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Crank
2 years ago
14

Un ciclista en una cuesta hacia abajo, marcha a una velocidad de 45km/h aplica los frenos y al cabo de 5s su velocidad se ha red

ucido a 15km/h, ¿cual es la distancia recorrida durante los 5 segundos?
Physics
1 answer:
nexus9112 [7]2 years ago
8 0

Answer:

Los 0.0416km

esto se debe a que transponemos la fórmula acelerada y obtenemos Distancia = velocidad × tiempo

también recuerda transponer los segundos a horas viendo que la velocidad es por hora

También tenga en cuenta que no hablo español, así que esto fue extremadamente difícil

culto

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a crate is being lifted into a truck. if it is moved with a 2470n force and 3650 j of work is done , then how far is the crate b
SVEN [57.7K]

Answer:

The crate was being lifted by a height of 1.48 meters.

Explanation:

In an attempt o move a crate;

Force applied = 2470 N

Work done by the force = 3650 J

We know that the work done is defined as the force used to move an object to a distance.

Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.

Work done is defined as:

Work = Force*distance covered in the direction of the force

3650 = 2470*distance

distance = 3650/2470

distance = 1.48 meters

4 0
3 years ago
A radioactive material has a count rate of 400 per minute. It has a half life of 40 years. How long will it take to decay to a r
cestrela7 [59]

Answer:

160 years.

Explanation:

From the question given above, the following data were obtained:

Initial count rate (Cᵢ) = 400 count/min

Half-life (t½) = 40 years

Final count rate (Cբ) = 25 count/min

Time (t) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Initial count rate (Cᵢ) = 400 count/min

Final count rate (Cբ) = 25 count/min

Number of half-lives (n) =?

Cբ = 1/2ⁿ × Cᵢ

25 = 1/2ⁿ × 400

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25 × 2ⁿ = 400

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2ⁿ = 16

Express 16 in index form with 2 as the base

2ⁿ = 2⁴

n = 4

Thus, 4 half-lives has elapsed.

Finally, we shall determine the time taken for the radioactive material to decay to the rate of 25 counts per minute. This can be obtained as follow:

Half-life (t½) = 40 years

Number of half-lives (n) = 4

Time (t) =?

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Cross multiply

t = 4 × 40

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7 0
2 years ago
Which describes what an outside observer at rest would observe about a spaceship that speeds up and approaches the speed of ligh
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The outside observer, at rest relative to the spaceship, would see the spaceship
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At the same time, the crew of the spaceship, looking back at the observer on
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3 years ago
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Answer:

4 A

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so the current is halved.

7 0
3 years ago
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