The coefficients of static and kinetic friction between a 50.0-kg box and a horizontal surface are 0.500 and 0.400 respectively. (a) What is the acceleration of the object if a 250-N horizontal force is applied to the box

Answers

Answer 1

Answer:

1.08 m/s^2

Explanation:

The maximum value of static friction F_s = μ_s mg

where μ_s = coefficient of static friction

m= mass of box= 50kg

g= 9.81 m/s^2 , acceleration due to gravity

F_s = 0.5×50×9.81 =245 N

Since, the applied force is greater than the maximum value of static friction F_s, the object must be in motion. Hence, kinetic friction must be taken into account and not static friction.

By equilibrium condition

ma = F - μ_k m g

where F = 250 N,  μ_k = coefficient of kinetic friction = 0.4, m= 50 Kg

a = acceleration of the box

50a = 250-0.4×50×9.81

Solving we get a=  1.08 m/s^2


Related Questions

Sound energy travels through matter in waves. Do sound waves travel faster through air or water? Explain your answer.

Answers

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wpazsebu

Question 2 of 25

Which symbol in the first law of thermodynamics represents the result of a

gas expanding against a constant pressure?

O A. a

B. V

C. U

D. W

Answers

Answer: The symbol representing  the result of a  gas expanding against a constant pressure is W

Explanation:

According to first law of thermodynamics, energy can neither be destroyed nor created but rather transferred to one place to another and converted to and from other energy forms.

The mathematical form of first law of thermodynamics is:

[tex]\Delta U=q+W[/tex]

where [tex]\Delta U[/tex] = change in internal energy

q = heat absorbed or released

W = work  [tex](W= P\Delta V)[/tex]

P = pressure

[tex]\Delta V[/tex] = volume change

Thus the changes in heat or work leads to change in internal energy.

Can someone please help

Answers

ummmmmmmmmmm thats confusing

tlo
A Cu rod whose length at 80 degree is 10.0 metres is heated at 50 degree find the new length take x-for cu as 0.000 to 17 k - 1​

Answers

Answer:

9.949 m

Explanation:

From the question,

L' = L+LαΔT................. Equation 1

Where L' = New length, L = Original length, α = linear expansivity of Cu, ΔT = change in temperature

Given: L = 10 m, α  = 0.00017 K⁻¹, ΔT = 50-80

L' = 10+10(0.00017)(50-80)

L' = 10-0.051

L' = 9.949 m

Hence the new legth of Cu is  9.949 m

when an object slides over a smooth horizontal surface, how does the force of friction depend on the surface area of blocks that's are in contact with the table?​

Answers

Answer: with a greater surface area, there will be a greater force of friction

Explanation:

An organ pipe is 1.6 m long and open at both ends. What are the first three harmonic frequencies of this pipe?

Answers

Answer:

first harmonic = 56.7 Hz

Third harmonic = 170 Hz

Fifth harmonic = 283.3 Hz

Explanation:

A balloon rubbed against a pair of jeans gains a charge of -6x10^-6 C. If the balloon is moved to a distance of 0.25 m away from the jeans, what is the size and direction of the electric force between the balloon and jeans ?

Answers

Answer: 5.184 N

Explanation:

Given

The charge on the balloon is [tex]q_1=-6\times 10^{-6}\ C[/tex]

distance between balloon and jeans is [tex]r=0.25\ m[/tex]

After rubbing jeans must acquire the same opposite charge i.e.

[tex]q_2=6\times 10^{-6}\ C[/tex]

The electrostatic force between them

[tex]\Rightarrow F=\dfrac{kq_1q_2}{r^2}\\\\\Rightarrow F=\dfrac{9\times 10^9\times 6\times 10^{-6}\times 6\times 10^{-6}}{0.25^2}\\\\ \Rightarrow F=4\times 4\times 9\times 36\times 10^{-3}=5.184\ N[/tex]

nature of force will be attractive i.e. Jeans will try to pull the balloon

The teacher asked the children to turn the scales sideways and adjust the scales to zero. Jason pushes on the top of the scales and Kerry pushes on the base. They hold the scales still. Jason’s push measures 80 N. What is the size of Kerry’s push? *

