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satela [25.4K]
2 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]2 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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when you drop a pebble from height h, it reaches the ground with kinetic energy k if there is no air resistance. from what heigh
marysya [2.9K]

Answer:

From the initial height h

Explanation:

When a material or substance is drop from a height h, it possesses potential energy, immediately it is dropped from that height, the potential energy is gradually converted to kinetic energy, it gets to a point where the potential energy equals the kinetic energy, as the material touches the ground, all potential energy has been converted to kinetic energy already

6 0
1 year ago
A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the sphere.
Airida [17]

Answer:

Part A :  E =   \frac{1}{4\pi}ε₀ Q₁/R₁² Volt/meter

Part B :  V =  \frac{1}{4\pi}ε₀ Q₁/R₁ Volt

Explanation:

Given that,

Charge distributed on the sphere is Q₁

The radius of sphere is R

₁

The electric potential at infinity is 0

<em>Part A</em>

The space around a charge in which its influence is felt is known in the electric field. The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.  

If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by

                          F = \frac{1}{4\pi}ε₀ Q₁/R₁²

Then the electric field at that point is

                                   E =  F/1

                            E =  \frac{1}{4\pi}ε₀ Q₁/R₁²  Volt/meter

Part B

The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.

Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation

                           V =  \frac{1}{4\pi}ε₀ Q₁/R₁  Volt

4 0
1 year ago
12*8A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 13 feet. The ball is started in
max2010maxim [7]

Answer:

See attached pictures.

Explanation:

See attachments for explanation.

6 0
1 year ago
During the 440, a runner changes his speed as he comes out of the curve onto the home stretch from 18 ft/sec to 38 ft/sec over a
Sloan [31]

Answer:

6.67ft/s^2

Explanation:

We are given that

Initial velocity=u=18ft/s

Final velocity,v=38ft/s

Time=t=3 s

We have to find the average acceleration over that 3 s period.

We know that

Average acceleration,a=\frac{v-u}{t}{t}

Using the formula

Average acceleration,a=\frac{38-18}{3}ft/s^2

Average acceleration,a=\frac{20}{3}ft/s^2

Average acceleration,a=6.67ft/s^2

Hence, the average acceleration=6.67ft/s^2

5 0
1 year ago
4. A trolley of mass 2kg rests next to a trolley of mass 3 kg on a flat
nydimaria [60]

Answer:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. The total momentum of the trolleys after separation is zero

c. The momentum of the 2 kg trolley after separation is 12 kg·m/s

d. The momentum of the 3 kg trolley is -12 kg·m/s

e. The velocity of the 3 kg trolley = -4 m/s

Explanation:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0

c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s

d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s

e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley

∴ The momentum of the 3 kg trolley = 3 kg × Velocity of the 3 kg trolley = -12 kg·m/s

The velocity of the 3 kg trolley = -12 kg·m/s/(3 kg) = -4 m/s

3 0
1 year ago
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