A mover slides a refrigerator weighing 650 N at a constant velocity across the floor a distance of 8.1 m. The force of friction between the refrigerator and the floor is 230 N. How much work has been performed by the mover on the refrigerator?​

Answer:

The displacement is 6.4m

Explanation:

Step one:

given

we are told that Lucy runs 4 meters to the east,

then 5 meters south.

let the distance east be the displacement in the x-direction, and south be the y-direction

Step two:

The resultant of the x and y displacement is the magnitude of the total displacement z

applying Pythagoras theorem we have

z=√x^2+y^2

z=√4^2+5^2

z=√16+25

z=√41

z=6.4m

Answer:

Fx = 467.65N

Fy = 270N

Explanation:

Given

Force = 540N

angle of inclination = 40 degree

Horizontal component Fx = Fcos 30

Fx = 540cos30

Fx = 540(0.8660)

Fx = 467.65N

Hence the horizontal component is 467.65N

Vertical component Fy = Fsin 30

Fy = 540sin30

Fy = 540(0.5)

Fy = 270N

Hence the vertical component is 270N

Answer:10000000

Explanation:

Answer:

1.898 × 10^27 kg

Explanation:

thats how much it ways

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

[tex]r= \sqrt{\frac{Gm1m2}{F} }[/tex]

[tex]r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m[/tex]

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required:

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

[tex]r= \sqrt{\frac{Gm1m2}{F} }[/tex]

[tex]r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m[/tex]

Answer:

The distance between the compact car and pickup truck is 0.96048 m

Explanation:

The gravitational force is directly proportional to the product of the masses of the interacting object, it is also inversely proportional to the square of the distance between them.  This is shown in equation 1;

[tex]F =G \frac{m_{1} X m_{2} }{d^{2} }[/tex]............ 1

Where F is the gravitational force = 0.0001 N

G is the gravitational constant = 6.673 x [tex]10^{-11} Nm^{2} kg^{-2}[/tex]

[tex]m_{1}[/tex]  is the mass of the compact car = 900kg

[tex]m_{2}[/tex] is the mass of the pickup truck = 1600kg

d is the distance and its unknown ?

Let us make d the subject formula in equation 1

[tex]d = \sqrt{G\frac{m_{1} m_{2} }{F } }[/tex] .... 2

Substituting into equation 2 we have

[tex]d = \sqrt{\frac{6.673x10^{-11} x 900 x 1600}{0.0001N} }[/tex]

d = 0.96048m

Therefore the distance between the compact car and pickup truck is 0.96048 m

Answer:

a) [tex]H_{in}=39.34 kJ[/tex]

b) Efficiency=76.77%

Explanation:

a)

In order to solve this problem, we can use the following formula:

[tex]H_{in}=H_{out}+W[/tex]

the problem provides us with all the necessary information so we can directly use the formula:

[tex]H_{in}=9.14kJ+30.2kJ[/tex]

[tex]H_{in}=39.34 kJ[/tex]

b) In order to find the efficiency, we can use the following formula:

[tex]Efficiency=\frac{W}{H_{in}}*100\%[/tex]

so we get:

[tex]Efficiency=\frac{30.2kJ}{39.34kJ}*100\%[/tex]

Efficiency=76.77%

The element named boron can be used to enhance the properties of semiconductors.

A semiconductor is a material that has electronic properties and has the value that falls in between a conductor. It can be a metallic copper or an insulator.

The rise in temperatures leads to a fall in resistivity. The element named boron can improve the electrical properties of the semiconductor as they form the impurities.

Find out more information about the element.

brainly.com/question/12389810

Answer:

OXYGEN

Explanation:brainlyist me

Answer:

Nitrogen

Explanation:

Oxygen is second

Answer:

724.3J/Kg.K

Explanation:

CHECK THE COMPLETE QUESTION BELOW

Compute the specific heat capacity at constant volume of nitrogen (N2) gas.and compare with specific heat of liquid water. The molar mass of N2 is 28.0 g/mol.

The specific heat capacity can be computed by using expression below

c= CV/M

Where c= specific heat capacity

M= molar mass

CV= molar hear capacity

Nitrogen is a diatomic element, the Cv can be related to gas constant with 5/2R

Where R= 8.314J/mol.k

Molar mass= 28 ×10^-3Kg/mol

If we substitute to the expression, we have

c= (5R/2)/(M)

=5R/2 × 1/M

=(5×8.314) /(2×28 ×10^-3)

=724.3J/Kg.K

Hence, the specific heat capacity at constant volume of nitrogen (N2) gas is

724.3J/Kg.K

The specific heat of liquid water is about 4182 J/(K kg) which is among substance with high specific heat, therefore specific heat of Nitrogen gas is 724.3J/Kg.K which is low compare to that of liquid water.

Answer:

Decreasing

Hope this helps! :)

Answer:

It’s a

Explanation:

Don’t actually put that i needed the points mb

Split up the forces into components acting parallel to and perpendicular to the slope. See the attached picture for the reference axes.

