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uranmaximum [27]
3 years ago
9

A 15 kg mass is lifted to a height of 2m. What is gravitational potential energy at this position

Physics
2 answers:
7nadin3 [17]3 years ago
7 0

Answer: 294J

Explanation:

U_g=mg(y_f-y_i)

U_g=(15)(9.8)(2-0)=294J

JulijaS [17]3 years ago
7 0

Answer:

294.3 J

Explanation:

U = mgh

U = energy

m = mass

g = gravity

h = height

so

U = 15kg * 9.81 m/s^2 * 2m

294.3 J (joules) [kg *m^2/s^2]

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I think its structural plasticity.

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Objects have a tendency to resist changing their motion. This property is called: *
Salsk061 [2.6K]

Answer:

The answer your looking for is option 2 - Inertia

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krok68 [10]

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5 0
3 years ago
A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1
bija089 [108]

1.

Answer:

Part a)

\rho = 1.35 \times 10^{-5}

Part b)

\alpha = 1.12 \times 10^{-3}

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

i = 18.7 A

V = 17.0 volts

now we will have

V = I R

17.0 = 18.7 R

R = 0.91 ohm

now we know that

R = \rho \frac{L}{A}

0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho = 1.35 \times 10^{-5}

Part b)

Now at higher temperature we have

V = I R

17.0 = 17.3 R

R = 0.98 ohm

now we know that

R = \rho \frac{L}{A}

0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho' = 1.46 \times 10^{-5}

so we will have

\rho' = \rho(1 + \alpha \Delta T)

1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))

\alpha = 1.12 \times 10^{-3}

2.

Answer:

Part a)

i = 1.55 A

Part b)

v_d = 1.4 \times 10^{-4} m/s

Explanation:

Part a)

As we know that current density is defined as

j = \frac{i}{A}

now we have

i = jA

Now we have

j = 1.90 \times 10^6 A/m^2

A = \pi(\frac{1.02 \times 10^{-3}}{2})^2

so we will have

i = 1.55 A

Part b)

now we have

j = nev_d

so we have

n = 8.5 \times 10^{28}

e = 1.6 \times 10^{-19} C

so we have

1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d

v_d = 1.4 \times 10^{-4} m/s

8 0
3 years ago
A certain superconducting magnet in the form of a solenoid of length 0.72 m can generate a magnetic field of 3.5 T in its core w
ira [324]

Answer:

55407

Explanation:

we have given that magnetic field B=3.5 T

current through the coil=90 A

Length of solenoid =0.72 m

we know the formula of magnetic field

B=\frac{\mu _0NI}{l}

so N=\frac{Bl}{\mu _0I}=\frac{3.5\times 0.72}{4\pi \times 10^{-6}\times 90}=55407.277=55407\ turns

so the number of turn in solenoid will be 55407

3 0
2 years ago
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