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weqwewe [10]
3 years ago
15

Add the following and give the answer using significant figures 5000+621

Physics
1 answer:
Alexxx [7]3 years ago
8 0

Answer:

;) hope it's helpful

Explanation:

5000 + 621 = 5621

Facts about 5621

Sig Figs

4

5621

Decimals

0

Scientific Notation

5.621 × 103

E-Notation

5.621e+3

Words

five thousand six hundred twenty-one

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Alicia está a punto de perder su bus. En un desesperado intento, corre a una velocidad constante de 5 m/s. Cuando está a 15 m de
pochemuha

Answer:

Si logra alcanzar el bus.

Explanation:

Para poder solucionar este problema debemos de tener en cuenta que Alicia corre a velocidad constante para poder alcanzar el bus. La formula de la cinematica que tiene en cuenta la velocidad constante es la siguiente:

x_{f} = x_{o}+(v*t)

donde:

Xf = Ubicacion del punto donde se encuentra el bus [m]

Xo = Ubicacion desde donde esta Alicia [m]

v = velocidad constante = 5 [m/s]

t = tiempo [s]

Xf - Xo = 15 [m]

15 = 5*t

t = 3 [s]

Ahora con el tiempo podemos encontrar la velocidad del bus por medio de la siguiente ecuacion de cinematica para la aceleracion constante:

v_{f} = v_{i}+(a*t)

donde:

Vf = velocidad del bus despues de los 3 [s]

Vi = velocidad inicial = 0

a = aceleracion = 0.5 [m/s^2]

Vf = 0 + (0.5*3)

Vf = 1.5 [m/s]

La velocidad del bus es menor que la velocidad de Alicia, por ende Alicia alcanzara el bus.

7 0
4 years ago
V
Darina [25.2K]
Resolving vectors question. See diagram enclosed.

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3 years ago
A figure shows a vertically moving block on the end of a cord. The graph next to the figure gives the vertical velocity componen
Sergio [31]
It should be 2/9/10 in functional form
3 0
3 years ago
9. A car driver brakes gently. Her car slows down front --
sleet_krkn [62]

Answer:

9) This is a case of deceleration

10)-0.8 ms-2

b) acceleration is the change in velocity with time

11)

a) 100 ms-1

b) 100 seconds

12) 10ms-1

13) more information is needed to answer the question

14) - 0.4 ms^-2

15) 0.8 ms^-2

Explanation:

The deceleration is;

v-u/t

v= final velocity

u= initial velocity

t= time taken

20-60/50 =- 40/50= -0.8 ms-2

11)

Since it starts from rest, u=0 hence

v= u + at

v= 10 ×10

v= 100 ms-1

b)

v= u + at but u=0

1000 = 10 t

t= 1000/10

t= 100 seconds

12) since the sprinter must have started from rest, u= 0

v= u + at

v= 5 × 2

v= 10ms-1

14)

v- u/t

10 - 20/ 25

10/25

=- 0.4 ms^-2

15)

a=v-u/t

From rest, u=0

8 - 0/10

a= 8/10

a= 0.8 ms^-2

7 0
3 years ago
What is the student's kinetic energy at the bottom of the hill if he is moving
soldi70 [24.7K]

Answer:

KE = 10530 J or 10.53 KJ

Explanation:

The formula for kinetic energy is KE = 1/2 mv^2

Let's apply the formula:

KE = 1/2 mv^2

KE = 1/2 (65kg) (18m/s)^2

KE = 10530 J or 10.53 KJ

5 0
3 years ago
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