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3 years ago

Please help!!!

Physics

2 answers:

Elena L [17]3 years ago

6 0

Answer:

0.66 psi

Explanation:

Lubov Fominskaja [6]3 years ago

5 0

question: Please help!!!

If a bottle is being squeezed with a force of 10 Newtons and the area of the bottle is 15

squared inches. What is the pressure??

Answer:

1025.64 N/m²

Explanation:

Pressure: This can be defined as the ratio of force to area. The si unit of pressure is N/m².

From the question,

P = F/A........................ Equation 1

Where F = Force, A = Area.

Given: F = 10 Newtons, A = 15 Squared Inches = (15×0.00065) = 0.00975 m²

Substitute these values into equation 1

P = 10/0.00975

P = 1025.64 N/m²

Hence the pressure of the bottle is 1025.64 N/m²

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A jogger runs at an average speed of 4.20 mi/h. (a) how fast is she running in m/s? (report your answer to the correct number of

PolarNik [594]

a) How fast is she running in m / s?
The first thing you should know for this case is the following conversion
1mi = 1609.34m
1h = 3600s
Applying the conversion
4.20 (mi / h) * (1609.34 / 1) (m / mi) * (1/3600) (h / s) = 1.88 m / s
answer
She is running 1.88 in m / s

(b) how many kilometers does she run in 33.0 min?
By definition, the distance equals the speed by time
d = v * t
Also,
1min = 60seconds
1 km = 1000 meters.
We have then
d = (1.88) * (33) * (60) * (1/1000) = 3.72 Km
answer
she run 3.72 Km in 33.0 min

(c) if she starts a run at 11:15 a.m., what time is it after she covers 4.84 × 104 ft? : p.m.
The first thing we need to know is the following conversion
1Km = 3280.84 feet
We have then
3.72 (Km) * (3280.84 / 1) (feet / Km) = 12204.73 feets
I have the following rule of three we can know how long she travels 4.84 × 104 ft
12204.73 feets ----> 33 min
4.84 * 10 ^ 4 feets ----> x
Clearing x:
x = ((4.84 * 10 ^ 4) / (12204.73)) * (33) = 131 min
131 min = 2:11:00 hours
she starts a run at 11:15 a.m.
she finishes at
1:26 p.m.
answer
it is 1:26 p.m when she covers 4.84 × 104 ft

5 0

4 years ago

A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimum coefficient of static friction

DanielleElmas [232]

Answer:

The minimum coefficient of static friction between the road and the car's tires is 0.816

Option "C"

Explanation:

Given;

velocity of the car, v = 40.0 m/s

radius of horizontal curve, r = 200 m

For a minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at the given speed without sliding, centripetal force must equal frictional force.

$F_{frictional} = F_{centripetal}\\\\\mu mg = \frac{mv^2}{r} \\\\\mu = \frac{v^2}{rg}$

where;

μ is the minimum coefficient of static friction

$\mu = \frac{40^2}{200*9.8} \\\\\mu = 0.816$

Thus, the minimum coefficient of static friction between the road and the car's tires is 0.816

Option "C"

3 0

3 years ago

A ladder is leaning against a vertical wall, and both ends of the ladder are at the point of slipping. The coefficient of fricti

zaharov [31]

Answer:

$\alpha =29.60$

Explanation:

Let the normal force of the wall on the ladder be N2 and the normal force of the ground on the ladder be N1.

Horizontal forces:

$N_{2}= (u1)(N_{1})$ [1]

Vertical forces:

$N_1 + (u2)(N_{2})=m*g$ [2]

Substitute [2] into [1]:

$N_2 = (u1)*[m*g - (u2)(N_2)]$

$N_2= \frac{(u1)m*g}{[1 + (u1)(u2)]}$ [3]

Torques about the point where the ladder meets the ground:

$m*g(\frac{L}{2})sin\alpha= (N_2)(L)cos\alpha+(u2)(N_2)(L)sin\alpha$

$(\frac{1}{2})*m*g=(N_2)*cot\alpha+(u2)(N_2)$

$[1 + (u1)(u2) - 2(u2)(u1)]/2 [1 + (u1)(u2)]= [(u1)/[1 + (u1)(u2)]]cot\alpha$

tanα = $\frac{2*(u1)}{1-u1*u2}$

$\alpha =tan^{-1}*(\frac{2*0.265}{1-0.265*0.253})$

$\alpha =tan^{-1}*(0.568)$

$\alpha =29.60$

4 0

3 years ago

A source containing a mixture of hydrogen and deuterium atoms emits light at two wavelengths whose mean is 540 nm and whose sepa

Ne4ueva [31]

Answer:

N=3176.5rulling

Explanation:

We were told that the source containing a mixture of hydrogen and deuterium atoms emits light with

wavelengths whose mean is 540 nm

Then λ= 540 nm, but we need to convert to metre which = (540× 10⁻⁹m)

Also whose separation is 0.170 nm, which mean the difference between the wavelength is 0.170 nm then

Δ λ = 0.170 nm the we convert to metre we have. Δλ= 0.170 nm= (0.170×10⁻⁹m)

the formular below can be used to can be used to calculate our minimum number of lines

N= λ /(m Δλ)

Where N is number of fillings i.e number of lines

λ= wavelength

Δλ= difference in wavelength

m=1

Then if we substitute the values we have

,N= (540× 10⁻⁹ m)/[(1)*(0.170× 10⁻⁹m)]

N =3176.5rulling

Therefore, minimum number of lines = =3176.5rulling

4 0

3 years ago

How long must a flute be in order to have a fundamental frequency of 262 Hz (this frequency corresponds to middle C on the evenl

spin [16.1K]

Answer:

L=0.654 m

Explanation:

Concepts and Principles

1- The speed of sound in air is expressed as a function of the temperature of air as follows:

v=(331 m/s)√(1+T_C/273°C) (1)

where 331 m/s is the speed of sound in air at temperature 0°C and Tc is the temperature of air in Celsius.

Standing Wave Patterns in Pipes:

A pipe open at both ends can have standing wave patterns with resonant frequencies:

f=v/λ=nv/2L n=1,2,3.........

where v is the speed of sound in air.

Given Data

f_1 (fundamental frequency of the flute) = 262 Hz

T (temperature of the air) = 20°C

The flute is open at both ends.

Required Data

We are asked to determine the length of the tube.

Solution

The speed of sound in air at temperature T = 20°C is found from Equation (1):

v=(331 m/s)√(1+T_C/273°C)

=342.91 m/s

The fundamental frequency of the flute is found by substituting n = 1 into Equation (2):

f=v/2L

Solve for L:

L=v/2f_1

L=0.654 m

7 0

3 years ago