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nataly862011 [7]
3 years ago
11

The wall is 500 sq. feet. A gallon of paint covers 160 sq. feet. What is an appropriate conversion factor to help determine how

many gallons will be needed to paint the wall?
Physics
1 answer:
lana66690 [7]3 years ago
4 0

Answer:

Explanation:

Given

Wall is 500 ft^2 in area

160\ ft^2 requires 1 gallon of paint

so using unitary method

1\ ft^2 requires \frac{1}{160}   gallon of paint

500 ft^2 wall will require =500\times \frac{1}{160} gallons of paint

=3.0125\ gallons

   

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If the earth can travel 1800 km in 1 minute what is it’s speed in km/s
Lady bird [3.3K]

Answer:

30km/s

Explanation:

1800/60=30

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4 years ago
Which element is found in all the acids
lana66690 [7]
Hydrogen .When acids touches all metal hydrogen gas is emitted .Strong acids is one that can produce a high concentration of hydrogen ions.Hope this helped!
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3 years ago
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Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.245 A/s ,
Airida [17]

Answer:

Complete question:

c.If the current in the second coil increases at a rate of 0.365 A/s , what is the magnitude of the induced emf in the first coil?

a.M= 6.53\times10^{-3} H

b.flux through each turn = Ф = 4.08\times10^{-4} Wb

c.magnitude of the induced emf in the first coil = e= 2.38\times10^{-3} V

Explanation:

a. rate of current changing = \frac{di}{dt}=[tex]M=\frac{1.60\times10^{-3} V}{0.240\frac{A}{s} }}[/tex]

  Induced emf in the coil =e= 1.60\times10^{-3} V

  For mutual inductance in which change in flux in one coil induces emf in the second coil given by the farmula based on farady law

     e=-M\frac{di}{dt}

     M=\frac{e}{\frac{di}{dt} }

     M=\frac{1.60\times10^{-3} }{-0.245}

   M= 6.53\times10^{-3} H

b.

  Flux through each turn=?

  Current in the first coil =1.25 A

   Number of turns = 20

       using   MI = NФ

     flux through each turn = Ф =  \frac{6.53\times10^{-3}\times1.25}{20}

   flux through each turn = Ф = 4.08\times10^{-4} Wb

c.

   second coil increase at a rate = 0.365 A/s

  magnitude of the induced emf in the first coil =?

 using   e=-M\frac{di_{2} }{dt}

            e= 6.53\times10^{-3} \times 0.365

magnitude of the induced emf in the first coil = e= 2.38\times10^{-3} V

4 0
3 years ago
At a processing facility outside of Detroit, a 7.0 kg box of goat cheese, initially at rest, is given a push up a smooth, inclin
Firlakuza [10]

Answer:

d = 5.9 m

Explanation:

During the initial push, there are two forces acting on the box along  the ramp: the component of gravity force along the ramp (directed to the bottom of the ramp), and the pushing force (up the ramp).

As we have as a given, the time during which the pushing force is applied, we can use Newton's 2nd Law, expressed in its original form, as follows:

Fnet *Δt = Δp = m* (vf-v₀) (1)

Fnet, in this case, is the difference between Fapp, and the projection of Fg along the ramp, which is equal to Fg times the sinus of  the angle of the ramp with respect to the horizontal.

We choose as positive, the direction up the ramp, so we can write the follwing equation:

⇒ Fnet = Fapp - Fg*sin θ = 140 N - (7kg*9.8m/s²*sin 30º) = 140 N - 34.3 N

⇒ Fnet = 105.7 N

Replacing in (1):

105.7 N * 0.5 s = 7 kg* vf (v₀=0, as the box starts from rest)

Solving for vf:

vf = 105.7 N* 0.5 s / 7 kg = 7.6 m/s

When the push ends, the only force remaining along  the ramp, is the component of Fg that we have already obtained, that will cause the box to have a deceleration, which we can find out aplying Newton's 2nd Law, as follows:

m*g*sin 30º = m*a

As we have defined as positive direction the one up the ramp, a will be negative (as it is slowing down the box) , and can be calculated as follows:

a = -34.3 N / 7 kg = -4.9 m/s²

As this value is constant, we can use any kinematic equation in order to get  the distance traveled, farther the point where it disappeared the influence of the pushing force:

vf² - v₀² = 2*a*d

As we know that finally the box will come momentarily at rest (before falling under the influence of  gravity) , we have vf =0:

⇒ -v₀² = 2*a*d

For this part, v₀, is just the value for vf, that we got above:

v₀= 7.6 m/s

⇒ -(7.6)² =2*(-4.9 m/s²)*d

Solving finally for d (the answer we are looking for):

d = (7.6)² (m/s)² / 2*4.9 m/s² = 5.9 m

8 0
3 years ago
The work energy principle states that the change in kinetic energy of an object is equal to the net work done on the object. If
sattari [20]

Answer:

v=\sqrt{2gh}\ m/s

Explanation:

From work energy theorem

Work done by all forces = Change in kinetic energy

Lets take

m= mass of object

h=height from the ground surface

initial velocity of object = 0 m/s

The final velocity of object is v

Work done by gravitational force = m g . h

The final kinetic energy = 1/2 m v²

So

Work done by all forces = Change in kinetic energy

m g h =  1/2 m v² - 0

v² = 2 g h

v=\sqrt{2gh}\ m/s

6 0
4 years ago
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