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2 years ago

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Sanjay is making plans for his Saturday afternoon. He can choose a health-enhancing activity like lifting weights, or he can cho

ose a sedentary activity such as watching TV. Sanjay weighs 185 pounds. In 30 minutes, how many more calories can Sanjay burn lifting weights than watching TV? Assume that he does not lift vigorously.

1 answer:

Bezzdna [24]2 years ago

7 0

Answer: Sanjay can burn 100 more calories every 30 minutes if he chooses to lift weights instead of watching tv

Explanation: 133-33= 100 calories (says in article and i just answered it)

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You fill a car with gasoline. The car now has...

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Answer:

Fuel or a provided source of energy for combustion

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2 years ago

A 3.50-meter length of wire with a cross-sectional area of 3.14 × 10-6 meter2 is at 20° Celsius. If the wire has a resistance of

maks197457 [2]

Answer:

$5.6\times 10^{-8}\ Ohm.m$

Explanation:

Resistivity is given by $\rho=\frac {AR}{L}$ where A is cross-sectional area, R is resistance, L is the length and $\rho$ is the reistivity. Substituting 0.0625 for R, 3.14 × 10-6 for A and 3.5 m for L then the resistivity is equivalent to

$\rho=\frac {3.14\times 10^{-6}\times 0.0625}{3.5}=5.60714285714285714285714285714285714285\times 10^{-8}\approx 5.6\times 10^{-8}\ Ohm.m$

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2 years ago

Quinn accelerates her skateboard along a straight path from 0 m/s to 4.0 m/s

umka2103 [35]

initial velocity=u=0m/s
Final velocity=v=4m/s
Time=t=2.5s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{4-0}{2.5}$

$\\ \sf\longmapsto Acceleration=\dfrac{4}{2.5}$

$\\ \sf\longmapsto Acceleration=1.6m/s^2$

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2 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:

$+4q$

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2 years ago

A circuit is built based on the circuit diagram shown.

julsineya [31]

The current in the 50 Ω resistor is A) 1.2 A

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2 years ago

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