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Sliva [168]
3 years ago
12

A wire is stretched right to its breaking point by a 5000 N force. A longer wire made of the same material has the same diameter

. Is the force that will stretch it right to its breaking point larger than, smaller than, or equal to 5000 N? Explain.
Physics
1 answer:
leva [86]3 years ago
4 0

Answer:

Equal to 5000N

Explanation:

The stress on the material is defined by force per unit of cross-sectional area. So it depends on the force and the diameter of the wire, which is the same for both wires. The material that defines the breaking point, is also the same. Therefore, both wires have their breaking point the same at 5000N. The wire length plays no role in here.

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What is the momentum of a ten ton truck which is sitting still?
ANTONII [103]

The momentum of the truck is given by:

p = mv

p = momentum, m = mass, v = velocity

There isn't too much work to do here; the truck is at rest, therefore its velocity is 0. This means the product of m and v is 0, giving the momentum as 0.

7 0
3 years ago
You're in a car that gets 37 miles per gallon of gas, driving it at a constant speed. If you took the gas from the car tank, and
Sergeu [11.5K]

Answer:

Required diameter of hose pipe = 0.2864 mm

Solution:

From the continuity eqn, the fluid flow rate is given by:

Av = \frac{V}{t}

where

A = cross-sectional area = \pi r^{2}

r = hose pipe radius

v = velocity of gas

Also, v = \frac{displacement, d}{time, t}

Using:

1 gallon = 3.854 l

1 mile = 1609.34 m

1 m^{3} = 1000 l

Therefore,

A\frac{d}{t} = \frac{V}{t}

\pi r^{2} = \frac{V}{d}

\pi r^{2} = \frac{(1 gal).(\frac{3.7854 l}{gal}).(\frac{10^{- 3} m^{3}}{l})}{37 miles(\frac{1609.34 m}{miles})}

6.357\times 10^{- 8} = \pi r^{2}

r^{2} = 2.024\times 10^{- 8}

r = 1.423\times 10^{- 4} m = 0.1423 mm

The diameter of the hose pipe = 2r = 2\times 1.423\times 10^{- 4}

The diameter of the hose pipe = 2.846\times 10^{- 4} m = 0.2846 mm

4 0
3 years ago
An astronomical source emits radio waves with a frequency of 450 MHz. a. If earth is 20 light years from the source, and we meas
hammer [34]

Answer:

Explanation:

a )

Frequency n = 450 x 10⁶ .

20 light years = 20 x 9.461 x 10¹⁵m

Let power of source be P

Intensity at distance R  =  \frac{P}{4\pi R^2}

Substituting the given values

8.5 x 10⁻¹⁰ = \frac{P}{4\pi ((20\times9.461\times10^{15})^2}

P = 3822457 x 10²⁰ W.

b )

Half the power will be from electric and half will be from magnetic field.

Total power = 8.5 x 10⁻¹⁰ W

Half = 4.25 x 10⁻¹⁰ W .

power of electric field

= \frac{1}{2}\epsilon\times E_0^2\times c

ε is permittivity , E₀ is amplitude of electric field , c is velocity of light .

Putting the values

4.25 x 10⁻¹⁰ = .5 x 8.85 x 10⁻¹² x E² x 3 x 10⁸

E₀² = .32 x 10⁻⁶

E₀ = .565 x 10⁻³ W / s .

6 0
3 years ago
Describe your acceleration if you ride your bike up a hill,then ride down the other side.
Ghella [55]

Answer:

first negative and then positve

5 0
2 years ago
A small 0.14 kg metal ball is tied to a very light (essentially massless) string that is 0.9 m long. The string is attached to t
Vsevolod [243]

Answer:

a) v=2.743m/s

b) a_c = 8.363m/s^2

c) T=2.543N

Explanation:

First, calculate the height of the ball at the starting point:

y' = 0.9cos(55)

y' = 0.516

At this point, just in the moment the ball is released, all the energy of the system is potencial gravitational energy. When it is at the bottom all the potencial energy is transformed into kinetic energy:

E_p=E_k\\mgh=\frac{mv^2}{2}

Solving for v:

v=\sqrt{2gh}

if h is the height loss: (l-y')

v=2.743m/s

The centripetal acceleration is the acceleration caused by the tension force exercised by the string, and is pointing outside of the trayectory path (at the lowest point, directly dawn):

a_c=\frac{v^2}{r}

a_c = 8.363m/s^2

To calculate tension, just make the free body diagram of forces in the ball, noticing the existence of the centripetal acceleration:

\sum{F_y}=ma_c=T-W\\T=ma_c+W\\T=m(a_c+g)\\T=0.14(8.363+9.8)\\T=2.543N

4 0
4 years ago
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