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3 years ago

11

If it requires 7.0 JJ of work to stretch a particular spring by 1.7 cmcm from its equilibrium length, how much more work will be

required to stretch it an additional 3.6 cm ?

1 answer:

Lynna [10]3 years ago

3 0

Answer:

Explanation:

First of all, well calculate the spring constant k

K = 2Ei/x^2

Where Ei = initial work required

x = initial stretch length

k = 2×7/0.017^2 = 48443J/m^2

Now work done in stretching it to 5.3cm (1.7 + 3.6) or 0.053m

EF = kx^2/2

48443 × 0.053^2/2 = 68J

Work done in stretching additional 3.6cm is equal to

68J-7J = 61J

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Bill and amy want to ride their bikes to school which is 14.4 kilometers away. It takes Amy 49 minutes to get to school and bill

Drupady [299]

3.4m/s

Explanation:

Given parameters:

Distance to school = 14.4km

Time taken by Amy = 49min

Time taken by bill = 20min after Amy = 20+49 = 69min

Unknown parameters:

How much faster is Amy's average speed = ?

Solution:

Average speed is the rate of change of total distance with total time taken.

Average speed = $\frac{total distance }{total time taken}$

convert units to meters and seconds

1000m = 1km

60s = 1min

Distance to school = 14.4 x 1000 = 14400m

Time taken by Amy = 49 x 60 = 2940s

Time taken by Bill = 69 x 60 = 4140s

Average speed of Amy = $\frac{14400}{2940}$ = 4.9m/s

Average speed of Bill = $\frac{4140}{2940}$ = 1.4m/s

Differences in speed = 4.9 - 1.5 = 3.4m/s

Amy was 3.4m/s faster than Bill

learn more:

Average speed brainly.com/question/8893949

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3 years ago

Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B caries a charge of-2q. Sphere C

miskamm [114]

<h2>20. How much charge is on sphere B after A and B touch and are separated?</h2><h3>Answer:</h3>

$\boxed{q_{B}=+2q}$

<h3>Explanation:</h3>

We'll solve this problem by using the concept of electric potential or simply called potential $V$ , which is <em>the energy per unit charge, </em>so the potential $V$ at any point in an electric field with a test charge $q_{0}$ at that point is:

$V=\frac{U}{q_{0}}$

The potential $V$ due to a single point charge q is:

$V=k\frac{q}{r}$

Where $k$ is an electric constant, $q$ is value of point charge and $r$ is the distance from point charge to where potential is measured. Since, the three spheres A, B and C are identical, they have the same radius $r$ . Before the sphere A and B touches we have:

$V_{A}=k\frac{q_{A}}{r_{A}} \\ \\ V_{B}=k\frac{q_{B}}{r_{A}} \\ \\ But: \\ \\ \ r_{A}=r_{B}=r$

When they touches each other the potential is the same, so:

$V_{A}= V_{B} \\ \\ k\frac{q_{A}}{r}=k\frac{q_{B}}{r} \\ \\ \boxed{q_{A}=q_{B}}$

From the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant. </em>So:

$q_{A}+q_{B}=q \\ \\ q_{A}=+6q \ and \ q_{B}=-2q \\ \\ So: \\ \\ \boxed{q_{A}+q_{B}=+4q}$

Therefore:

$(1) \ q_{A}=q_{B} \\ \\ (2) \ q_{A}+q_{B}=+4q \\ \\ (1) \ into \ (2): \\ \\ q_{A}+q_{A}=+4q \therefore 2q_{A}=+4q \therefore \boxed{q_{A}=q_{B}=+2q}$

So after A and B touch and are separated the charge on sphere B is:

$\boxed{q_{B}=+2q}$

<h2>21. How much charge ends up on sphere C?</h2><h3>Answer:</h3>

$\boxed{q_{C}=+1.5q}$

<h3>Explanation:</h3>

First: A and B touches and are separated, so the charges are:

$q_{A}=q_{B}=+2q$

Second: C is then touched to sphere A and separated from it.

Third: C is to sphere B and separated from it

So we need to calculate the charge that ends up on sphere C at the third step, so we also need to calculate step second. Therefore, from the second step:

Here $q_{A}=+2q$ and C carries no net charge or $q_{C}=0$ . Also, $r_{A}=r_{C}=r$

$V_{A}=k\frac{q_{A}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

Applying the same concept as the previous problem when sphere touches we have:

$k\frac{q_{A}}{r} =k\frac{q_{C}}{r} \\ \\ q_{A}=q_{C}$

For the principle of conservation of charge:

$q_{A}+q_{C}=+2q \\ \\ q_{A}=q_{C}=+q$

Finally, from the third step:

Here $q_{B}=+2q \ and \ q_{C}=+q$ . Also, $r_{B}=r_{C}=r$

$V_{B}=k\frac{q_{B}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

When sphere touches we have:

$k\frac{q_{B}}{r} =k\frac{q_{C}}{r} \\ \\ q_{B}=q_{C}$

For the principle of conservation of charge:

$q_{B}+q_{C}=+3q \\ \\ q_{A}=q_{C}=+1.5q$

So the charge that ends up on sphere C is:

$q_{C}=+1.5q$

<h2>22. What is the total charge on the three spheres before they are allowed to touch each other.</h2><h3>Answer:</h3>

$+4q$

<h3>Explanation:</h3>

Before they are allowed to touch each other we have that:

$q_{A}=+6q \\ \\ q_{B}=-2q \\ \\ q_{C}=0$

Therefore, for the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant, </em>then this can be expressed as:

$q_{A}+q_{B}+q_{C}=+6q -2q +0 \\ \\ \therefore q_{A}+q_{B}+q_{C}=+4q$

Lastly, the total charge on the three spheres before they are allowed to touch each other is:

$+4q$

8 0

3 years ago

Anon25 [30]

I thinks its period because they are on the last column to the right

6 0

3 years ago

Read 2 more answers

A car slams on its breaks,producing friction between the tires and the road.Into which type of energy is the mechanical energy o

Step2247 [10]

The answer is Heat Energy

4 0

3 years ago

Read 2 more answers

Block 1 (mass 2.00 kg) is moving rightward at 10.0 m/s and block 2 (mass 5.00 kg) is moving rightward at 3.00 m/s. The surface i

DaniilM [7]

Answer:

a) 0.25m

b) 5 m/s

Explanation:

When the spring is compressed both boxes are moving with the same velocity, so applying the principle of linear momentum conservation:

$m1*v_{o1}+m2*v_{o2}=(m1+m2)*v\\v=5m/s$

Now applying the principle of energy conservation:

$K1+K2+U_{g1}-U_e=Kf+U_{g2}\\K1+0-U_e=K2+0\\U_e=K1+K2-kf\\\frac{1}{2}*k*x^2+=\frac{1}{2}*m1*v1^2+\frac{1}{2}*m1*v1^2-\frac{1}{2}*(m1+m2)*v^2\\\\x=\sqrt{\frac{2.00kg*(10m/s)^2+5.00kg*(3.00m/s)^2-7.00kg*(5m/s)^2}{1120N/m}}\\x=0.25m$

We got that the maximum compression is 0.25m.

5 0

3 years ago