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3 years ago

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A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.0 m/s. The car is a distanc

e away. The bear is 26m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible values for d?

1 answer:

Brums [2.3K]3 years ago

4 0

For this specific problem, the maximum value for d is 52m. I am hoping that this answer has satisfied your query about and it will be able to help you, and if you’d like, feel free to ask another question.

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Spin dynamics during chirped pulses: applications to homonuclear decoupling and broadband excitation

Likurg_2 [28]

Swept-frequency pulses have found use in a variety of fields, including spectroscopic methods where effective spin control is necessary.

To find more, we have to study about the spectroscopic methods.

<h3>What is homonuclear decoupling and broadband excitation?</h3>

A thorough understanding of the evolution of spin systems during these pulses is crucial for many of these applications since it not only helps to describe how procedures work but also makes new methodologies possible.
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7 0

2 years ago

How much work is needed to stretch this spring a distance of 5. 0 cm , starting with it unstretched?

densk [106]

The work done by the unstretched spring is 5 J.

<h3>What is the Hooke's law?</h3>

Hooke's law states that the extension of a given material is directly proportional to the force applied as log as the elastic limit is not exceeded. Thus we have to know that; F = Ke

F = force applied

K = force constant

e = extension

Using the graph;

K = F/e

F = 200N

e = 5 cm or 5 * 10^-2 m

K = 200N/ 5 * 10^-2 m

K = 4000 N/m

Now;

Work = 1/2Ke^2

Work = 0.5 * 4000 N/m * (5 * 10^-2 m)^2

Work = 5 J

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2 years ago

A train travels 70 kilometers in 1 hours, and then 63 kilometers in 2 hours. What is its average speed?

lozanna [386]

The answer is: " 44 $\frac{1}{3}$ km " ;
or; write as: " 44.333 km " .
___________________________________________________________
Explanation:
___________________________________________________________
(70 km + 63 km) ÷ (2 + 1 ) = 133 km ÷ 3 = " 44 $\frac{1}{3}$ km " ;
or; write as: " 44.333 km " .
___________________________________________________________

3 0

4 years ago

: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

b

$F_{No}= 7771.125 \ N$

c

$F_p = 2.2*10^{6} N$

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

So

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

So

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

6 0

4 years ago

Which of these statements describes a constraint imposed on scientific

valina [46]

Answer:

The answer is C.

4 0

3 years ago