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3 years ago

Fig. 2.1 shows the extension-load graph for a spring.

Physics

1 answer:

trapecia [35]3 years ago

3 0

Answer:

(1) Hooke's law

(2) a) Extension is directly proportional to the applied load

b) The starting point of the graph is the origin (0, 0) or absence of load, no extension

Explanation:

(1) The law obeyed by the spring is known as Hooke's law which states that the extension or compression, x, of a spring proportional to the applied force, F

F = -k × x

Where;

k = The spring constant

(2) Given that the law mathematically is F = -k × x

The two features of the graph that show that the law is obeyed are;

a) The extension increases as the load is increased

b) The extension is zero when the there is no applied load.

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Firdavs [7]

I think its a im not sure

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3 years ago

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A 1420-kg car is traveling with a speed of 12.4 m/s. What is the magnitude of the horizontal net force that is required to bring

zzz [600]

Answer:

1419.01436 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-12.4^2}{2\times 78}\\\Rightarrow a=-0.98564\ m/s^2$

The force on the car

$F=ma\\\Rightarrow F=1420\times -0.98564\\\Rightarrow F=-1419.01436\ N$

Magnitude of the horizontal net force that is required to bring the car to a halt is 1419.01436 N

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3 years ago

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A"boat"is"moving"to"the"right"at"5"m/s"with"respect"to"the"water."A"wave"moving"to"the"left,"opposite"the"motion"of"the"boat."Th

s2008m [1.1K]

Answer:

0.99m

Explanation:

Firs you calculate the relative velocity between the boat and the wave. The velocity of the boat is 5m/s and the velocity of the wave is given by:

$v=\lambda f=\lambda\frac{1}{T}=(110m)\frac{1}{8.3s}=13.25\frac{m}{s}$

the relative velocity is:

$v'=13.25m/s-5m/s=8.25\frac{m}{s}$

This velocity is used to know which is the distance traveled by the boat after 20 seconds:

$x'=v't=(8.25m/s)(20s)=165m$

Next, you use the general for of a wave:

$f(x,t)=Acos(kx-\omega t)=Acos(\frac{2\pi}{\lambda}x-\omega t)$

you take the amplitude as 2.0/2 = 1.0m.

$\omega=\frac{2\pi}{T}=\frac{2\pi}{8.3s}=0.75\frac{rad}{s}$

by replacing the values of the parameters in f(x,t) you obtain the vertical displacement of the boat:

$f(165,20)=1.0m\ cos(\frac{2\pi}{110m}(165)-(0.75\frac{rad}{s})(20s))\\\\f(165,20)=0.99m$

6 0

4 years ago

When mass m is tied to the bottom of a long, thin wire suspended from the ceiling, the wire's second-harmonic frequency is 180 h

Katena32 [7]

The frequency of the nth-harmonic of a string is given by
$f_n = \frac{n}{2L} \sqrt{ \frac{T}{\mu} }$
where n is the number of the harmonic, L is the length of the string, T the tension and $\mu$ the linear density.

In our problem, since the mass m is tied to the string, the tension is equal to the weight of the object tied:
$T=mg$
Substituting into the first formula, we have
$f_n = \frac{n}{2L} \sqrt{ \frac{mg}{\mu} }$

In our problem we have n=2 (second harmonic). In the previous equation, the only factor which is not constant between the first and the second part of the problem is m, the mass. So, we can rewrite everything as
$f_2 = K \sqrt{m}$
where we called
$K= \frac{2}{2L} \sqrt{ \frac{g}{\mu} }$

In the first part of the problem, the mass of the object is m and $f_2 = 180 Hz$ . So we can write
$180 Hz = K \sqrt{m}$

When the mass is increased with an additional 1.2 kg, the relationship becomes
$270 Hz = K \sqrt{(m+1.2 Kg)}$

By writing K in terms of m in the first equation, and subsituting into the second one, we get
$180 Hz \sqrt{ \frac{m+1.2 Kg}{m} }=270 Hz$
and solving this, we find
$m=0.96 kg$

5 0

3 years ago

A bullet with mass 0.04000 kg, traveling at a speed of 310.0 m/s, strikes a stationary block of wood having mass 6.960 kg. The b

ch4aika [34]

Answer:

1.77 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m')............................ Equation 1

Where m = mass of the bullet, m' = mass of the block, u = initial velocity of the bullet, u' = initial velocity of the block, V = Velocity of the bullet-wood combination immediately after collision

make V the subject of the equation

V = (mu+m'u')/(m+m').................. Equation 2

Given: m = 0.04 kg, m' = 6.96 kg, u = 310 m/s, u' = 0 m/s (stationary)

Substitute into equation 2

V = (0.04×310+6.96×0)/(0.04+6.96)

V = 12.4/7

V = 1.77 m/s

Hence the speed, in m/s, of the bullet-plus-wood combination immediately after the collision is 1.77 m/s

6 0

4 years ago