How do I find the specific heat?
I know that
Q= m x c x difference in T
Q= 1.61
M= 476
C=?
Difference in temperature = 20

Answer:

i). (a) in the same direction as , (d) lower

ii). (f) opposite the direction of, (g) higher

Explanation:

An proton may be defined as a sub atomic particle and it has a positive electrical charge. Its mass is slightly less than that of a neutron. When a proton is placed in an electrical field that is uniformly charged, it is at rest. When the proton first moves out from rest from the uniform electric field, it will move in a direction which is same as that of the electric field and it will move to a region of higher potential.

An electron is defined as the subatomic particle having negative electric charge. When an electron is released form rest from an uniform electric field, it will move in the opposite direction of the uniform electric field and will move to the region of lower electric potential.

Answer:

v = 4.0 m/s

Explanation:

       [tex]p_{f} = m_{w} * v_{w} + m_{g+w} *v_{g+w} = 0 (1)[/tex]

       [tex]v_{g+w} =- \frac{m_{w} *v_{w}}{m_{g+w} } = - \frac{36.0kg *6.5m/s}{58.0kg } = -4.0m/s (2)[/tex]

Answer:

t = 25.0 s

Explanation:

       [tex]v_{f}^{2} - v_{o}^{2} = 2*a* \Delta x (1)[/tex]

       [tex]v_{o} = 84.1 km/h*\frac{1h}{3600s} *\frac{1000m}{1km} = 23.4 m/s (2)[/tex]

      [tex]v_{f} = 15.8 km/h*\frac{1h}{3600s} *\frac{1000m}{1km} = 4.4 m/s (3)[/tex]

       [tex]a = \frac{(4.4m/s)^{2} - (23.4m/s)^{2}}{2*350m} = -0.76 m/s2 (4)[/tex]

       [tex]t =\frac{v_{f}-v_{o}}{a} = \frac{(4.4m/s)-(23.4m/s)}{-0.76m/s2} = 25.0 s (5)[/tex]

Answer:

Explanation:

Answer:

4units

Explanation:

To calculate the total distance the beam will travel along this path, you will use the formula for calculating the distance between two coordinates expressed as;

D = √(x2-x1)²+(y2-y1)²

Given the coordinate points

(3,5) and (7,5)

Substitute

D = √(7-3)²+(5-5)²

D = √(7-3)²+0²

D = √4²

D = √16

D = 4

Hence the total distance the beam will travel along this path is 4units

Answer:

Explanation:

Power, by definition, is the amount of work per unit of time. We arent given work in this question, but we can find it because work is how much force per unit of distance.

[tex]P=\frac{W}{t_f-t_i} =\frac{F*d}{t_f-t_i}[/tex]

Plug in all the values, and its algebra at this point

[tex]500=\frac{F*96}{12}[/tex]

6000 = 96F

F = 62.5 Newtons

The force which is being applied to the object with the power of 500 Watt is 62.5 N.

Power can be defined as the rate of doing work. It is the work done in unit time. The SI unit of power is Watt (W) which is equal to joules per second (J/s). Sometimes, the power of motor vehicles and other machines is given in terms of Horsepower (hp), This unit is approximately equal to 745.7 watts of power.

The power of a system can be calculated as the product of force applied and distance travelled by the system per unit time taken.

Therefore, the force of the object with power 500W can be calculated as:

P = f × d/ t

where, P = Power of the object,

f = Force applied,

d = distance travelled,

t = time taken to cover the distance

f = (P × t)/ d

f = (500 × 12) / 96

f = 6000/ 96

f = 62.5 N

Therefore, the force which is being applied to the object is 62.5N (Newton).

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Answer:

particles

Explanation:

in liquids, particle are close together

Answer:

The added mass will mean a longer period of oscillation.

Explanation:

The period of oscillation here is given by the formula;

T = 2π√(m/k)

Where m is mass and k is spring constant

From the equation of oscillation period above, it's obvious that when we increase the mass, the oscillation period will also increase.

Thus, the added mass will mean a longer period of oscillation.

