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Ne4ueva [31]
1 year ago
15

Based on the graph, how would you describe the net forces acting on the moving

Physics
1 answer:
ladessa [460]1 year ago
3 0

Given the velocity-time graph of an object.

The slope of a velocity-time graph gives the acceleration acting on the object.

From the graph, we can see that the slope of the graph is zero. That is, the velocity of the object is constant and hence the net acceleration acting on the object is zero.

From Newton's second law, the net force acting on an object is given by the product of the mass of the object and its velocity. Therefore when the acceleration of the object is zero, the net force on the object is also zero.

Therefore the net force acting on the given object is zero. Hence, the correct answer is option A.

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A spring that is compressed 14.5 cm from its equilibrium position stores 2.99 J of potential energy. Determine the spring consta
strojnjashka [21]

Answer:

284.4233 N/m

Explanation:

k = Spring constant

x = Compression of spring = 14.5 cm

U = Potential energy = 2.99 J

The potential energy of a spring is given by

U=\dfrac{1}{2}kx^2

Rearranging to get the value of k

\\\Rightarrow k=\dfrac{2U}{x^2}\\\Rightarrow k=\dfrac{2\times 2.99}{0.145^2}\\\Rightarrow k=284.4233\ N/m

The spring constant is 284.4233 N/m

7 0
3 years ago
Find an equation relating the rate of change of kinetic energy to the rate of change of velocity
alexdok [17]

Answer:

djjdjfjfntjfjjfjxuxidie

Explanation:

hjdjfjjfjf urfjfjfuuijjjjjjjjjkjsjjsj

4 0
3 years ago
Matter is made of small particles to small to be seen. Which of these best describe evidence of this statement? 1. Tara’s crayon
son4ous [18]

Answer:

Explanation:

I think the answer is statement no 3.

Hope it helps.

5 0
2 years ago
Read 2 more answers
Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi
Orlov [11]

The number converted is 0.0467 \frac{(kg)(s)}{m^3}

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

1 kg = 1000 g = 10^6 mg

1 min = 60 s

1 m^3 = 1000 L

The original unit that we have is

\frac{mg\cdot min}{L}

Therefore, it can be rewritten as:

=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot  60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}

Therefore, since the initial number was 0.779, the final value is

0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}

#LearnwithBrainly

5 0
3 years ago
An electron is accelerated by a 3.6 kv potential difference. the charge on an electron is 1.60218 × 10−19 c and its mass is 9.10
katen-ka-za [31]
By definition, the potential energy is:
 U = qV
 Where,
 q: load
 V: voltage.
 Then, the kinetic energy is:
 K = mv ^ 2/2
 Where,
 m: mass
 v: speed.
 As the power energy is converted into kinetic energy, we have then:
 U = K
 Equating equations:
 qV = mv ^ 2/2
 From here, we clear the speed:
 v = root (2qV / m)
 Substituting values we have:
 v = root ((2 * (1.60218 × 10 ^ -19) * 3600) /9.10939×10^-31))
 v = 3.56 × 10 ^ 7 m / s
 Then, the centripetal force is:
 Fc = Fm
 mv ^ 2 / r = qvB
 By clearing the magnetic field we have:
 B = mv / qr
 Substituting values:
 B = (9.10939 × 10 ^ -31) * (3.56 × 10 ^ 7) / (1.60218 × 10 ^ -19) * 0.059
 B = 3.43 × 10 ^ -3 T
 Answer:
 
A magnetic field that must be experienced by the electron is:
 
B = 3.43 × 10 ^ -3 T
6 0
2 years ago
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