Answers

Answer: J

Explanation:

Jason pushes on the top of the scales and Kerry pushes on the base. They hold the scales still. Jason's push measures 80N. What is the size of Kerry's push?

he pushes zero

A man with a weight of 450 N climbs a ladder to a height of 5.2 meters. How much work did he do?​

Answers

Answer:

the work done by him is 2,340 joules

Explanation:

The computation of the work done by him is shown below:

= Weight of man × height

= 450 N × 5.2 meters

= 2,340 joules

Hence, the work done by him is 2,340 joules

We simply multiply the man weight with the height

Score
A 4400 Kg Track Travelling at Intial speed 52m\s can be stopeed in 42 sec. By Gently Break, also the track can be stoped
in 7.6 m\s if the driver hit the wall.
a. What is the Impulse Excerted on the Vehicle?
b. Whta is the Averge force is exeted on the Track in each stops ?

Answers

Answer:

(a) J = 10560  kg-m/s (b) 251.42 N

Explanation:

Given that,

Mass, m = 4400 kg

Initial speed, u = 5.2 m/s

Final speed, v = 7.6 m/s

Time, t = 42 s

(a) Let J be the impulse exerted on the vehicle. Impuse is equal to the change in momentum such as :

J = m(v-u)

J = 4400 (7.6-5.2)

J = 10560  kg-m/s

(b) Impulse = Force × t

[tex]F=\dfrac{J}{t}\\\\F=\dfrac{10560}{42}\\F=251.42\ N[/tex]

Hence, this is the required solution.

Which type of biological molecule would contain fats?
A) Amino Acids.
C) Nucleic Acids.
B) Lipids.
D) Carbohydrates.

Answers

B

Explanation:

lipids contains fat

hope it helps

How much Work is done when a 5 kg object is picked up a distance of 8 m.

Answers

Answer:

392J?

Explanation:

what happens to the loudness of the sound as the amplitude increases

Answers

it increases and gets louder

and it amplifies the noise

A 21 kg mountain lion carries a 3kg cub in it's mouth as it jumps from rest on the ground to the top of a 2 m talk rock. It takes 1 seconds for the mountain lion to jump and reach the top. How much power did the mountain lion exert? I need help to solve for power

Answers

Answer:

The power exerted by the mountain lion is 1,472.35 W.

Explanation:

Given;

mass of mountain, m₁ = 21 kg

mass of the cub, m₂ = 3 kg

height jumped by the mountain lion, h = 2 m

time taken for the mountain lion to jump, t = 1 s

Determine the weight of the lions on the top rock;

W = F = (m₁ + m₂)g

F = (21 + 3) x 9.8

F = (24) x 9.8

F = 235.2 N

Determine the final velocity of the mountain rock as it jumped to the top;

v² = u² + 2gh

where;

u is the initial velocity = 0

h is the height jumped = 2 m

v² = 0 + 2 x 9.8 x 2

v² = 39.2

v = √39.2

v = 6.26 m/s

The power exerted by the mountain lion is calculated as;

P = Fv

P = 235.2 x 6.26

P = 1,472.35 W

Therefore, the power exerted by the mountain lion is 1,472.35 W.

Fifty grams of ice at 0◦ C is placed in a thermos bottle containing one hundred grams of water
at 6◦ C. How many grams of ice will melt? The heat of fusion of water is 333 kJ/kg and the
specific heat is 4190 J/kg · K.Immersive Reader

Answers

Answer:

7.55 g

Explanation:

Given that:

Heat of fusion = 333kj/kg

Heat capacity, c = 4190 j/kg /k

The Number of grams of ice that will melt can be represented as y:

Number of grams of ice that will melt * heat of fusion = specific heat capacity * temperature change

y * 333 * 10^3 J = (4190) * (6 - 0)

333000y = 25140

y = 25140 / 333000

y = 0.0754954 kg

y = 0.0754954 * 100

y = 7.549 g

Hence, Number of grams of ice that will melt = 7.55 g

Suppose you are standing at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward. You have a compass that is free to swivel in any direction. Which way does your compass point? Suppose you are standing at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward. You have a compass that is free to swivel in any direction. Which way does your compass point? It would point up. It would point east. It would point down. It would point west.

