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3 years ago

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Two small nonconducting spheres have a total charge of Q = Q1 +Q2 = 91.0 pC, Q1 < Q2. When placed 32.0 cm apart, the force ea

ch exerts on the other is 12.0 N and is repulsive.
a) what is the charge of Q1
b) what is the charge of Q2
c) what would Q1 be if forces attract
d) what would Q2 be if forces attract

1 answer:

stiv31 [10]3 years ago

3 0

When a product fails to perform as warranted, this is called a) contractual liability. b) product malfunction. c) malicious manufacture. d) breach of warranty.

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A 2.35-kg rock is released from rest at a height of 21.4 m. Ignore air resistance and determine (a) the kinetic energy at 21.4 m

kvasek [131]

Explanation:

Given that,

The mass of rock, m = 2.35-kg

It was released from rest at a height of 21.4 m.

(a) The kinetic energy is given by : $E_k=\dfrac{1}{2}mv^2$

As the rock was at rest initially, it means, its kinetic energy is equal to 0.

(b) The gravitational potential energy is given by : $E_p=mgh$

It can be calculated as :

$E_p=2.35\times 9.8\times 21.4\\\\E_p=492.84\ J$

(c) The mechanical energy is equal to the sum of kinetic and potential energy such that,

M = 0 J + 492.84 J

M = 492.84 J

Hence, this is the required solution.

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3 years ago

Water at the top of a slope has potential energy. true or false

ASHA 777 [7]

<span>The statement is TRUE. Water does have potential energy at the top of a slope. The reason why is that potential energy is energy possessed by a body based on its position relative to a zero point. In this case, water at the top of the slope is at an elevation above ground (zero point). The energy is not kinetic (moving) energy since the water is not moving.</span>

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3 years ago

Read 2 more answers

What would be the best design for an experiment that tests how much water expands when frozen?

galben [10]

B. Purchase a small plastic container and mark 1-ounce increments on the outside to determine volume. Pour 5 ounces of water into the container, and place in the freezer for 8 hours. Compare the frozen or ending volume with the liquid or beginning volume.

<h3>How much water expands when frozen?</h3>

Ice is less denser than the liquid form. Water is the only known non-metallic substance that expands when it freezes because it is the unique property of water. Water density decreases and it expands approximately about 9% by volume. For calculating the expansion of water, plastic container is the best option. We know that water expands when the water freezes because it is a unique property of water which allows the survival of aquatic organisms.

So we can conclude that option B is the right answer.

Learn more about water here: brainly.com/question/1313076

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1 year ago

1. An object on Earth and the same object on the Moon would have a difference in

Feliz [49]

Answers: (1) a. weight, (2)b. Force changes by 2/9, (3)b. movement, (4)a. 40,000 Joules, (5)c. the soil will be 5°C.

<h2>Answer 1: a. weight</h2>

Mass and weight are very different concepts.

Mass is the amount of matter that exists in a body, which only depends on the quantity and type of particles within it. This means mass is an intrinsic property of each body and remains the same regardless of where the body is located.

On the other hand, weight is a measure of the gravitational force acting on an object and is directly proportional to the product of the mass $m$ of the body by the acceleration of gravity $g$ :

$W=m.g$

Then, since the Earth and the Moon have different values of gravity, t<u>he weight of an object in each place will vary</u>, but its mass will not.

<h2>Answer 2: b. Force changes by 2/9</h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

$F=G\frac{m_{1}m_{2}}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

If we double the mass of one object (for example $2m_{1}$ ) and triple the distance between both (for example $3r$ ). The equation (1) will be rewritten as:

$F=G\frac{2m_{1}m_{2}}{(3r)^2}$ (2)

$F=\frac{2}{9}G\frac{m_{1}m_{2}}{r^2}$ (3)

If we compare (1) and (2) we will be able to see the force changes by 2/9.

<h2>Answer 3: b. movement</h2>

The Work $W$ done by a Force $F$ refers to the release of potential energy from a body that is <u>moved</u> by the application of that force to overcome a resistance along a path.

When the applied force is constant and <u>the direction of the force and the direction of the movement are parallel,</u> the equation to calculate it is:

$W=(F)(d)$

Now, <u>when they are not parallel, both directions form an angle</u>, let's call it $\alpha$ . In that case the expression to calculate the Work is:

$W=Fdcos{\alpha}$

Therefore, pushing on a rock accomplishes no work unless there is movement (independently of the fact that movement is parallel to the applied force or not).

<h2>Answer 4: a. 40,000 Joules</h2>

The Kinetic Energy is given by:

$K=\frac{1}{2}mV^{2}$ (4)

Where $m$ is the mass of the body and $V$ its velocity

For the first case (kinetic energy $K_{1}=10000J$ for a car at $V_{1}=30 mph=13.4112m/s$ ):

$K_{1}=\frac{1}{2}mV_{1}^{2}$ (5)

Finding $m$ :

$m=\frac{2K_{1}}{V_{1}^{2}}$ (6)

$m=\frac{2(10000J)}{(13.4112m/s)^{2}}$ (7)

$m=111.197kg$ (8)

For the second case (unknown kinetic energy $K_{2}$ for a car with the same mass at $V_{2}=60 mph=26.8224m/s$ ):

$K_{2}=\frac{1}{2}mV_{2}^{2}$ (9)

$K_{2}=\frac{1}{2}(111.197kg)(26.8224m/s)^{2}$ (10)

$K_{2}=40000J$ (11)

<h2>Answer 5: c. the soil will be 5°C</h2>

The formula to calculate the amount of calories $Q$ is:

$Q=m. c. \Delta T$ (12)

Where:

$m$ is the mass

$c$ is the specific heat of the element. For water is $c_{w}=1 kcal/g\°C$ and for soil is $c_{s}=0.20 kcal/g\°C$

$\Delta T$ is the variation in temperature (the amount we want to find for both elements)

This means we have to clear $\Delta T$ from (12) :

$\Delta T=\frac{Q}{m.c}$ (13)

For Water:

$\Delta T_{w}=\frac{Q_{w}}{m_{w}.c_{w}}$ (14)

$\Delta T_{w}=\frac{1kcal}{(1kg)(1 kcal/g\°C)}$ (15)

$\Delta T_{w}=1\°C)}$ (16)

For Soil:

$\Delta T_{s}=\frac{Q_{s}}{m_{s.c_{s}}$ (17)

$\Delta T_{s}=\frac{1kcal}{(1kg)(0.20 kcal/g\°C)}$ (18)

$\Delta T_{s}=5\°C)}$ (19)

Hence the correct option is c.

5 0

3 years ago

Read 2 more answers

A low C (f=65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and

WITCHER [35]

Answer:

T = 676 N

Explanation:

Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g $/m^{2}$ = 0.005 kg

A stationary wave that is set up in the string has a frequency of;

f = $\frac{1}{2L}$ $\sqrt{\frac{T}{M} }$

⇒ T = 4 $L^{2}$ $f^{2}$ M

Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.

But M = L × ρ = (2 × 0.005) = 0.01 kg/m

T = 4 × $2^{2}$ × $65^{2}$ × 0.01

= 4 × 4 ×4225 × 0.01

= 676 N

Tension of the wire is 676 N.

4 0

3 years ago