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In-s [12.5K]
1 year ago
6

A cart moves along a track at a velocity of 3.5 cm/s. When a force is applied to the cart, its velocity increases to 8.2 cm/s. I

f it takes the cart 1.5 seconds to reach 8.2 cm/s, what is the acceleration of the cart? Round your answer to the nearest tenth. cm/s2
Physics
2 answers:
zheka24 [161]1 year ago
5 0
The first thing we are going to do for this case is write the equations of movement of the car.
 We have then:
 vf = a * t + vo
 Substituting values:
 8.2 = a * (1.5) + (3.5)
 Clearing the acceleration we have:
 a = (8.2-3.5) / (1.5)
 a = 3.1 m / s ^ 2
 Answer: 
 the acceleration of the cart is: 
 a = 3.1 m / s ^ 2
9966 [12]1 year ago
5 0

Answer:

3.1cm/s^2

Explanation:

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You stand on a bathroom scale in a moving elevator. what happens to the scale reading if the cable holding the elevator suddenly
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A bathroom scales works due to gravity. Under normal conditions, a reading can be obtained when your body is pushing some force on the scale. However in this case, since you and the scale are both moving downwards, so your body is no longer pushing on the scale. Therefore the answer is:

<span>The reading will drop to 0 instantly</span>

7 0
2 years ago
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A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, take
Ad libitum [116K]

Answer:

F=(3i+3.6j)\ N

Explanation:

It is given that,

Mass of the puck, m = 4.8 kg

Initial velocity of the puck, u=(1i+0j)\ m/s

After 8 seconds, final velocity of the puck, v=(6i+6j)\ m/s

Let the x and y component of force is given by F_x\ and\ F_y.

x component of force is given by :

F_x=m\times \dfrac{v-u}{t}

F_x=4.8\times \dfrac{6-1}{8}

F_x=3\ N

y component of force is given by :

F_y=m\times \dfrac{v-u}{t}

F_y=4.8\times \dfrac{6-0}{8}

F_y=3.6\ N

So, the component of the force is F=(3i+3.6j)\ N. Hence, this is the required solution.

7 0
2 years ago
An airplane flying parallel to the ground undergoes two consecutive dis- placements. The first is 75 km 30.0° west of north, and
torisob [31]

Answer: displacement of airplane is 172 km in direction 34.2 degrees East of North

Explanation:

In constructing the two displacements it is noticed that the angle between the 75 km vector and the 155 km vector is a right angle (90 degrees).

 

Hence if the plane starts out at A, it travels to B, 75 km away, then turns 90 degrees to the right (clockwise) and travels to C, 155 km away from B. Angle ABC is 90 degrees, hence we can use Pythagoras theorem to solve for AC

 

AC2 = AB2 + BC2 ; AC^2 = 752 + 1552  ; from this we get AC = 172 km (3 significant figures)

 

Angle BAC = Tan-1(155/75) ; giving angle BAC = 64.2 degrees

 

Hence AC is in a direction (64.2 - 30) = 34.2 degrees East of North

 

Therefore the displacement of the airplane is 172 km in a direction 34.2 degrees East of North

5 0
1 year ago
The earth's magnetic field, with a magnetic dipole moment of 8.0×1022Am2, is generated by currents within the molten iron of the
Fed [463]

To solve this problem it is necessary to apply the concepts related to the magnetic dipole moment in terms of the current and the surface area, as well as the current density, as a function of the current over the area.

Part A) By definition we know that magnetic dipole moment is

m = IS

Where,

I = Current

S = Area

m = IA \rightarrow m= I( \pi r^2)

Replacing with our values we have that,

8*10^{22} = I \pi(\frac{3000*10^3}{2})^2

Re-arrange to find I,

I = 1.1317*10^{10} A

Part B) To find the Current density we need to find the cross sectional area of the Wire:

A = \pi r^2 \\A = \pi (\frac{1000*10^3}{2})^2\\A = 7.854*10^{11}m^2

Finally the current density is simply J

J = \frac{I}{A}\\J = \frac{1.1317*10^{10}}{7.854*10^{11}}\\J = 0.0144A/m^2

PART C) Finally to make the comparison with the given values we have to cross-sectional area would be

A = \pi (10-3)^2 \\A = 49\pi

Therefore the current density would be

J = \frac{I}{A}\\J = \frac{30A}{49\pi}\\J = 9.549*10^5 A/m^2

Comparing the two values we can see that the 2mm wire has a higher current density.

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1 year ago
read the excerpt below and answer the question. "no roving foot shall crush thee here, no busy hand provoke a tear." what type o
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The type of figurative language that Freneau employ in these lines from "The Wild Honeysuckle" is personification. The correct answer would be option D. Why is it personification? Personification is a figure of speech that uses human attributes to something that is not living. In this line, the author attributes the words "roving" and "busy" to foot and hand, respectively. 
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