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2 years ago

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A car drives off a cliff next to a river at a speed of 30 m/s and lands on the bank on theother side. The road above the cliff i

s horizontal and 8.3 m above the other shore wherethe car lands. The tires on the car all hit at once and the air resistance is insignificant.How long is the car in the air?

1 answer:

dezoksy [38]2 years ago

7 0

Answer:1.301 s

Explanation:

Given

Initial Velocity(u)=30 m/s

Height of cliff=8.3 m

Time taken to cover 8.3 m

$h=ut+\frac{at^2}{2}$

here Initial vertical velocity is 0

$8.3=\frac{gt^2}{2}$

$t^2=1.69$

$t=1.301 s$

Horizontal distance

$R=u\times t$

$R=30\times 1.301=39.04 m$

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You, Archimedes, suspect that the king’s crown is not solid gold but is instead gold-plated lead. To test your theory, you weigh

hjlf

Answer:

a) 16675.75 Kg/m³ b) 77.6%

Explanation:

the weight of the crown = 60 N, density of gold = 19300 Kg/m^3, density of lead = 11340 kg/m^3, density of water = 1000kg/m^3 and acceleration due to gravity = 9.8 m/s^2

upthrust on the crown = weight in air - weight when fully submerged in water = 60 - 56.4 = 3.6 N

mass of water displaced = 3.6 / 9.8 since weight = mass × g

mass of water displaced = 0.367 Kg

density of water = mass / volume

1000 = 0.367 / volume

cross multiply and find volume

volume of the crown = 0.367 / 1000 = 0.000367 m³ since the crown will displace water of equal volume according to Archimedes principle

Let V1 represent the volume of Gold and let V2 represent the volume of lead

Total volume of the crown = V1 + V2

also

density of gold = mass of gold / V1 and density of lead = mass of lead / V2

19300 = mass of gold in the crown / V1 and 11340 = mass of lead in the crown / V2

19300 V1 = mass of gold and 11340 V2 = mass of lead

add the two together

19300 V1 + 11340 V2 = weigth of the crown / 9.8

19300 V1 + 11340 V2 = 6.12 also

V1 + V2 = 0.000367

make V1 subject of the formula in equation 2

V1 = 0.000367 - V2

substitute for V1 in equation 1

19300 (0.000367 - V2) + 11340 V2 = 6.12

open the bracket

7.083 - 19300 V2 + 11340 V2 = 6.12

rearrange the equation

-7960 V2 = 6.12 - 7.083

-7960 V2 = -0.963

V2 = -0.963 / -7960 = 0.000121 (volume of lead in the crown)

substitute V2 into equation 2

V1 + 0.000121 = 0.000367m³

V1 = 0.000367 - 0.000121 = 0.000246m³ (volume of gold in the crown)

so mass of gold in the crown = 19300 × 0.000246 = 4.748 kg

and mass of lead = 11340 × 0.000121 = 1.372 kg

average density of the crown = (mass of gold + mass of lead) / total volume = 6.12 / 0.000367 = 16675.75 kg/ m³

b) percentage make of gold = mass of gold / total mass × 100 = 77.6 % approx

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2 years ago

Researchers interested in studying stress gave 150 high school seniors a very difficult math exam. After the test, the researche

belka [17]

Answer:

<u>Informed Consent- The study didn't conform this ethical standard.</u>

<u>Debriefing- The study conforms this ethical standard.</u>

<u>Confidentiality- The study conforms this ethical standard.</u>

<u>Protection from harm- The study conforms this ethical standard.</u>

Explanation:

<u>Informed consent</u> is described as a procedure whereby researchers tend to provide details about specific research they are going to conduct, the risk & benefits involved in the study, different alternatives involved in the procedure, etc.

<u>Debriefing </u>is described as a process that is being conducted in any of the different psychological research encompassing human participants after specific research is completed. The researcher tends to describe the details of the research to the participants.

<u>Confidentiality</u> is described as one of the different code of ethics followed by health workers or psychologists. While practicing confidentiality, a psychologist tends to promise his or her participants that he or she will keep everything a secret whatever is being discussed or shared between both of them.

<u>Protection from harm</u> determines that the psychologists conducting research follows the ethics in which he or she has to protect all the participants from any kind of harm.

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2 years ago

If a sound with frequency fs is produced by a source traveling along a line with speed vs. If an observer is traveling with spee

Alexus [3.1K]

Answer:

457.81 Hz

Explanation:

From the question, it is stated that it is a question under Doppler effect.

As a result, we use this form

fo = (c + vo) / (c - vs) × fs

fo = observed frequency by observer =?

c = speed of sound = 332 m/s

vo = velocity of observer relative to source = 45 m/s

vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).

fs = frequency of sound wave by source = 459 Hz

By substituting the the values to the equation, we have

fo = (332 + 45) / (332 - (-46)) × 459

fo = (377/ 332 + 46) × 459

fo = (377/ 378) × 459

fo = 0.9974 × 459

fo = 457.81 Hz

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2 years ago

Calculate the amount of work done to draw a current of 8A from a point at 100V to a point at 120V in 2 seconds?

Morgarella [4.7K]

Given:
I=8A
t=2second
Potential difference,V=120-100=20volt
Workdone=V×i×t
=20×8×2
=320 joule.

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2 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

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

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

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2 years ago