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Nutka1998 [239]
4 years ago
12

A"car"initially"at"rest"experiences"a" constant"acceleration"along"a"horizontal" road."the"position"of"the"car"at"several" succe

ssive"equal"time"intervals"is" illustrated"here. between"which"adjacent"positions"is"the" change"in"kinetic"energy"of"the"car"the" greatest
Physics
1 answer:
marysya [2.9K]4 years ago
7 0

In the process of peppering the question with those forty (40 !) un-necessary quotation marks, you neglected to actually show us the illustration.  So we have no information to describe the adjacent positions, and we're not able to come up with any answer to the question.

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A soccer player icks a rock horizontally off a 40m high cliff into a pool f water if the player hears the sound of the splash s
Semenov [28]

Answer:

v = 9.936 m/s

Explanation:

given,

height of cliff = 40 m

speed of sound = 343 m/s

assuming that time to reach the sound to the player = 3 s

now,

time taken to fall of ball

t = \sqrt{\dfrac{2s}{g}}

t = \sqrt{\dfrac{2\times 40}{9.8}}

t = 2.857 s

distance

d = v  x t

d = v x 2.875

time traveled by the sound before reaching the player

t_0 = t - t_{fall}

t_0 = 3 - 2.875

t_0 = 0.143 s

distance traveled by the wave in this time'

r = 0.143 x 343

r= 49.05 m

now,

we know.

d² + h² = r²

d² + 40² = 49.05²

d =28.387 m

v x 2.875=28.387 m

v = 9.936 m/s

7 0
3 years ago
Air in a 120 km/h wind strikes head on the face of abuilding 45m wide by 75m high and is brought to rest. if air has a mass of 1
AlekseyPX
From conservation of momentum, the ram force can be calculated similarly to rocket thrust:
 F = d(mv)/dt = vdm/dt.
 <span>In other words, the force needed to decelerate the wind equals the force that would be needed to produce it. 
</span><span> v = 120/3.6 = 33.33 m/s 
</span><span> dm/dt = v*area*density
</span> dm/dt = (33.33)*((45)*(75))*(1.3) 
 dm/dt = <span> 146235.375 </span><span>kg/s 
</span><span> F = v^2*area*density
</span> F = (33.33)^2*((45)*(75))*(1.3) = <span> <span>4874025 </span></span><span>N 
</span> This differs by a factor of 2 from Bernoulli's equation, which relates velocity and pressure difference in reference not to a head-on collision of the fluid with a surface but to a fluid moving tangentially to the surface. Also, a typical mass-based drag equation, like Bernoulli's equation, has a coefficient of 1/2; however, it refers to a body moving through a fluid, where the fluid encountered by the body is not stopped relative to the body (i.e., brought up to its speed) like is the case in this problem.
4 0
3 years ago
Read 2 more answers
A moon orbits a planet every 42 hours with a mean orbital radius of .002819 AU. The mass of the moon is 8.932 x 1022 kg. Using N
Pepsi [2]

Answer:

The mass of the planet  is 1.9407\times10^{27}\ kg

Explanation:

Given that,

Time period = 42 hours = 151200 sec

Orbital radius = 0.002819 AU = 421716397.5 m

Mass of moon m=8.932\times10^{22}\ kg

We need to calculate the mass of the planet

Using Kepler’s third law

T^2\propto a^3

T^2=\dfrac{4\pi^2}{G(M+m)}\times a^3

Where, a = orbital radius

T = time period

G = gravitational constant

M = mass of moon

m = mass of planet

Put the value into the formula

(151200)^2=\dfrac{4\pi^2}{6.673\times10^{-11}(8.932\times10^{22}+m)}\times(421716397.5)^3

(8.932\times10^{22}+m)=\dfrac{4\pi^2}{6.673\times10^{-11}}\times\dfrac{(421716397.5)^3}{(151200)^2}

(8.932\times10^{22}+m)=1.94087\times10^{27}

m=1.94087\times10^{27}-8.932\times10^{22}

m=1.9407\times10^{27}\ kg

Hence, The mass of the planet  is 1.9407\times10^{27}\ kg

8 0
4 years ago
Describe the relationship between joules, meters, and newtons
Naily [24]
Newton is a unit of force. Joules is an amount of work which is equal to the force times distance, or newton meters. So the product of newtons and meters makes joules.
5 0
3 years ago
Mary’s mass is 57 kg and Elizabeth’s weight is 624 N. Each one carries a 9 kg box to the library. Who has the greater normal for
stepan [7]
Elizabeth, weight formula is 624 divided by 9.8 which is greater than 57kg
5 0
3 years ago
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