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lord [1]
2 years ago
6

A car is traveling at the bottom of a 9.00-meter-radius circular hill with a constant speed v. The moment the car is at the bott

om of the hill, it is noted that a person sitting on a scale in the car reads a value off the scale that is 80% more than normally when the scale is at rest in a bathroom. With what speed is the car traveling
Physics
1 answer:
Rina8888 [55]2 years ago
3 0

Answer:

Explanation:

reading of scale = reaction force of surface R

centripetal force = R - mg = m v² / R , m is mass , v is velocity and R is radius of the circular path .

R = mg + m v² / R

given ,

m v² / R = .80 mg

v² = .80 x g x R

= .8 x 9.8 x 9 = 70.56

v = 8.4 m /s

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To solve this problem we will use the basic concept given by the Volume of a sphere with which the atom approaches. The fraction in percentage terms would be given by the division of the total volume of the nucleus by that of the volume of the atom, that is,

\% Percent = \frac{V_{nucleus}}{V_{atom}}*100

\% Percent = \frac{4/3 \pi (d_{nucleus}/2)^3}{4/3 \pi (d_{atom}/2)^3}*100

\% Percent = \frac{(d_{nucleus}/2)^3}{ (d_{atom}/2)^3}*100

\% Percent =\frac{(1.0*10^{-14}/2)^3}{ (1.1*10^{-10}/2)^3}*100

\% Percent = 7.51*10^{-13}*100

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2 years ago
You have a series circuit powered by a 9V battery. If you double the amount of
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Distinguish between nonsampling error and sampling error. Choose the correct answer below.
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Answer:

The correct answer is D.

Non-sampling error is the error that results from​ under-coverage, non-response​ bias, response​ bias, or​ data-entry errors. Sampling error is the error that results because a sample is being used to estimate information about a population.

Explanation:

Sampling error is related to the variation between the true values of the sample and the population. If occurred, it is always random depending upon the sample chosen.

Non-sampling error can be random as well as non-random. Non-sampling error can occur irrespective of the sample chosen. It is related to the inappropriate analysis of the data.

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Which part of a stream's sediment load moves the slowest?
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1. An object on Earth and the same object on the Moon would have a difference in
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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.

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3 years ago
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