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2 years ago

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It takes 87 j of work to stretch an ideal spring, initially 1.4 m from equilibrium, to a position 2.9 m from equilibrium. what i

s the value of the spring constant (force constant) of this spring?

1 answer:

gayaneshka [121]2 years ago

7 0

Given:

Work done = 87 Joule

$x_{1}$ = 1.4 m

$x_{2}$ = 2.9 m

To find:

Spring constant = ?

Formula used:

W = $\frac{1}{2} k \triangle x^{2}$

Solution:

Work done required to stretch a spring from 1.4 m to 2.9 m is 87 Joule. Work done is given by following formula.

W = $\frac{1}{2} k \triangle x^{2}$

where W = work done

k = spring constant or force constant

$\triangle x$ is change in length i. e $\triangle x=x_{2} x_{1}$

k = $\frac{2W}{\triangle x^{2}}$

k= $\frac{87 \times 2}{[x_{2} - x_{1} ]^{2}}$

k = 77.33 N/m

Thus, spring constant of the given spring is 77.33 N/m.

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A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest. A small amount of gunpowder h

Anestetic [448]

Answer:

Speed of 1.83 m/s and 6.83 m/s

Explanation:

From the principle of conservation of momentum

$mv_o=m(v_1 + v_2)$ where m is the mass, $v_o$ is the initial speed before impact, $v_1$ and $v_2$ are velocity of the impacting object after collision and velocity after impact of the originally constant object

$5m=m(v_1 +v_2)$

Therefore $v_1+v_2=5$

After collision, kinetic energy doubles hence

$2m*(0.5mv_o)=0.5m(v_1^{2}+v_2^{2})$

$2v_o^{2}=v_1^{2} + v_2^{2}$

Substituting 5 m/s for $v_o$ then

$2*(5^{2})= v_1^{2} + v_2^{2}$

$50= v_1^{2} + v_2^{2}$

Also, it’s known that $v_1+v_2=5$ hence $v_1=5-v_2$

$50=(5-v_2)^{2}+ v_2^{2}$

$50=25+v_2^{2}-10v_2+v_2^{2}$

$2v_2^{2}-10v_2-25=0$

Solving the equation using quadratic formula where a=2, b=-10 and c=-25 then $v_2=6.83 m/s$

Substituting, $v_1=-1.83 m/s$

Therefore, the blocks move at a speed of 1.83 m/s and 6.83 m/s

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1 year ago

Two wires are stretched between two fixed supports and have the same length. One wire A there is a second-harmonic standing wave

lina2011 [118]

(a) Greater

The frequency of the nth-harmonic on a string is an integer multiple of the fundamental frequency, $f_1$ :

$f_n = n f_1$

So we have:

- On wire A, the second-harmonic has frequency of $f_2 = 660 Hz$ , so the fundamental frequency is:

$f_1 = \frac{f_2}{2}=\frac{660 Hz}{2}=330 Hz$

- On wire B, the third-harmonic has frequency of $f_3 = 660 Hz$ , so the fundamental frequency is

$f_1 = \frac{f_3}{3}=\frac{660 Hz}{3}=220 Hz$

So, the fundamental frequency of wire A is greater than the fundamental frequency of wire B.

(b) $f_1 = \frac{v}{2L}$

For standing waves on a string, the fundamental frequency is given by the formula:

$f_1 = \frac{v}{2L}$

where

v is the speed at which the waves travel back and forth on the wire

L is the length of the string

(c) Greater speed on wire A

We can solve the formula of the fundamental frequency for v, the speed of the wave:

$v=2Lf_1$

We know that the two wires have same length L. For wire A, $f_1 = 330 Hz$ , while for wave B, $f_B = 220 Hz$ , so we can write the ratio between the speeds of the waves in the two wires:

$\frac{v_A}{v_B}=\frac{2L(330 Hz)}{2L(220 Hz)}=\frac{3}{2}$

So, the waves travel faster on wire A.

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1 year ago

When carrying extra weight, the space formed between the top of your head and the two axles of the motorcycle is referred to as

Dafna11 [192]

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1 year ago

1 lb equals how many grams

shusha [124]

Answer:

1 lb. is 453.592 grams

Explanation:

1lb is 453.592 grams

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2 years ago

Read 2 more answers

A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.

Pie

#1

Volume of lead = 100 cm^3

density of lead = 11.34 g/cm^3

mass of the lead piece = density * volume

$m = 100 * 11.34 = 1134 g$

$m = 1.134 kg$

so its weight in air will be given as

$W = mg = 1.134* 9.8 = 11.11 N$

now the buoyant force on the lead is given by

$F_B = W - F_{net}$

$F_B = 11.11 - 11 = 0.11 N$

now as we know that

$F_B = \rho V g$

$0.11 = 1000* V * 9.8$

so by solving it we got

V = 11.22 cm^3

(ii) this volume of water will weigh same as the buoyant force so it is 0.11 N

(iii) Buoyant force = 0.11 N

(iv)since the density of lead block is more than density of water so it will sink inside the water

#2

buoyant force on the lead block is balancing the weight of it

$F_B = W$

$\rho V g = W$

$13* 10^3 * V * 9.8 = 11.11$

$V = 87.2 cm^3$

(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight = 11.11 N

(iii) Buoyant force = 11.11 N

(iv) since the density of lead is less than the density of mercury so it will float inside mercury

#3

Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid

3 0

1 year ago