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user100 [1]
3 years ago
7

¿Qué distancia recorrió un avión que viajaba a 750 km/h después de 2 h y media de vuelo?

Physics
1 answer:
arsen [322]3 years ago
7 0

Answer:

mehhvbhhhhhhhehshrbeheherhhehehthsjrjjrhrn

Explanation:

bdbsbdbi

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If you double the net force on an object what is the result on the acceleration?
melamori03 [73]

Answer:the acceleration will double

Explanation:

3 0
3 years ago
2. Lenses such as those in microscopes and telescopes depend on which property of<br><br> light?
Marizza181 [45]

Answer:

The eyepiece comprises a converging lens that is a magnifying lens. The lens has a short focal length,

This lens magnifies this image.

Explanation:

In lenses such as those in microscopes and telescopes, the objective forms an image with the following features:

1. Image is real

2. Image is diminished in size

3. Also, the image formed is upside-down.

The eyepiece comprises a converging lens that is a magnifying lens. The lens has a short focal length,

This lens magnifies this image.

5 0
3 years ago
Pada setiap tumbukan berlaku hukum
sweet-ann [11.9K]
Hij ane les spoona sudhea alldor oakie
6 0
3 years ago
There are (one can say) three coequal theories of motion for a single particle: Newton's second law, stating that the total forc
PtichkaEL [24]

Answer:

vf = 14.2176 m/s

Explanation:

Given

m = 4 Kg

viy = 7.00 ĵ m/s

Fx = 11.0 î N

t = 4.5 s

vf = ?

Using the Impulse - Momentum Theorem, we have

F*Δt = m*Δv    ⇒  F*Δt = m*(vf - vi)

⇒    vf = (F*Δt + m*vi) / m

⇒    vf = (F*Δt + m*vi) / m

For <em>x-component</em>

⇒    vfx = (Fx*Δt + m*vix) / m = (11 N*4.5 s + 4 Kg*0 m/s) / (4 Kg)

⇒    vfx = 12.375 î m/s

For <em>y-component</em>

⇒    vfy = (Fy*Δt + m*viy) / m = (0 N*4.5 s + 4 Kg*7 m/s) / (4 Kg)

⇒    vfy = 7 ĵ m/s

Finally:

vf = √(vfx² + vfy²)

⇒   vf = √((12.375 m/s)² + (7 m/s)²)

⇒   vf = 14.2176 m/s

8 0
3 years ago
Find the minimum kinetic energy in MeV necessary for an α particle to touch a 39 19 K nucleus that is initially at rest, assumin
SIZIF [17.4K]

Answer:

KE = 9.6MeV

Explanation:

Given the relationships we understand that,

q_1 = 19e, q_2 = 2e

The Potential at point p, is given by the following formula

V_ {p} = \frac {Kq_1} {r}

According to the graphic designed, you have,

V_ {p} = \frac {Kq_1} {(r_1 + r_2)}

V_ {p} = \frac {(9 * 10 ^ 9) (19 * 1.6 * 10 ^ {- 15})} {(3.7 + 2) * 10 ^{- 15}}}

V_ {p} = 4.8 * 10 ^ 6V

The kinetic energy of the particle would be given by

KE = q_2 * V_ {p} = 2e * 4.8 * 10 ^ 6V

KE = 9.6MeV

4 0
3 years ago
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