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ddd [48]
2 years ago
6

An airplane accelerat s down a runway at 3.20 m/s(squared) for 32.8s until is finally lifts off the ground. Presume to he plane

starts from the rest. Determine the distance traveled before takeoff.
The answer is 1721.3 but I want to know how to get it
Physics
1 answer:
rusak2 [61]2 years ago
5 0
SORRRY not this smart

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A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0° east of
koban [17]

<u>Answer:</u>

 Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

<u>Explanation:</u>

  Let the east point towards positive X-axis and north point towards positive Y-axis.

  Speed of truck = 25 m/s north = 25 j m/s

  Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

                          = (1.43 i + 1.00 j) m/s

    Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

    Magnitude of velocity = 26.04 m/s

    Angle from positive horizontal axis = 86.85⁰

 So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.

4 0
3 years ago
A 2.6 kg mass attached to a light string rotates on a horizontal,
Ainat [17]

The maximum speed the mass can have before it breaks is 2.27 m/s.

The given parameters:

  • <em>maximum mass the string can support before breaking, m = 17.9 kg</em>
  • <em>radius of the circle, r = 0.525 m</em>

The maximum speed the mass can have before it breaks is calculated as follows;

T = ma_c\\\\Mg = \frac{Mv^2}{r} \\\\v^2 = rg\\\\v = \sqrt{rg} \\\\v_{max} = \sqrt{0.525 \times 9.8} \\\\v_{max} = 2.27 \ m/s

Thus, the maximum speed the mass can have before it breaks is 2.27 m/s.

Learn more about maximum speed of horizontal circle here:brainly.com/question/21971127

8 0
2 years ago
I NEED HELP WITH NUMBER 7 I WILL GIVE U BRAINLIEST
grigory [225]

Answer:

the blue one

Explanation:

3 0
3 years ago
A passenger in a train accelerating smoothly away from a station observes that a child’s yo-yo hanging by its string from a lugg
olchik [2.2K]

Answer:

1.73 m/s²

3.0 cm

Explanation:

Draw a free body diagram of the yo-yo.  There are two forces: weight force mg pulling down, and tension force T pulling up 10° from the vertical.

Sum of forces in the y direction:

∑F = ma

T cos 10° − mg = 0

T cos 10° = mg

T = mg / cos 10°

Sum of forces in the x direction:

∑F = ma

T sin 10° = ma

mg tan 10° = ma

g tan 10° = a

a = 1.73 m/s²

Draw a free body diagram of the sphere.  There are two forces: weight force mg pulling down, and air resistance D pushing up.  At terminal velocity, the acceleration is 0.

Sum of forces in the y direction:

∑F = ma

D − mg = 0

D = mg

½ ρₐ v² C A = ρᵢ V g

½ ρₐ v² C (πr²) = ρᵢ (4/3 πr³) g

3 ρₐ v² C = 8 ρᵢ r g

r = 3 ρₐ v² C / (8 ρᵢ g)

r = 3 (1.3 kg/m³) (100 m/s)² (0.47) / (8 (7874 kg/m³) (9.8 m/s²))

r = 0.030 m

r = 3.0 cm

3 0
3 years ago
Technician A says a short to voltage may or may not blow a fuse and may or may not affect more than one circuit. Technician B sa
Tomtit [17]

Answer:

C. Both A and B

Explanation:

Fuses are rated by the amperage they can carry before heat melts the element. The fuse is ideal for protection against short circuits. Short circuits produce enough amperage to vaporize a fuse element and break connection in one cycle of a 60-cycle system.

Specifically, the voltage rating determines the ability of the fuse to suppress the internal arcing that occurs after a fuse link melts and an arc is produced.

6 0
3 years ago
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