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satela [25.4K]
4 years ago
14

The weights of a large number of miniature poodles are approximately normally distributed with a mean of 99 kilograms and a stan

dard deviation of 0.90.9 kilogram. If measurements are recorded to the nearest tenth of a​ kilogram

Physics
1 answer:
Karo-lina-s [1.5K]4 years ago
8 0

Answer:

See explanation below

Explanation:

As I say in the comments, the question is incomplete, however, I will try to answer this by using data that I found on another site.

This is the part of the question that is not here:

If measurements are recorded to the nearest

tenth of a kilogram, find the fraction of these poodles

with weights

(a) over 9.5 kilograms;

(b) of at most 8.6 kilograms;

So, assuming a mean of 8 kg, and 0.9 of standard deviation, let X represents the weight of the poodles

The expression to calculate the fraction of poodle needed is:

Z = X - u / d

u: weight of the large number of poodle

d: standard deviation

Replacing data of a) wer have:

Z = 9.5 - 8 / 0.9

Z = 1.67

With this value, we need to take the value of Z, and see the area under the curve of standard deviation (see table attached)

Therefore:

P (X > 9.5) = P(Z > 1.67) = 0.5 - P (Z < 1.67) = 0.5 - 0.4525 = 0.0475

b) In this part, is the same as part a) so:

Z = 8.6 - 8 / 0.9 = 0.67

The value for area in the curve is 0.2486 so:

P = 0.5 + 0.2486 = 0.7486

Hope this helps

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A spider begins to spin a web by first hanging from a ceiling by his fine, silk fiber. He has a mass of 0.025 kg and a charge of
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Answer:

a) (5.59 × 10³) N/C

b) 0.226 N directed away from the spider.

Explanation:

a) Electric field, E, felt as a result of point charge, Q, at a distance, d away is given by

E = kQ/d²

So, magnitude of the electric field due to the charge on the second spider at the position of the first spider

Q = 4.2 µC = 4.2 × 10⁻⁶ C

k = Coulomb's constant = 8.99 × 10⁹ Nm²/C

d = 2.6 m

E = (8.99 × 10⁹ × 4.2 × 10⁻⁶)/2.6²

E = 5.59 × 10³ N/C

b) Tension in the silk fiber above the spider is the net force due to the weight of spider one and the force of repulsion due the two charges.

Force due to the two charges = Eq

where q now represents the charge of the first spider at the first point, feeling the electric field calculated in (a)

F = 5.59 × 10³ × 3.4 × 10⁻⁶ = 0.01901 N directed upwards. (That is, F = + 0.019 N)

Weight of the spider = mg = 0.025 × 9.8 = 0.245 N directed downwards. (That is, W = -0.245 N)

Net force, T = mg - F = 0.245 - 0.019 = 0.226 N (that is, 0.226 N, directed upwards, away from the spider).

8 0
4 years ago
1.5-m length of straight wire experiences a maximum force of 1.2 N when in a uniform magnetic field that is 1.8 T. 1) What curre
Iteru [2.4K]

Answer:

 I = 0.44 A

Explanation:

The magnetic force on a conductor is given by the expression

       F = I L x B

Where bold letters indicate vectors, I is the current, L is the vector in the direction of the current, and B is the magnetic field

Since the force is maximum, the wire must be perpendicular to the magnetic field, therefore

        F = I L B sin 90

        I = F / L B

Let's calculate

        I = 1.2 / 1.5 1.8

        I = 0.44 A

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3 years ago
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The force acting between two charged particles a and b is
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Answer: The correct answer is B. trust me, I just took that test. :)

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3 years ago
An observation which is descriptive is considered quantitative, creative, qualitive
Vesnalui [34]
A descriptive observation may very well be a mixture of both quantitative and qualitative as it can utilize elements of both types. Qualitative deals with the kinds of observations that cannot be measured in numerical form. Quantitative data is just that.
3 0
4 years ago
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Hi, so i have to find T1, can some1 help?
iragen [17]

30.1 N

Explanation:

Given:

W_1 = 16\:\text{N}

W_2 = 8\:\text{N}

Let's write the components of the net forces at the intersections. Note that the system is equilibrium so all the net forces are zero.

<u>Forces</u><u> </u><u>involving</u><u> </u><u>W1</u><u>:</u>

x:\:\:\:-T_1 + T_3\cos \alpha = 0\:\: \\ \text{or}\:\:T_2 = T_3\cos \alpha\:\:\:\:\:(1)

y:\:\:\:T_3\sin \alpha - W_1 = 0\:\:\: \\ \text{or}\:\:\:T_3\sin \alpha = W_1\:\:\:\:\:\:(2)

<u>Forces</u><u> </u><u>involving</u><u> </u><u>W2</u><u>:</u>

x:\:\:\:T_1\sin 53 - T_3\sin \alpha = W_2\:\:\:\:\:\:\:(3)

y:\:\:\:T_4 - T_1\cos 53 - T_3\cos \alpha = 0\:\:\:\;(4)

Substitute (2) into (3) and we get

T_1\sin 53 - W_1 = W_2

Solving for T_1,

T_1 = \dfrac{W_1 + W_2}{\sin 53} = 30.1\:\text{N}

7 0
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