The box stays on the surface of the plane, so that the net force acting perpendicular to it is 0, and the only acceleration is applied in the parallel direction.

Let m be the mass of the box, θ the angle the plane makes with the ground, and a the acceleration of the box. By Newton's second law, we have

• net parallel force

∑ Force (//) = W (//) - F = m a

(that is, the net force in the parallel direction is the sum of the parallel component of the weight W and the friction F which acts in the negative direction)

• net perpendicular force

∑ Force (⟂) = W (⟂) + N = 0

Notice that

W (//) = W sin(θ) … … … which is positive since it points down the plane

W (⟂) = -W cos(θ) … … … which is negative since it points opposite the normal force N

So the equations become

W sin(θ) - F = m a

-W cos(θ) + N = 0

Solving for a gives

a = (W sin(θ) - F ) / m

which is good enough if you know the magnitude of the friction force.

If you don't, you can write F in terms of the coefficient of kinetic friction between the box and plane, µ, as

F = µ N

so that

a = (W sin(θ) - µ N ) / m

and the normal force itself has a magnitude of

N = W cos(θ)

so that

a = (W sin(θ) - µ W cos(θ) ) / m

The weight W has magnitude m g, where g is the magnitude of the acceleration due to gravity, so

a = (m g sin(θ) - µ m g cos(θ) ) / m

a = g (sin(θ) - µ cos(θ))

Answer:

options B,C,D are true

Explanation:

Answer:

Velocity (m/s) over time (s) graph

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

Explanation:

What this formula is telling us is that if we know the acceleration of an object, and the ... we can plug in our acceleration of 12.5 m/s2 for a, and 4 seconds for t.

Velocity (m/s) over time (s) graph

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

Answer:

a)t  = 1,43 s

b) V = 10,49 m/s

c) V₀ₓ = 10,49 m/s   ;    V₀y = 14,01 m/s

d) Vf = 17,5 m/s

Explanation:

According to the problem statement

V₀ = V₀ₓ    and  V₀y = 0

And at the end of the movement t = ?  the distance y = 10 m

Therefore as

h = V₀y - (1/2)*g*t²

Vertical distance y = h = 10 = V₀y*t - 0,5 (-9,8)*t²

10 = 4,9*t²

t² = 10/4,9    ⇒  t² = 2,04 s

t  = 1,43 s

a) 1,43 s is the time of movement

b) V₀ = V₀ₓ        V₀y = 0     and  V₀ₓ = Vₓ     ( constant )

Just before touching the ground, the horizontal distance is

hd = 15 = Vₓ * t

Then  15 /1,43 = Vₓ = V₀ₓ

Vₓ = 10,49 m/s

Then initial speed is V = 10,49 m/s    since V₀y = 0

Vf² = Vₓ² + Vy²

Vyf = V₀y - g*t

Vyf =  0 - 9,8 *1,43

Vyf = - 14,01 m/s

And finally the speed when the projectile strike the ground is:

Vf² = Vₓ² + Vy²

Vf = √ (10,49)² + (14,01)²

Vf = 17,50 m/s

Answer:

The value is  [tex]p = 0.7556 c[/tex]

Explanation:

From the question we are told that

   The speed at which galaxy B moves away from galaxy A is  [tex]v = 0.577c[/tex]

Here c is the speed of light with value  [tex]c = 3.0 *10^{8} \ m/s[/tex]

     The speed at which galaxy C moves away from galaxy B is  [tex]u = 0.731 c[/tex]

Generally from the equation of  relative speed we have that  

     [tex]u = \frac{p - v}{ 1 - \frac{ p * v}{c^2} }[/tex]

Here p is the velocity at which galaxy C recede from galaxy A so

     [tex]0.731c = \frac{p - 0.577c }{ 1 - \frac{ p * 0.577c}{c^2} }[/tex]

=>   [tex]0.731c [1 - \frac{ p * 0.577}{c}] = p - 0.577c[/tex]

=>   [tex]0.731c - 0.4218 p = p - 0.577c[/tex]

=>   [tex]0.731c + 0.577c = p + 0.4218 p[/tex]

=>   [tex]1.308 c = 1.731 p[/tex]

=>    [tex]p = 0.7556 c[/tex]

Answer:

in a spontaneous process, the entropy of the universe increases.

Explanation:

Entropy is a measure of of the degree of  randomness or disorderliness in a system.

The second law of thermodynamics can be stated as follows; "in any spontaneous process, the entropy of the universe increases."  

The universe here refers to the system's disorder and the disorder of the surroundings.  Therefore, a spontaneous process can occur, in which the entropy of the system decreases, only if the entropy increases in the surroundings.

For instance, when ice freezes, the entropy of liquid water decreases, that is, the entropy of the system decreases. However, heat is given off to the surroundings and the entropy of the surroundings increases. This is an obvious expression of this law.

Answer:98

Explanation:hope this helps!