Answer:

you mean this halo right

Explanation:

Halo is an American military science fiction media franchise managed and developed by 343 Industries and published by Xbox Game Studios. The franchise and its early main installments were originally developed by Bungie. The central focus of the franchise builds off the experiences of Master Chief John-117, one of a group of supersoldiers codenamed Spartans, and his artificial intelligence (AI) companion, Cortana.

The original trilogy centers on an interstellar war between humanity and an alliance of aliens known as the Covenant. The Covenant, led by their religious leaders called the Prophets, worship an ancient civilization known as the Forerunners, who perished while defeating the parasitic Flood. The eponymous Halo Array are a group of immense, habitable, ring-shaped superweapons that were created by the Forerunners to destroy the Flood, but which the Covenant mistake for religious artifacts that, if activated, would transport them on a Great Journey to meet the Forerunners. They are similar to the Orbitals in Iain M. Banks' Culture novels, and to a lesser degree to author Larry Niven's Ringworld concept.[1][2][3][4]

The games in the series are critically acclaimed, with the original considered the Xbox's "killer app".[5] This led to the term "Halo killer" being used to describe console games that aspire, or are considered, to be better than Halo.[6] Fueled by the success of Halo: Combat Evolved, and by marketing campaigns from publisher Microsoft, its sequels went on to record-breaking sales.[7][8][9] The games have sold over 65 million copies worldwide, with the games alone grossing almost $3.4 billion.[10][11][12] Halo has since become one of the highest-grossing media franchises of all time. These strong sales led to the franchise's expansion to other media; the Halo Universe now spans multiple best-selling novels, graphic novels, comic books, short movies, animated movies and feature films, as well as other licensed products.

T

Answer:

1.8 m/s

Explanation:

Distance = 2*50 +2*40 [m] = 180 [m]

Time = 100 [s]

Velocity = Distance/Time = 1.8 m/s

Average velocity for the lap is 1.8 m/s

GIven:

Length of rectangular track = 50 m

Width of rectangular track = 40 m

Time taken to cover a lap = 100 seconds

Find:

Average velocity for the lap

Computation:

Perimeter of rectangle = Length of one lap

So,

Perimeter of rectangle = 2(l + b)

So,

Length of one lap = 2[50 + 40]

Length of one lap = 2[90]

Length of one lap = 180 meter

Average velocity = Distance / Time

Average velocity for the lap = 180 / 100

Average velocity for the lap = 1.8 m/s

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The maximum height : 81.63 m

Given

0.4 kg mass

vo = initial speed = 40 m/s

Required

the maximum height

Solution

We can use the law of conservation energy(ME=PE+KE) or use parabolic motion

For parabolic motion :

h max = (vo²sin²θ)/2g

θ = 90°(straight upward)

Input the value :

h max = (40²sin²90°)/2 x 9.8

h max = 81.63 m

Answer:

The answer is below

Explanation:

Let us assume that the slower truck which is the FedEx truck travels d mile. The pass started when the Walmart truck was 45 ft behind the FedEx truck and ended when the Walmart truck is 45 ft in front of the FedEx truck.

The additional distance covered by the truck Walmart = 45 ft + 60 ft (length of FedEx truck) + 45 ft = 150 ft

1 mile = 5280 ft

150 ft = 150 ft * 1 mile/5280 ft = 0.0284 mile

Therefore the Walmart truck covered an additional 0.0284 mile. Hence the distance covered by the walmart truck = d + 0.0284

Let us say it took t hours for the pass to be completed. Hence:

For FedEx; 53 mph = d / t.   t =d / 53

For Walmart; 61 mph = (d + 0.0284) / t.   t = (d + 0.0284) / 61

Since it took the trucks the same time. Hence:

d / 53 = (d + 0.0284) / 61

61d = 53d + 1.506

8d = 1.506

d = 1.506/8

d = 0.1882 mile = 993.75 ft

The FedEx truck traveled 993.75 mile while the Walmart truck traveled 1143.75 ft (993.75 + 150 ft)

The time (t) = d / 53 = 0.1882 / 53

t=0.00355 h = 12.8 seconds

Answer:

In my physics class, something that helps connect better with kids is keeping connected with them always make sure to ask if they understand what you're teaching if they are following because sometimes most kids are to afraid to admit that they are lost, another way to connect with kids is maybe posting surveys to be able to check in with each student especially during this hard times :)

Explanation:

1. start with fun activities.

