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Shtirlitz [24]
3 years ago
8

Can someone say for me ost right or wrong if it wrong say for me what is the answer plrase

Physics
1 answer:
saveliy_v [14]3 years ago
8 0

Answer:

It is correct

Explanation:

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(1 point) Suppose a spring with spring constant 1 N/m is horizontal and has one end attached to a wall and the other end attache
fredd [130]

Answer:

Explanation:

Given that,

The spring constant

K = 1N/m

Frequency of motion

f = 14Hz

We want calculate the mass m?

The frequency of spring system is related to the mass by

From w = √k/m

Where w = 2πf

f = 1/2π √k/m

Where,

w is angular frequency in rad/s

m is mass of object attached in kg

k is the spring constant in N/m

f Is the frequency in Hz

Then, make m subject of formula

Multiply both sides by 2π

2πf = √k/m

Square both sides

4π²f² = k/m

Then, k= 4π²f² × m

m = k / 4π²f²

m = 1 / (4π² × 1.4)

m = 1 / 55.27

m = 0.0181 kg

m= 18.1 g

The mass of the object attached in 18.1 g or 0.0181 kg

4 0
3 years ago
Friction is defined as *
den301095 [7]

Answer:

O a force that opposes motion

5 0
3 years ago
Transfer of heat through conduction can occur in which of the following?
lianna [129]

Answer:d I think

Explanation:

8 0
3 years ago
Read 2 more answers
A cylinder of radius r=10.0 cm and height h=20.0 cm
Olenka [21]
D I think is the correct answer
If the cylinder is slightly
7 0
2 years ago
An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch
irinina [24]

Answer:

a) \sigma_{\rm in} = -2.18~{\rm \mu C/m^2}

b) \sigma_{\rm out}= 1.12~{\rm \mu C/m^2}

c) E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}

Explanation:

Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:

Q_{\rm in} = \sigma A_{\rm in} = \sigma(2\pi r_1 h) = (-0.35)(2\pi (0.08) h) = -0.175h~{\rm \mu C}

Q_{\rm out} = \sigma A_{\rm out} = \sigma(2\pi r_2 h) = (-0.35)(2\pi (0.1) h) = -0.22h~{\rm \mu C}

Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.

When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.

Q_{\rm wire} = \lambda h = 1.1h~{\rm \mu C}

a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:

\sigma_2 = \frac{Q_{\rm in}}{2\pi r_1h} = \frac{-1.1h}{2\pi r_1 h} = \frac{-1.1}{2\pi(0.08)} = -2.18~{\rm \mu C/m^2}

b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.

The new surface charge density can be calculated as follows:

\sigma_{\rm out}= \frac{Q_{\rm out}}{2\pi r_2h} = \frac{0.705h}{2\pi r_2 h} = \frac{0.705}{2\pi(0.1)} = 1.12~{\rm \mu C/m^2}

c) The electric field outside the cylinder can be found by Gauss' Law:

\int{\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}

We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.

E(2\pi r h) = \frac{Q_{\rm enc}}{\epsilon_0}\\E2\pi rh = \frac{\sigma 2\pi (r_1 + r_2)h}{\epsilon_0}\\E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}

4 0
3 years ago
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