Answers

Answer:

It would point up.

Explanation:

Since I am at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward, the north pole of the compass would also point towards the earth's geographic north magnetic pole, since all other compasses point toward there.

Since the compass is free to swivel in any direction, the compass would point up, since it is at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward.

So, the compass would point up.

The filament in the bulb is moving back and forth, first pushed one way and then the other. What does this imply about the current in the filament? Select the best answer from the choices provided. View Available Hint(s) Select the best answer from the choices provided. The current is turning on and off. The magnitude of the current is varying with time. The current switches direction, going one way and then the other.

Answers

Answer:

the correct answer is: The magnitude of the current varies with time.

Explanation:

The movement of a filament by a current is due to the electric force when electrons accumulate on one side attract positively charged nuclei.

In this case the voters in a period of time are on one side, let's call it forward and in another period of time it accumulates on the other side, let's call it back, this explains the very small movement of the filament, therefore we have a current that varies over time.

The other two options are not correct because turning off the current the filament goes to the central position and does not go back, the other explanation the current goes in one direction or another, the voters are always on the same side therefore the filament only goes in a sense.

Consequently the correct answer is: The magnitude of the current varies with time.

Which type of land usage would increase as the population increases?

Answers

Answer:

If the population were to INCREASE land use would also INCREASE, if the human population DECREASED so would land use because there wouldn’t be as many people to using it. hope i helped

Answer:

D. Urban use

Explanation:

It shows here on this graph that overtime urban use is rising as the population is increasing! The other answers that are on this graph does not show it increasing!

why do fishermen like fishing during a full moon?

Answers

Answer:

The full moon attracts more fish towards the surface of the water.

Explanation:

It is a proven fact that fish are attracted to shiny/glowing items. for example, a metal fishing lure. Or, an angler fish light-bulb.

An object with mass 100 kg moved in outer space. When it was at location <8, -30, -4> its speed was 5.5 m/s. A single constant force <220, 460, -200> N acted on the object while the object moved from location <8, -30, -4> m to location <14, -21, -7> m. Then a different single constant force <100, 260, 210> N acted on the object while the object moved from location <14, -21, -7> m to location <17, -27, -3> m. What is the speed of the object at this final location

Answers

Answer:

v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s

Explanation:

We can solve this problem using the kinematic relations, we have a three-dimensional movement, but we can work as three one-dimensional movements where the only parameter in common is time (a scalar).

X axis.

They indicate the initial position x = 8 m, its initial velocity v₀ = 5.5 m / s, the force Fx₁ = 220 N x₁ = 14 m, now the force changes to Fx₂ = 100 N up to the point xf = 17 m. The final speed is wondered.

As this movement is in three dimensions we must find the projection of the initial velocity in each axis, for this we can use trigonometry

the angle fi is with respect to the in z and the angle theta with respect to the x axis.

               sin φ = z / r

                Cos φ = r_p / r

               z = r sin φ

               r_p = r cos φ

the modulus of the vector r can be found with the Pythagorean theorem

               r² = (x-x₀) ² + (y-y₀) ² + (z-z₀) ²

               r² = (14-8) 2 + (-21 + 30) 2+ (-7 +4) 2

               r = √126

               r = 11.23 m

Let's find the angle with respect to the z axis (φfi)

                φ = sin⁻¹ z / r

                φ = sin⁻¹ ( [tex]\frac{-7+4}{11.23}[/tex] )

                φ = 15.5º

Let's find the projection of the position vector (r_p)

                r_p = r cos φ

                r_p = 11.23 cos 15.5

                r_p = 10.82 m

This vector is in the xy plane, so we can use trigonometry to find the angle with respect to the x axis.