2. Encourage single-tasking.

3. Designate a learning playing field.

Answer:

Distance, S = 440 meters.

Explanation:

Given the following data;

Initial velocity, u = 17m/s

Time, t = 20 seconds

Final velocity, v = 27m/s

To find the distance;

First of all, we would determine the acceleration of the truck.

Acceleration = (v-u)/t

Substituting the given values into the equation, we have;

Acceleration = (27 - 17)/20

Acceleration = 10/20

Acceleration = 0.5m/s²

Now, we would use the second equation of motion to find the distance traveled.

S = ut + ½at²

S = 17*20 + ½*0.5*20²

S = 340 + 0.25*400

S = 340 + 100

S = 440m

The equations of motion can be used to obtain the distance covered as 440 m.

We have to use of the equations that are used for uniformly accelerated motion in solving the problem. The chosen equation must be;

v^2 = u^2 + 2as and v = u + at

v = final velocity

u = initial velocity

a = acceleration

s = distance

To obtain the acceleration;

27 = 17 + 20(a)

27 - 17 = 20a

a = 0.5 ms-2

Now, to obtain the distance;

v^2 = u^2 + 2as

v^2 - u^2/as = s

s = (27)^2 - (17)^2/2(0.5)

s = 440 m

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Answer:

8.5 hours

Explanation:

Answer:

Explanation:

Atomic mass is nuetrons + protons.

Basically it's

? nuetrons + protons = 64

I did process of elimination:

b) It cannot be "b" because 29 + 64 doesn't equal 64.

c) It cannot be "c" because in a regular molecule the amount of protons equals to amount of electron(s).

d) 29 + 29  doesn't = 64.

Therefore, the answer is "A"

Answer: A

Explanation: Just took the test on a.pex

Answer:

If A + B = C for the vector equation then substituting for B gives

A + (-A) = C = 0

The only thing that can be said about "C" is that it has zero magnitude.

Answer:

distance travelled by the block is 0.796 m

{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }

Explanation:

Given that;

mass of block m = 0.200 kg

distance travelled d = 0.796 m

time t = 2.00 s

m₂ = 0.400 kg

If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s?

Now, using the second equation of motion;

d = ut + ([tex]\frac{1}{2}[/tex] × at²)

as the object started from rest, u=0

so, we substitute

0.796  = 0×2 + ([tex]\frac{1}{2}[/tex] × a(2)²)

0.796  = 0 + ([tex]\frac{1}{2}[/tex] × 4a)

0.796  = 2a

a = 0.796 / 2

a = 0.398 m/s²

using first equation of motion

[tex]V_{f}[/tex] = u + at

we substitute

[tex]V_{f}[/tex] = 0 + 0.398 × 2

[tex]V_{f}[/tex] = 0.796 m/s

now, average velocity is given as;

[tex]V_{avg}[/tex] = ( 0.796 m/s  + 0 ) / 2

[tex]V_{avg}[/tex] = ( 0.796 m/s  + 0 ) / 2

now, distance as the block moves in 2s will be;

D = [( 0.796 m/s  + 0 ) / 2 ] × 2

D = 0.796 m

Therefore, distance travelled by the block is 0.796 m

{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }

The distance traveled by the object with the uniform motion can be given by the second equation of the motion.

The distance traveled by the big block with weight in 2 seconds is.

What is second equation of motion?

The distance traveled by the object with the uniform motion can be given by the second equation of the motion. It can be given as,

[tex]s=ut+\dfrac{1}{2} at^2[/tex]

Given information-

The mass of the small block is 0.200 kg.

The total distance traveled by the block is 0.796.

Initial velocity of small block is zero.

Total time taken by the block to travel this distance is 2 seconds.

