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3 years ago

Which of these devices does NOT contain a magnet or electromagnet?

Physics

2 answers:

Black_prince [1.1K]3 years ago

7 0

I think it is D a light bulb

andrey2020 [161]3 years ago

3 0

The answer is D.
A lightbulb has no magnet or electromagnet.

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How large a force is necessary to stretch a 4.0-mm-diameter steel wire from its original length by 1.0%?

jekas [21]

The force needed to stretch the steel wire by 1% is 25,140 N.

The given parameters include;

diameter of the steel, d = 4 mm

the radius of the wire, r = 2mm = 0.002 m

original length of the wire, L₁

final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)

extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁

the Youngs modulus of steel, E = 200 Gpa

The area of the steel wire is calculated as follows;

$A = \pi r^2\\\\ A= 3.142 \times (0.002)^2\\\\ A= 1.257 \times 10^{-5} \ m^2$

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;

$E = \frac{stress}{strain} = \frac{F/A}{e/L} = \frac{FL}{Ae} \\\\F = \frac{EAe}{L}$

$F = \frac{200 \times 10^9\ \times\ 1.257\times 10^{-5}\ \times \ 0.01l_1}{l_1} \\\\F = 25,140\ N$

Thus, the force needed to stretch the steel wire by 1% is 25,140 N.

Learn more here: brainly.com/question/21413915

4 0

2 years ago

Strontium 3890Sr has a half-life of 28.5 yr. It is chemically similar to calcium, enters the body through the food chain, and co

patriot [66]

Answer:

Thus the time taken is calculated as 387.69 years

Solution:

As per the question:

Half life of $^{3890}Sr\, t_{\frac{1}{2}}$ = 28.5 yrs

Now,

To calculate the time, t in which the 99.99% of the release in the reactor:

By using the formula:

$\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$

where

N = No. of nuclei left after time t

$N_{o}$ = No. of nuclei initially started with

$\frac{N}{N_{o}} = 1\times 10^{- 4}$

(Since, 100% - 99.99% = 0.01%)

Thus

$1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}$

Taking log on both the sides:

$- 4 = \frac{t}{28.5}log\frac{1}{2}$

$t = \frac{-4\times 28.5}{log\frac{1}{2}}$

t = 387.69 yrs

5 0

3 years ago

physics A river flows at a speed vr = 5.37 km/hr with respect to the shoreline. A boat needs to go perpendicular to the shorelin

vova2212 [387]

Answer: Vb is the vector (-5.37m/s, 8.59 m/s), with a module 10.13m/s

then the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9

Explanation:

We can use the notation (x, y) where the river flows in the x-axis and the pier is on the y-axis.

We have Vr = (5.37m/s, 0m/s)

Now, if the boat wants to move only along the y-axis (perpendicularly to the shore).

The velocity of the boat Vb will be:

Vb = (-c*sin(32). c*cos(32))

Then we should have that:

5.37 m/s - c*sin(32) = 0

c = (5.37/sin(32))m/s = 10.13 m/s

the velocity in the y-axis is:

10.13m/s*cos(32) = 8.59 m/s

So Vb = (-5.37m/s, 8.59 m/s)

the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9 where i used Vb as the module of the boat's velocity.

7 0

3 years ago

PLEASE PLEASE HELP AND PUT A REAL ANSWER ;-;. ALSO WILL GIVE BRAINLEST!!

Xelga [282]

Answer:

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2 years ago

An object is accelerating at a constant rate due to a force being applied.what can be done to the force to decrease the objects

Sliva [168]

Answer:

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2 years ago