                 cos θ = x / r_p

                 θ = cos⁻¹ x / r_p

                 θ = cos⁻¹ ( [tex]\frac{14-8}{10.82}[/tex])  

                 θ = 56.3º

taking the angles we can decompose the initial velocity.

               sin φ = v_z / v₀

               cos φ = v_p / v₀

               v_z = v₀ sin φ

               v_z = 5.5 sin 15.5 = 1.47 m / z

               v_p = vo cos φ

               v_p = 5.5 cos 15.5 = 5.30 m / s

                 

               cos θ = vₓ / v_p

                sin θ = v_y / v_p

                vₓ = v_p cos θ

                v_y = v_p sin θ

                vₓ = 5.30 cos 56.3 = 2.94 m / s

                v_y = 5.30 sin 56.3 = 4.41 m / s

 

                 

we already have the components of the initial velocity

                v₀ = (2.94 i ^ + 4.41 j ^ + 1.47 k ^) m / s

let's find the acceleration on this axis (ax1) using Newton's second law

                Fₓx = m aₓ₁

                aₓ₁ = Fₓ / m

                aₓ₁ = 220/100

                aₓ₁ = 2.20 m / s²

Let's look for the velocity at the end of this interval (vx1)

Let's be careful if the initial velocity and they relate it has the same sense it must be added, but if the velocity and acceleration have the opposite direction it must be subtracted.

                 vₓ₁² = v₀ₓ² + 2 aₓ₁ (x₁-x₀)

                 

let's calculate

                 vₓ₁² = 2.94² + 2 2.20 (14-8)

                 vₓ₁ = √35.04

                 vₓ₁ = 5.92 m / s

to the second interval

they relate it to xf

                   aₓ₂ = Fₓ₂ / m

                   aₓ₂ = 100/100

                   aₓ₂ = 1 m / s²

final speed

                    v_{xf}²  = vₓ₁² + 2 aₓ₂ (x_f- x₁)

                    v_{xf}² = 5.92² + 2 1 (17-14)

                    v_{xf} =√41.05

                    v_{xf} = 6.41 m / s

We carry out the same calculation for each of the other axes.

Axis y

acceleration (a_{y1})

                      a_{y1} = F_y / m

                      a_{y1} = 460/100

                      a_[y1} = 4.60 m / s²

the velocity at the end of the interval (v_{y1})

                      v_{y1}² = v_{oy}² + 2 a_{y1{ (y₁ -y₀)

                      v_{y1}2 = 4.41² + 2 4.60 (-21 + 30)

                      v_{y1} = √102.25

                       v_{y1} = 10.11 m / s

second interval

acceleration (a_{y2})

                      a_{y2} = F_{y2} / m

                      a_{y2} = 260/100

                      a_{y2} = 2.60 m / s2

final speed

                     v_{yf}² = v_{y1}² + 2 a_{y2} (y₂ -y₁)

                     v_{yf}² = 10.11² + 2 2.60 (-27 + 21)

                      v_{yf} = √ 71.01

                      v_{yf} = 8.43 m / s

here there is an inconsistency in the problem if the body is at y₁ = -27m and passes the position y_f = -21 m with the relationship it must be contrary to the velocity

z axis

 

first interval, relate (a_{z1})

                      a_{z1} = F_{z1} / m

                      a_{z1} = -200/100

                      a_{z1} = -2 m / s

the negative sign indicates that the acceleration is the negative direction of the z axis

the speed at the end of the interval

                    v_{z1}² = v_{zo)² + 2 a_{z1} (z₁-z₀)

                    v_{z1}² = 1.47² + 2 (-2) (-7 + 4)

                    v_{z1} = √14.16

                    v_{z1} = -3.76 m / s

second interval, acceleration (a_{z2})

                    a_{z2} = F_{z2} / m

                    a_{z2} = 210/100

                    a_{z2} = 2.10 m / s2

final speed

                    v_{fz}² = v_{z1}² - 2 a_{z2} | z_f-z₁)

                    v_{fz}² = 3.14² - 2 2.10 (-3 + 7)

                     v_{fz} = √6.94

                     v_{fz} = 2.63 m / s

speed is     v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s

Which two options are forms of kinetic energy?
A. Thermal energy
OO
O B. Nuclear energy
c. Sound energy
O D. Gravitational energy
O E. Chemical energy

