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9 days ago

Compare and contrast the strength of the forces between two objects with a mass of 1 kg each, a charge of 10

1 answer:

3 0

Let's explore the commonalities between the two forces: * They are proportional to the product of either mass or charge. * Their relationship to distance is inverse, meaning the forces weaken with the square of the distance. * Both forces are long-range, as a distance of zero does not exist up to infinity. Gravitational force is always attractive, while electric force can be either attractive or repulsive. Now, let's look at the distinctions: * The electric force is significantly stronger than the gravitational force. * The gravitational force is perpetually attractive, in contrast to the electric force, which can either attract or repel.

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A coffee company wants to make sure that their coffee is being served at the right temperature. if it is too hot, the customers

Maru [2360]

Response:

The population mean is parameter = 65 c

Explanation:

In the analysis of samples and inferring population behavior, two key elements are essential.

To ascertain the population mean, we typically extract various samples and calculate their average. The average of all these means will serve as an estimate for the population mean. According to the central limit theorem, as sample sizes increase, the average of a sample tends to follow a normal distribution with an estimated mean being the sample mean.

A statistic pertains to a sample, while a parameter refers to the whole population.

In this case, 65 degrees C represents the entire population; thus, it constitutes a parameter.

4 0

11 days ago

Read 2 more answers

Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of t

Softa [2035]

Answer:

1/7 kg

Explanation:

Refer to the attached diagram for enhanced clarity regarding the question.

One of the blocks weighs 1.0 kg and accelerates downward at 3/4g.

g denotes the acceleration due to gravity.

Let M represent the block with known mass, while 'm' signifies the mass of the other block and 'a' refers to the acceleration of body M.

Given M = 1.0 kg and a = 3/4g.

By applying Newton's second law; $\sum fy = ma_y$

For the body with mass m;

T - mg = ma... (1)

For the body with mass M;

Mg - T = Ma... (2)

Combining equations 1 and 2 gives;

+Mg -mg = ma + Ma

Ma-Mg = -mg-ma

M(a-g) = -m(a+g)

Substituting M = 1.0 kg and a = 3/4g into this equation leads to;

3/4 g-g = -m(3/4 g+g)

3/4 g-g = -m(7/4 g)

-g/4 = -m(7/4 g)

1/4 = 7m/4

Multiplying gives: 28m = 4

m = 1/7 kg

Hence, the mass of the other box is 1/7 kg

3 0

27 days ago

The acceleration of segment D is m/s2. Rank segments A, B , C from least accelerations to greatest acceleration. Least

Ostrovityanka [2208]

Answer:

D, C, B, A

Explanation:

The derivative from a velocity-time graph provides the acceleration value.

Segment A

$\frac{dy}{dx} = \frac{15m/s}{1s} = 15m/s^2$

Segment B

$\frac{dy}{dx} = \frac{5m/s}{1s} = 5m/s^2$

Segment C

$\frac{dy}{dx} = \frac{0m/s}{2s} = 0m/s^2$

Segment D

$\frac{dy}{dx} = \frac{-20m/s}{1s} = -20m/s^2$

Sorted from the lowest to the highest acceleration:

D, C, B, A

8 0

12 days ago

Read 2 more answers

A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect

Softa [2035]

Answer:

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

Explanation:

Consider the following:

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field measured at point P

$E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta$

$sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}$

$r^2=d^2+x^2$

Thus,

$E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}$

Now, by applying integration to the equation above

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

4 0

25 days ago

Violet light (λ = 400 nm) passing through a diffraction grating for which the slit spacing is 6.0 μm forms a pattern on a screen

Maru [2360]

Answer:

The third-order dark fringe

Explanation:

y = Distance from central bright fringe = 204 mm

λ = Wavelength = 400 nm

L = Distance from screen to source = 1 m

d = Slit spacing = 6 μm

$tan\theta =\frac{y}{L}\\\Rightarrow tan\theta =\frac{204}{1000}=0.2012^{\circ}$

$dsin\theta=m\lambda\\\Rightarrow m=\frac{dsin\theta}{\lambda}\\\Rightarrow m=\frac{6\times 10^{-6}sin0.2012}{400\times 10^{-9}}=2.9982\approx 3$

The order of the fringe is 3

Thus, it’s identified as a dark fringe.

3 0

18 days ago