Put the values in the above equation as,

[tex]0.796=0\times 2+\dfrac{1}{2} a\times 2^2\\0.796=2a\\a=0.398[/tex]

Thus the acceleration of the small block is 0.398 meter per second.

Now the mass is doubled which is, 0.400 kg. As the acceleration does not depends on the mass, thus the acceleration for both cases is 0.398.

The velocity of the big block can be given as,

[tex]V=u+0.398\times2\\V=0+0.796\\V=0.796[/tex]

The velocity of the big block is 0.796.

The average velocity of the big block is given by,

[tex]V_{avg}=(\dfrac{0.796+0}{2} })\\V_{avg}=0.398[/tex]

The distance traveled by the object is the ratio of the velocity of the body to the time taken by it. Thus the distance traveled by the big block in 2 seconds is,

[tex]d={{0.398} \times2}\\d=0.796[/tex]

Thus the distance traveled by the big block with weight in 2 seconds is.

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Time needed = t = 20.83 s

Given

car speed = 85 km/h

truck speed = 73 km/h

Required

the time it takes for the car to reach the truck

Solution

When the car reaches the truck, the distance between them will be the same

x car - 250 m = x truck

General formula for distance (d) :

d = v.t

So the equation becomes :

85t-250 = 73t

12t=250

t = 20.83 s

Answer:

Explanation:

When two north magnetic poles are  placed close to each other , they will repel each other . If distance between them is increased due to mutual repulsion from x to 12 x , work is done by the magnetic field . This results in decrease of magnetic field energy .

Hence , when the distance between the poles becomes 12x from x , the magnitude of the field energy between them decreases .

Answer:

k = 7.84 N/m

Explanation:

We are given;

Mass hanging object; m = 0.2 kg

Extension; Δx = 0.25 m

Now, formula for the force is;

F = k•Δx

Where k is the spring constant

Since we have mass, then F = W = mg = 0.2 × 9.8 = 1.96 N

Thus;

1.96 = k × 0.25

k = 1.96/0.25

k = 7.84 N/m

Answer:

simple carbohydrate

Explanation:

Simple carbs are like cakes, donuts, and candy, etc.

Hope this helps :D

Answer:

blue

Explanation:

nswer the following about two objects, A and B, whose motion produced ihe following ... "A" starts with a greater (t) position. Since they ... since you are moving away from the origin

Answer:

Explanation:

To make this problem the easiest way possible, draw a picture and choose the side between the charges. The field will be zero at that point, and I'll prove it in just a second.

[tex]E=\frac{kq}{r^2}[/tex]

k is a constant with a value of [tex]8.99*10^9Nm^2/C^2[/tex]

q is the magnitude of the charge producing the field

r is the distance from the source charge to the test charge

So first determine the electric field from the charge on the bottom left corner, then we'll determine the electric from the bottom right corner.

Convert the centimeters to meters and nano-Coulombs to Coulombs

5cm = 0.05m

2 nC = 2 x 10^-9 C

[tex]E=(8.99*10^9)(2*10^-^9)/(0.025)^2=2.8768*10^4N/C[/tex]

This is pointing to the right because electric field lines point away from positive charges.

[tex]E=(8.99*10^9)(2*10^-^9)/(0.025)^2=-2.8768*10^4N/C[/tex]

This is pointing to the left because of the same reason. Field lines point away from positive charges.

You are able to sum them up because they are both in the x-direction. Their sum will be a net field value of zero.

Answer:

a) Remember the relationship:

v = f*λ

where:

v = velocity of the wave, remember that for an electromagnetic wave, this always is equal to the speed of light, then:

v = c = 3*10^8 m/s

f is the frequency of the wave

λ is the wavelength.

