A person walks 40. m east, then 20. m west. He then

Sleep is a difficult problem for dolphins because dolphins __________. A. need more sleep than their predators B. must be awake

Alex17521 [72]

Need more sleep than their predators

5 0

3 years ago

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A thin, light wire 75.2 cm long having a circular cross section 0.560 mm in diameter has a 25.2 kg weight attached to it, causin

blsea [12.9K]

Answer:

The stress is calculated as $1.003\times 10^{9}\ Pa$

Solution:

As per the question:

Length of the wire, l = 75.2 cm = 0.752 m

Diameter of the circular cross-section, d = 0.560 mm = $0.560\times 10^{- 3}\ m$

Mass of the weight attached, m = 25.2 kg

Elongation in the wire, $\Delta l = 1.10\ mm = 1.10\times 10^{- 3}\ m$

Now,

The stress in the wire is given by:

$Stress,\ \sigma = \frac{Force,\ F}{Area,\ A}$ (1)

Now,

Force is due to the weight of the attached weight:

F = mg = $25.2\times 9.8 = 246.96\ N$

Cross sectional Area, A = $\pi (\frac{d}{2})^{2} = \pi (\frac{0.560\times 10^{- 3}}{2})^{2} = 2.46\times 10^{- 7}\ m^{2}$

Using these values in eqn (1):

$\sigma = \frac{246.96}{2.46\times 10^{- 7}} = 1.003\times 10^{9}\ Pa$

8 0

3 years ago

We can expect the force of friction to be greater for an object moving on a surface if the surfaces in contact are

nikitadnepr [17]

Answer:

c

Explanation:

Friction is caused by how rough and object is, and how much the object weighs, as this causes it to drag more.

8 0

2 years ago

1. In a(n) ______________________ circuit, all parts are connected one after another along one path.

poizon [28]

Answer:

1. A series circuit is a closed circuit which has all loads connected in a row and there is only one path for the current to flow.

2. The Ohm's Law state that a current flow through a resistor is directly proportional to the voltage across it $R = \frac{V}{I}$

3. A parallel circuit is a closed circuit divided into branches that it has two o more paths for the current to flow and the loads are parallel to each other which mean the voltage across them is the same for all loads.

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3 years ago

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The total electric flux from a cubical box 26.0 cm on a side is 1840 N m2/C. What charge is enclosed by the box?

vitfil [10]

The expression of the electric flux is

$\Phi = \frac{Q}{\epsilon_0}$

Here,

Q = Total charge enclosed in the closed surface

$\epsilon_0$ = Permittivity due to free space

Rearranging to find the charge,

$Q = \epsilon_0 \Phi$

Replacing with our values we have finally

$Q = (8.85*10^{-12}F\cdot m^{-1})(1.84*10^3 N\cdot m^2/C)$

$Q = 1.6284*10^{-8} C$ $(\frac{10^9nC}{1C})$

$Q = 0.1684nC$

The charge enclosed by the box is 0.1684nC

The sign of the charge can be decided by using the direction of the flux. The charge enclosed by the cube can be calculated by using the electric flux and the permitivity of free space.

7 0

3 years ago