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4 years ago

The inner solar system contains the

1 answer:

3 0

Jovian planets and dwarf planets like pluto are in the outer part of our solar system. so, terrestrial planets ,which means earth like planets, live in the inner part of our solar system. or B. is the answer

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Suppose a soccer ball is kicked from the ground at an angle 20.0º above the horizontal at 8.00 m/s. The y-velocity is determined

julia-pushkina [17]

initial speed of the ball by which it is kicked is 8 m/s

now the angle at which it is kicked is 20 degree

here we will have

$v_y = v sin20$

$v_y = 8 sin20 = 2.74 m/s$

now we will have

$\Delta y = v_y t + \frac{1}{2}at^2$

so if the ball again land on the ground at same level then we have

$\Delta y = 0$

$0 = 2.74 t - \frac{1}{2}(9.8) t^2$

$0 = 2.74 - 4.9 t$

$t = 0.56 s$

so total time will be 0.56 s

8 0

4 years ago

(MICROWAVE) <br><br><br>Describe the device and its function.

andreev551 [17]

Answer: You use it to cook food so you can eat. put whatever kind of food you like in there and heat it up for how long you want and then it will beep when done. then you can eat!

8 0

3 years ago

3. Specify the wrong sentences.

soldi70 [24.7K]

Answer:

a: false

b: True

c: i dont know

Explanation:

4 0

2 years ago

Una barra de plata de 335.2 g con una temperatura de 100 ºC se introduce un calorímetro de aluminio de 60 g de masa que contiene

sdas [7]

Respuesta:

0,0560 cal / gºC.

Explicación:

Cantidad de calor; (Q)

Q = mcΔt; Δt = t2 - t1

m = masa, c = capacidad calorífica específica; Δt = cambio de temperatura

c de agua = 1 cal / gºC

c de aluminio = 0,22 cal / gºC

QTotal = Q de agua + Q de aluminio

Q de agua = 450 * 1 * (26 - 23) = 1350 cal

Q de aluminio = 60 * 0.22 * (26 - 23) = 39.6 cal

QTotal = 1350 + 39,6 = 1389,6 cal

Calor perdido = calor ganado

QTotal = calor perdido

- 1389,6 = 335,2 * c * (26 - 100)

-1389,6 = −24804,8 * c

c = 1389,6 / 24804,8

c = 0,056021 cal / gºC.

Capacidad calorífica específica de la plata = 0,0560 cal / gºC.

8 0

3 years ago

A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb

NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

$\sum \tau = F*d$

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

$F*(27-6)= 6*600$

$F = \frac{6*600}{21}$

$F= 171.42 lb$

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

4 0

3 years ago