So if we want to compute the frequency, we have the equation:

f = c/λ

then:

We know that UVA has the range from 320 nm to 400nm, converting these to meters we hav:

1nm = 1m*10^(-9)

then:

320mm = (320*10^(-9)) m = 0.00000032 m

400nm = (400*10^(-9)) m =  0.00000040 m

Replacing these in our equation we can find that the range of frequencies is:

f = (3*10^8 m/s)/( 0.00000032 m) = 9.375*10^(14) Hz

f =  (3*10^8 m/s)/( 0.00000040 m) = 7.5*10^(14) Hz

Then the range of frequencies is:

( 7.5*10^(14) Hz to 9.375*10^(14) Hz)

Similar for the case of UVB:

the wavelengths are:

280 nm to  320 nm

Rewriting these in meters we get:

280nm = 280*10^(-9) m = 0.00000028m

320nm = 320*10^(-9) m = 0.00000032 m

Then the frequencies are:

f = (3*10^8 m/s)/( 0.00000032 m) = 9.375*10^(14) Hz

f = (3*10^8 m/s)/( 0.00000028 m) = 1.07*10^(15) Hz

The range of frequencies for UVB is:

(9.375*10^(14) Hz to 1.07*10^(15) Hz)

B) We want to know the range of wavenumbers for both types of waves.

The wave number is calculated as:

1/λ

Which represents the number of waves in a given unit of length.

Then the range for UVA will be:

1/320nm = 0.0031 nm^-1

1/400nm = 0.0025 nm^-1

Then the range is:

( 0.0025 nm^-1, 0.0031 nm^-1)

And for the case of UVB we will have:

1/320 nm = 0.0031 nm^-1

1/280 nm = 0.0036 nm^-1

Then the range is:

(0.0031 nm^-1, 0.0036 nm^-1)

Answer:

the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N

Explanation:

Given that;

Length L = 10 m

mass of beam m_b = 150 kg; weight W_beam = 150×9.8

mass of sign m = 75 kg

distance of sign hung from the beam from the wall d = 2.50 m

angle ∅ = 60°

g = 9.8 m/s²

Now,

Torque acting at one end of the beam will be;

[tex]T_{net}[/tex] = Tsin∅ × L - mg(d)-W × (L/2)

for equilibrium, [tex]T_{net}[/tex] = 0

therefore, 0 = Tsin∅ × L - mg(d)-W × (L/2)

so we substitute

Tsin(60°) × 10 - 75×9.8(2.50) - 150 × 9.8× (10/2) = 0

Tsin(60°) × 10 - 1837.5 - 7350 = 0

Tsin(60°) × 10 - 9187.5 = 0

Tsin(60°) × 10 = 9187.5

divide both side by 10

Tsin(60°) = 918.75

T × 0.8660 = 918.75

T = 918.75 / 0.8660

T = 1060.9 N

Therefore, the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N

The magnitude of the tension on the cable that keep the system in equilibrium is 1060.9 N.

Torque acting at one end of the beam,  

= Tsin∅ × L - mg(d)-W × (L/2)

When equilibrium  = 0  

Tsin∅ × L - mg(d)-W × (L/2) = 0

Where,

L - Length  =  10 m

m - mass of sign bord= 75 kg

g- gravitational accelaration = 9.8 m/s²

W - weight of beam =  150×9.8 = 1470 kg

Put the values in the formula,

Tsin(60°) × 10 - 75×9.8(2.50) - 150 × 9.8× (10/2) = 0  

Tsin(60°) × 10 - 1837.5 - 7350 = 0  

Tsin(60°) × 10 - 9187.5 = 0  

Tsin(60°) × 10 = 9187.5    

Tsin(60°) = 918.75  

T × 0.8660 = 918.75  

T = 918.75 / 0.8660  

T = 1060.9 N

Therefore, the magnitude of the tension on the cable that keep the system in equilibrium is 1060.9 N.

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Answer:The 50-kg has the greater potential energy because it has greater mass

Explanation:

The average velocity of the car is 37.27 km/h.

The given parameters;

The average velocity of the car is calculated as follows;

[tex]average \ velocity = \frac{Total \ displacement}{Total \ time}[/tex]

The total displacement of the car is calculated as follows;

[tex]d = \sqrt{x^2 + y^2} \\\\d = \sqrt{100^2 \ + \ 50^2} \\\\d = 111.803 \ km[/tex]

The average velocity of the car is calculated as follows;

[tex]v = \frac{111.803}{3} \\\\v = 37.27 \ km/h[/tex]

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