m_2v_2]_b=[(m_1+m_2)v]_a[/tex]  Filling in:

[tex][1200(4.5)+2100(0)]=[(1200+2100)v][/tex] which simplifies to

5400 + 0 = 3300v

so v = 1.6 m/s to the east, choice B