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Darina [25.2K]
3 years ago
6

The velocity is zero but the speed is a nonzero quantity?

Physics
1 answer:
jenyasd209 [6]3 years ago
5 0

The starting and ending points of the motion are the same.. . . . .

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If two objects have the same volume but one has a greater mass, the one with greater mass
11Alexandr11 [23.1K]

Explain more. I can't understand

7 0
3 years ago
HELP ME PLEASE I NEED IT NOW
tia_tia [17]

Answer:

the angle of incident is 40°

Explanation:

NQ is the normal to the mirror, therefore

angle NQA =90°

PQA = 50°

incident angle = NQA - PQA

90°- 50° = 40°

note that the angle of reflection is equal to the angle of incident

8 0
3 years ago
a plane is flying due east in still air at 395 km/h. suddenly, the plane is hit by wind blowing at 55km/h toward the west. what
Sphinxa [80]
Let's be clear:  The plane's "395 km/hr" is speed relative to the
air, and the wind's "55 km/hr" is speed relative to the ground.

Before the wind hits, the plane moves east at 395 km/hr relative
to both the air AND the ground.

After the wind hits, the plane still maintains the same air-speed.
That is, its velocity relative to the air is still 395 km/hr east.
But the wind vector is added to the air-speed vector, and the
plane's velocity <span>relative to the ground drops to 340 km/hr east</span>.

6 0
3 years ago
It has been suggested that rotating cylinders several miles in length and several miles in diameter be placed in space and used
stepladder [879]

Answer:

the required revolution per hour is 28.6849

Explanation:

Given the data in the question;

we know that the expression for the linear acceleration in terms of angular velocity is;

a_{c} = rω²

ω² = a_{c} / r

ω = √( a_{c} / r )

where r is the radius of the cylinder

ω is the angular velocity

given that; the centripetal acceleration equal to the acceleration of gravity a a_{c}  = g = 9.8 m/s²

so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m

Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m

so we substitute

ω = √( 9.8 m/s² / 3909.87 m )

ω = √0.002506477 s²  

ω = 0.0500647 ≈ 0.05 rad/s  

we know that; 1 rad/s = 9.5493 revolution per minute

ω = 0.05 × 9.5493 RPM

ω = 0.478082 RPM  

1 rpm = 60 rph  

so  

ω = 0.478082 × 60

ω = 28.6849  revolutions per hour  

Therefore, the required revolution per hour is 28.6849

7 0
3 years ago
Consider the graph below that represents the variation of the velocity with respect to time of an object moving along the x - ax
vazorg [7]

In a velocity-time graph, the area under the curve represents the distance.

The distance traveled from 10s to 18 s is

\begin{gathered} d\text{ = }\frac{1}{2}\times10\times(18-10) \\ =\text{ 40 m} \end{gathered}

Final Answer: The distance traveled is 40 m from time 10 s to 18 s.

3 0
1 year ago
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