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Furkat [3]
3 years ago
14

In comparison to radio waves, visible light has:

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

Answer:

Visible light has a shorter wavelength than radio waves

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Determina el volumen de 450 g de alcohol
Rufina [12.5K]

Luego de utilizar la fórmula del volumen de un líquido encontramos que si el acohol tiene 450 gramos entonces su volumen es de

La fórmula del volumen de la densidad de un líquido es la siguiente:

Densidad= Masa / Volumen

Volumen =Masa / Densidad

Es conocido que la densidad del alcohol es 789 kg/m³, y como sabemos por dato que tiene 450gr el siguiente paso es la sustitución:

Volumen=0.45/ 789

Volumen = 0,00057 m³    

6 0
2 years ago
Plz di all plz i will give brainest and thanks to best answer do it right
Inga [223]

Answer:

area 4

Explanation:

area 4 has low pressure

7 0
3 years ago
An object has a mass of 25 kg. What is the force of gravity acting on the object?
aleksandrvk [35]

Answer:

25

Explanation:

Because u don't know the gravity so u take that force and just put it in the gravity spot

5 0
3 years ago
An Airbus A380 airliner can takeoff when its speed reaches 80 m/s. Suppose its engines together can produce an acceleration of 3
Tomtit [17]

Answer:

(a). The time is 26.67 sec.

(b). The distance traveled during this period is 1066.9 m.

Explanation:

Given that,

Speed = 80 m/s

Acceleration = 3 m/s

Initial velocity = 0

(a). We need to calculate the time

Using equation of motion

v = u+at

t = \dfrac{v-u}{a}

Put the value into the formula

t = \dfrac{80-0}{3}

t =26.67\ sec

The time is 26.67 sec.

(b). We need to calculate the distance traveled during this period

Using equation of motion

s = ut+\dfrac{1}{2}at^2

s = 0+\dfrac{1}{2}\times3\times(26.67)^2

s =1066.9\ m

The distance traveled during this period is 1066.9 m.

Hence, This is the required solution.

7 0
3 years ago
I've got an energy and work problem. The premise of the problem is:
Alenkasestr [34]
Refer to the diagram shown below.

μ =  the coefficient of dynamic friction between the crate and the ramp.

1. The applied force of F acts over a distance, d.
    The work done is F*d.

2. The component of the weight of the crate acting down the ramp is
    mg sin(30) = 0.5mg. 
    The work done by this force is 0.5mgd.

3. The normal force is N = mgcos(30) = 0.866mg.
     This force is perpendicular to the ramp, therefore the work done is zero.

4. The frictional force is μN = μmgcos(30) = 0.866μmg.
    The work done by the frictional force is 0.866μmgd.

5. The total force acting on the crate up the ramp is
     F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ) 

6. The work done on the crate by the total force is
    d*(F - 0.5mg - 0.866μmg)

7 0
3 years ago
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