Answers

Answer:

Themal energy is the best answer from you

Sound EnergyThermal Energy

Light undergoes refraction as it travels from water to glass. The refractive index of water = 1.33 and refractive index of glass = 1.5. Calculate the angle of refraction inside glass if angle of incidence in the water is 32 degrees

Answers

Answer:

28°

Explanation:

Snell's Law and equation

Which three terms are needed to describe the energy a BASE Jumper has as
she falls toward the ground?
A. Potential
B. Kinetic
C. Gravitational
D. Electromagnetic

Answers

A Potential (as she is off the ground she has gravitational potential energy)
B kinetic (because she’s moving down)
and
C gravitational (as gravity is pulling her back down)

Answer:

A.Potential

B.Kinetic

C.Gravitational

Explanation:

I took the test

How many electron flow through a light bulb each second if the current flow through the light bulb 0.75A.The electric charge of one electron is 1.6 x 10-19C

Answers

Answer:

[tex]n=4.68\times 10^{18}[/tex]

Explanation:

The current through the bulb, I = 0.75 A

We need to find the number of electrons flowing per second. We know that the electric current is given by :

[tex]I=\dfrac{ne}{t}\\\\n=\dfrac{It}{e}\\\\n=\dfrac{0.75\times 1}{1.6\times 10^{-19}}\\\\n=4.68\times 10^{18}[/tex]

So, there are [tex]4.68\times 10^{18}[/tex] electrons flowing per second.

There are "[tex]4.68\times 10^{18}[/tex]" electron flowing per second.

Electric charge:

Whenever retained inside an electric as well as the magnetic field, this same basic physical attribute of matter which thus permits it to perceive a force, is considered as Electric charge.

According to the question,

Current flow, I = 0.75 A

Electric charge of one electron, e = 1.6 x 10⁻¹⁹ C

As we know the relation,

Current, I = [tex]\frac{ne}{t}[/tex]

or,

→ n = [tex]\frac{It}{e}[/tex]

By substituting the given values in the above formula, we get

      = [tex]\frac{0.75\times 1}{1.6\times 10^{-19}}[/tex]

      = [tex]4.68\times 10^{18}[/tex]

Thus the approach above is appropriate.

Find out more information about electric charge here:

https://brainly.com/question/14372859

3. The car's mass is 400 kg. It moves at a velocity of 20 m/s. Calculate the car's momentum. *
(10 Points)
0.05 kg.m/s
8000 kg.m/s
80,000 kg.m/s
20 kg.m/s

Answers

Answer:

momentum=mass×velocity

momentum =400kg×20m/s=8000kg.m/s

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 ×102 T, what is the number of turns per meter for this solenoid?

Answers

Correct question:

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?

Answer:

the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

Explanation:

Given;

length of solenoid, L= 0.35 m

diameter of the solenoid, d = 0.04 m

current through the solenoid, I = 5.0 A

magnetic field in the center of the solenoid, 2.8 x 10⁻² T

The number of turns per meter for the solenoid is calculated as follows;

[tex]B =\mu_o I(\frac{N}{L} )\\\\B = \mu_o I(n)\\\\n = \frac{B}{\mu_o I} \\\\n = \frac{2.8 \times 10^{-2}}{4 \pi \times 10^{-7} \times 5.0} \\\\n = 4.5 \times 10^3 \ turns/m[/tex]

Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

Where do animals such as snakes and frogs, which do not make their own body heat, usually get their heat?

fuel
the sun
hot lava
friction

plssssssss answer correctly

Answers

The answer to this question is the Sun.

As im writing this, 12,826 people vote the Sun.

animals such as snakes and frogs, which do not make their own body heat. They usually get their heat from the sun. Hence option B is correct.

What is Cold-blooded animals ?

