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qwelly [4]
3 years ago
15

What are food source of calcium?

Physics
2 answers:
bearhunter [10]3 years ago
5 0

Answer:

Milk

Explanation:

It can be milk and other foods similar to milk like ice cream, cheese, yogurt, etc. This is why people say milk is good for your bones cause it has calcium

emmasim [6.3K]3 years ago
5 0

Answer:

The main foods rich in calcium are dairy products like milk, cheese and yogurt. However, many non-dairy sources are also high in this mineral. These include seafood, leafy greens, legumes, dried fruit, tofu and various foods that are fortified with calcium.

Explanation:

hope dis helps mark me as brainliest pls!!!!!

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Blocks A (mass 2.00 kg) and B (mass 10.00 kg, to the right of A) move on a frictionless, horizontal surface. Initially, block B
olga2289 [7]

Answer:dont care

Explanation:

5 0
3 years ago
A certain unfiltered full-wave rectifier with 120 V, 60 Hz input produces an output with a peak of 15 V. When a capacitor-input
KATRIN_1 [288]

Answer:

The peak-to-peak ripple voltage = 2V

Explanation:

120V and 60 Hz is the input of an unfiltered full-wave rectifier

Peak value of  output voltage = 15V

load connected = 1.0kV

dc output voltage = 14V

dc value of the output voltage of capacitor-input filter

where

V(dc value of output voltage) represent V₀

V(peak value of output voltage) represent V₁

V₀ = 1 - ( \frac{1}{2fRC})V₁

make C the subject of formula

V₀/V₁ = 1 - (1 / 2fRC)

1 / 2fRC = 1 - (v₀/V₁)

C = 2fR ((1 - (v₀/V₁))⁻¹

Substitute  for,

f = 240Hz , R = 1.0Ω, V₀ = 14V , V₁ = 15V

C = 2 * 240 * 1 (( 1 - (14/15))⁻¹

C = 62.2μf

The peak-to-peak ripple voltage

= (1 / fRC)V₁

= 1 /  ( (120 * 1 * 62.2) )15V

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The peak-to-peak ripple voltage = 2V

3 0
3 years ago
What would happen if the poles of the magnet were reversed?
erica [24]
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6 0
3 years ago
A jetliner, traveling northward, is landing with a speed of 71.9 m/s. Once the jet touches down, it has 675 m of runway in which
Leviafan [203]

Answer:

The value is  a =  -3.7 \  m/s^2

Explanation:

From the question we are told that

   The  landing speed is  u =  71.9 \  m/s

   The  distance traveled is  d =  675 \  m

    The velocity it is reduced to is  v  =  11.3 \  m/s

   

Generally the average acceleration is mathematically represented as

      a =  \frac{ v^2  -  u^2 }{ 2 * d }

=>  a =  \frac{ 11.3^2 - 71.9^2 }{ 2 * 675 }

=>   a =  -3.7 \  m/s^2

5 0
3 years ago
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