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Ugo [173]
3 years ago
7

From the frame of reference of car 2, what is the velocity of car

Physics
1 answer:
Alborosie3 years ago
5 0
It’s is the last one d
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A man is at a car dealership, looking for a car to buy. He looks at the sticker on the driver’s window of a car and sees that th
maxonik [38]

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he will see the sticker because its behind a window bruh and thats a big daddy stack of greens

Explanation:

3 0
3 years ago
A)topirea ghetii este un fenomen...
timofeeve [1]
What language is this?? So I can translate it and answer your question
6 0
4 years ago
La superficie de unas botas suman 400 cm2 y la persona que las usa tiene 45 kg de masa, calcula la presión que ejerce sobre el p
liubo4ka [24]

Answer:

11025 N / m²

Explanation:

Los siguientes datos se obtuvieron de la pregunta:

Área (A) = 400 cm²

Masa (m) = 45 Kg

Aceleración por gravedad (g) = 9,8 m / s²

Presión (P) =?

A continuación, determinaremos la fuerza aplicada. Esto se puede obtener de la siguiente manera:

Masa (m) = 45 Kg

Aceleración por gravedad (g) = 9,8 m / s²

Fuerza (F) =.?

F = m × g

F = 45 × 9,8

F = 441 N

A continuación, convertiremos 400 cm² a m². Esto se puede obtener de la siguiente manera:

1 cm² = 0,0001 m²

Por lo tanto,

400 cm² = 400 cm² × 0,0001 m² / 1 cm²

400 cm² = 0,04 m²

Por tanto, 400 cm² equivalen a 0,04 m².

Finalmente, determinaremos la presión ejercida de la siguiente manera:

Área (A) = 0.04 m².

Fuerza (F) = 441 N

Presión (P) =?

P = F / A

P = 441 / 0,04

P = 11025 N / m²

Por tanto, la presión ejercida es 11025 M / m²

4 0
3 years ago
A roller coaster car starts from rest at the top of a hill 15 m high and rolls down to ground level. From there it starts into a
Softa [21]

Answer:

955.5N

Explanation:

The normal force is given by the difference between the centripetal force and gravity at the top of the loop:

F_N = F_C - F_G = m\frac{v^{2} }{r} - mg

mass m = 65kg

radius of the loop r = 4m

velocity v = ?

g = 9.8 m/s²

To find the centripetal force, you need to find the velocity of the car at the top of the loop.

Use energy conservation:

E_{tot}=mgh + \frac{1}{2} mv^{2}

At the top of the hill:

E_{tot}= mgh_{hill}

At the top of the loop:

E_{tot}=mgh_{loo}_p +\frac{1}{2} m v^{2}

Setting both energies equal and canceling the mass m gives:

gh_{hill} = gh_{loo}_p + \frac{1}{2} v^{2}

Solving for v:

v^{2} = 2g(h_{hill}-h_{loo}_p)

Using v in the first equation:

F_N = \frac{2mg(h_{hill}-h_{loo}_p)}{r} - mg

F_N = 955.5N

7 0
3 years ago
A student walks 3 blocks east, 4 blocks north, and 3 blocks west. What is the displacement of the student?
12345 [234]
4 blocks north because he is it not asking for north east 
5 0
3 years ago
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