A body temperature that is only slightly higher than the ambient temperature. This condition differentiates cold-blooded, or homoiothermic, animals from fishes, amphibians, reptiles, and invertebrates (birds and mammals). Due of their reliance on external warmth for metabolic activity, terrestrial cold-blooded species are restricted to locations with temperatures ranging from 5-10° to 35-40° C (41-50° to 95-104° F).

Cold-blooded creatures cannot create their own body heat, but they may control it by modifying their surroundings. Alligators and other reptiles frequently lay in the sun to warm up. They, on the other hand, cool off by swimming, going into a burrow in the earth, or moving inside the sade of a rock.

Hence option B is correct.

To know more about Cold-blooded animals :

https://brainly.com/question/30914929

#SPJ2.

d. Two point charges, q1 = +25 nC and q2 = -75 nC, are separated by a distance of 3.0 cm. Find the magnitude and direction of; i. the electric force q1 exerts on q2 [5] ii. the force that q2 exerts on q1 [4] (take k = 9.0 x 109 N.m2 /C2 )

Answers

Answer:

a) F₂₁ = 0.02 N, attracting.

b) F₁₂ = 0.02 N, attracting.

Explanation:

a)

The magnitude of the force that q₁ exerts on q₂ (F₂₁) is given by Coulomb's Law, as follows:

      [tex]F_{21} = k * \frac{q_{1} *q_{2}}{r_{12}^{2} } = 9e9 N.m2/C2 * \frac{(25e-9C)*(75e-9C)}{(0.03m)^{2}} = 0.02 N (1)[/tex]

Since q₁ and q₂ have opposite signs, the force between them will be always attractive, i.e., from q₂ towards q₁, along the line that joins both charges.

b)

The magnitude of the force on q₁ due to q₂ can be obtained applying Newton's 3rd Law, or using (1), because all parameters are the same, so F₁₂ (in  magnitude) = F₂₁ = 0.02 NAs we have already said, it must be opposite to the one found in a) so it must go from q₁ towards q₂, it is an attracting force also.

A runner taking part in the 200-m dash must run around the end of a track that has a circular arc with a radius of curvature of 29.5 m. The runner starts the race at a constant speed. If she completes the 200-m dash in 24.4 s and runs at constant speed throughout the race, what is her centripetal acceleration as she runs the curved portion of the track

Answers

Answer: [tex]2.27\ m/s^2[/tex]

Explanation:

Given

Length of the race track [tex]L=200\ m[/tex]

the radius of curvature of the track [tex]r=29.5\ m[/tex]

time taken to run on track is [tex]t=24.4\ s[/tex]

Speed of runner is

[tex]\Rightarrow v=\dfrac{L}{t}=\dfrac{200}{24.4}\\\\\Rightarrow v=8.196\ m/s[/tex]

Centripetal acceleration is

[tex]\Rightarrow a_c=\dfrac{v^2}{r}=\dfrac{8.196^2}{29.5}\\\\\Rightarrow a_c=2.27\ m/s^2[/tex]

A sphere of uniform density with mass 15 kg and radius 0.6 m is spinning, making one complete revolution every 0.3 s. The center of mass of the sphere has a speed of 3 m/s. (a) What is the rotational kinetic energy of the sphere

Answers

Answer:

[tex]473.56\ \text{J}[/tex]

Explanation:

T = Time taken to complete one revolution = 0.3 s

m = Mass of sphere = 15 kg

r = Radius of sphere = 0.6 m

I = Moment of inertia of sphere = [tex]\dfrac{2}{5}mr^2[/tex]

Angular speed of the sphere is

[tex]\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{0.3}\\\Rightarrow \omega=20.94\ \text{rad/s}[/tex]

Rotational kinetic energy is given by

[tex]K_r=\dfrac{1}{2}I\omega^2=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2\\\Rightarrow K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 15\times 0.6^2\times 20.94^2\\\Rightarrow K_r=473.56\ \text{J}[/tex]

The rotational kinetic energy of the sphere is [tex]473.56\ \text{J}[/tex].

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