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Mazyrski [523]
2 years ago
5

Why does the closed top of a convertible bulge when the car is riding along a highway? Why does the closed top of a convertible

bulge when the car is riding along a highway? The air pressure is greater outside the car than inside. The volume of air inside the car increases. The air blows into the front part of the roof, lifting the back part. The air pressure inside the car is greater than the pressure outside.
Physics
1 answer:
iren2701 [21]2 years ago
4 0

Answer:

Option D

The air pressure inside the car is greater than the pressure outside.

Explanation:

When considering airflow over and around a surface, from Bernoulli's equation, air flow regions with higher velocity have a lower pressure, and regions with lower velocity have a higher pressure.

The air outside the convertible is moving faster than the air inside the convertible. This leads to a higher pressure zone just below the surface of the roof (inside the car) causing the roof of the convertible to bulge upwards

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What is the mass and density of 237 mL of water
oee [108]

Answer:

<h2><em>V(water)= 237 mL=237×10^-6 m^3</em></h2><h2><em>ρ(water)=1000 kg/m^3</em></h2><h2><em>m=</em><em>ρ×V=(1000)×(237×10^-6)</em></h2><h2><em>m= 237×10^-3 = 0.237 kg</em></h2><h2><em>m= 237 gram.</em></h2>
8 0
2 years ago
In the Atwood machine shown below, m1 = 2.00 kg and m2 = 6.05 kg. The masses of the pulley and string are negligible by comparis
Rus_ich [418]
M1 descending
−m1g + T = m1a 

m2 ascending
m2g − T = m2a

this gives :
(m2 − m1)g = (m1 + m2)a 

a = (m2 − m1)g/m1 + m2
   = (5.60 − 2)/(2 + 5.60) x 9.81 
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5 0
2 years ago
A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag
klio [65]

Answer:

57.94°

Explanation:

we know that the expression of flux

\Phi =E\times S\times COS\Theta

where Ф= flux

           E= electric field

           S= surface area

        θ = angle between the direction of electric field and normal to the surface.

we have Given Ф= 78 \frac{Nm^{2}}{sec}

                          E=1.44\times 10^{4}\frac{Nm}{C}

                          S=\pi \times 0.057^{2}

                         COS\Theta =\frac{\Phi }{S\times E}

 =   \frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}

 =0.5306

 θ=57.94°

4 0
2 years ago
If Earth were twice as far from the sun, the force of gravity attracting the Earth to the Sun would be
zalisa [80]
If Earth was twice as far from the sun, the force of gravity attracting the Earth to the sun would be only one-quarter as strong. The correct answer will be C. 
6 0
2 years ago
Read 2 more answers
A person wants to lose weight by "pumping iron". The person lifts an 80 kg weight 1 meter. How many times must this weight be li
statuscvo [17]

Answer:

37357 sec  

or 622 min

or 10.4 hrs

Explanation:

GIVEN DATA:

Lifting weight 80 kg

1 cal = 4184 J

from information given in question we have

one lb fat consist of 3500 calories = 3500 x 4184 J

= 14.644 x 10^6 J  

Energy burns in 1 lift = m g h

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lifts required = \frac{(14.644 x 10^6)}{784}

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from the question,

1 lift in 2 sec.

so, total time = 18679 x 2 = 37357 sec  

or 622 min

or 10.4 hrs

3 0
2 years ago
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