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MAXImum [283]
3 years ago
13

Using two spring scales, students pull on opposite sides of a dynamic cart at rest. 7 Newtons of force is pulling left and 5 New

tons force is pulling right. Describe the motion of the cart?
Physics
1 answer:
kirill [66]3 years ago
7 0

Answer:

The cart will accelerate to the left

Explanation:

The cart is initially at rest. In order to understand its motion, we should consider the net force acting on the cart. We have two forces acting on it:

- A force of 7 N to the left

- A force of 5 N to the right

Therefore, the net force is

F=7 N -5N=2 N

in the direction of the stronger force (so, to the left).

According to Newton's second law:

F=ma

a net force different from zero produces an acceleration of the object, in the same direction as the net force. Therefore, the cart will accelerate to the left.

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The SI unit of length or distance is the meter.

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What is the equation for force
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The equation is F=ma :)
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What is the temperature 32 degrees Fahrenheit in degrees Celsius? A. 0°C B. 20°C C. 10°C D. –10°C
omeli [17]
The answer would 0. The reasoning of this is because freezing point in celsius is always 0 degrees but in fahrenheit the freezing point is 32 degrees.
8 0
3 years ago
A sound wave has a frequency of 425 hz. what is the period of this wave? 0.00235 seconds 0.807 seconds 425 seconds 850 seconds
77julia77 [94]

The period of the sound wave at the given frequency is determined as 0.00235 second.

<h3>Period of the sound wave</h3>

The period of the sound wave at the given frequency is calculated as follows;

Period is reciprocal of frequency.

T = 1/f

T = 1/425

T = 0.00235 second

Thus, the period of the sound wave at the given frequency is determined as 0.00235 second.

Learn more about period here: brainly.com/question/10428039

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3 0
2 years ago
A positively charged particle Q1 = +35nC is held fixed at the origin. A second charge of mass m = 3.5ug is floating a distance d
cupoosta [38]

Answer:

Explanation:

Q1 = 35 nC = 35 x 10^-9 C

m = 3.5 micro gram = 3.5 x 10^-9 Kg

d  = 35 cm = 0.35 m

(a) The electrostatic force between the two charges is balanced by the weight of another charge.

F = m g

\frac{1}{4\pi \epsilon _{0}}\frac{Q_{1}Q_{2}}{d^{2}}=mg

Q_{2}=\frac{4\pi \epsilon _{0}mgd^{2}}{Q_{1}}

(b) By substituting the values

Q_{2}=\frac{3.5\times 10^{-9}\times 9.8\times 0.35\timees 0.35}{9\times 10^{9}\times 35\times 10^{9}}

Q2 = 13.34 x 10^-12 C

Q2 = 0.0134 nC

4 0
4 years ago
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