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3 years ago

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A rubber band has potential energy of 5 J. If the spring constant of the rubber band is 50 N/m, what is the displacement of the

rubber band?

2 answers:

Valentin [98]3 years ago

7 0

To determine the displacement, since we are given the potential energy, we use the equation for potential energy. For a spring, it is one-half the product of the spring constant and the square of the displacement. We do as follows:

PE = kx^2/2
5 Nm = 50N/m (x^2)
x = 0.32 m

Therefore, the displacement would be 0.32 m.

Allisa [31]3 years ago

7 0

The correct answer is 0.4 m on Edge.

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A woman can row her canoe at 2.5 m/s. If she faces an opposing current of 3.0 m/s, how fast will she go forward?

taurus [48]

Answer:

V = - 0.5 [m/s]

Explanation:

In order to solve this problem, we must use the principle of relative speeds. This is for an observer who is on the edge of the river he can see how the river moves to the left and the woman tries to move to the right but can not since:

$V_{total}=-3+2.5\\V_{total}=-0.5 [m/s]$

That is, the person sees how the woman moves to the left but with avelocity of 0.5 [m/s] to the left

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2 years ago

A stone is dropped from a bridge and hits the pavement below in two seconds. What is the velocity of the stone when it hits the

Helen [10]

We have: a = v/t
Here, t = 2 s [ Given ]
a = 9.8 m/s² [constant value for earth system ]

Substitute their values into the expression:
9.8 = v/2
v = 9.8 × 2
v = 19.6 m/s

In short, Your Answer would be Option B

Hope this helps!

4 0

3 years ago

Read 2 more answers

Is friction used when buttering a cake pan?

Elena L [17]

Answer: friction is used when 2 and 2 things mix and come together but normally friction slows things down but when buttering a pan it doesnt slow down

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3 years ago

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A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 29.4 m/s2 m /

Sidana [21]

Answer:

The maximum height is 2881.2 m.

Explanation:

Given that,

Acceleration = 29.4 m/s²

Time = 7.00 s

We need to calculate the distance

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times29.4\times7^2$

$s=720.3\ m$

We need to calculate the velocity

Using formula of velocity

$v=a\times t$

Put the value into the formula

$v=29.4\times7$

$v=205.8\ m/s$

We need to calculate the height

Using formula of height

$H=\dfrac{v^2}{2g}$

Put the value into the formula

$H=\dfrac{(205.8)^2}{2\times9.8}$

$H=2160.9\ m$

We need to calculate the maximum height

Using formula for maximum height

$H'=H+s$

Put the value into the formula

$H'=2160.9+720.3$

$H'=2881.2\ m$

Hence, The maximum height is 2881.2 m.

4 0

3 years ago

Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina

Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

$\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}$

where

$E_f = 131 GPa$ is the fiber elasticity module

$E_m = 2.4 GPa$ is the matrix elasticity module

$V_f=0.3$ is the fraction of volume of the fiber

$V_m=0.7$ is the fraction of volume of the matrix

Substituting,

$\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4$ (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

We can rewrite (1) as

$F_m = \frac{F_f}{23.4}$

And inserting this into (2):

$F=F_f + \frac{F_f}{23.4}$

Solving the equation, we find the actual load carried by the fiber phase:

$F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N$

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

Using

F = 46500 N

$F_f = 44594 N$

We can immediately find the actual load carried by the matrix phase:

$F_m = F-F_f = 46,500 N - 44,594 N=1,906 N$

(d) 437 MPa

The cross-sectional area of the fiber phase is

$A_f = A V_f$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_f=0.3$ , we have

$A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2$

And the magnitude of the stress on the fiber phase is

$\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa$

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

$A_m = A V_m$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_m=0.7$ , we have

$A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2$

And the magnitude of the stress on the matrix phase is

$\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa$

(f) $3.34\cdot 10^{-3}$

The longitudinal modulus of elasticity is

$E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa$

While the total stress experienced by the composite is

$\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa$

So, the strain experienced by the composite is

$\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}$

3 0

3 years ago