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3 years ago

4- Protein foods?? pls answer these i really need these

Physics

1 answer:

vampirchik [111]3 years ago

8 0

The #1 answer would be meats and eggs. But, as a veggan, I present you with meatless protien-rich food!

1 - <u>Pea protein</u> (used in Beyond Meat, which tastes just like meat btw!)

2 - <u>Soy protein</u> such as tofu

3 - <u>Nuts</u>

4 - <u>Beans</u>! oml so many beans have MORE protien than meat!! ikr!?!?

may I have Brainliest pls =)

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What process change earths surface during Precambrian time?

andreyandreev [35.5K]

The three eons of precambrian time i.e hadean eon ,Archean eon and proyerozoic eon change earth's surface during Precambrian time.

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4 years ago

Match each key term with the correct definition.

oksano4ka [1.4K]

Answer:

Radioisotope -> An atom with an unstable nucleus

Radioactivity -> The spontaneous discharge of energy from an unstable nucleus

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Strong Nuclear Force -> Binds protons and neutrons together in the nucleus

Explanation:

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8 0

3 years ago

A uniform rod is 2. 0 m long and has mass 15 kg. What is most nearly the rod's mass moment of inertia?

trapecia [35]

The rod's mass moment of inertia is 5kgm².

<h3>Moment of Inertia:</h3>

The "sum of the product of mass" of each particle with the "square of its distance from the axis of rotation" is the formula for the moment of inertia.

The Parallel axis Theorem can be used to compute the moment of inertia about the end of the rod directly or to derive it from the center of mass expression. I = kg m². We can use the equation for I of a cylinder around its end if the thickness is not insignificant.

If we look at the rod we can assume that it is uniform. Therefore the linear density will remain constant and we have;

or = M / L = dm / dl

dm = (M / L) dl

$I = \int\limits^M_0 {r^2} \, dm$

$I = \int\limits^\frac{L}{2} _\frac{-L}{2} {I^2 (M/L)} \, dl$

Here the variable of the integration is the length (dl). The limits have changed from M to the required fraction of L.

$I = \int\limits^\frac{L}{2} _\frac{-L}{2} {I^2 (M/L)} \, dl$

$I = \frac{M}3L}[(\frac{L^3}{2^3} - \frac{-L^3}{2^3} )]\\\\I = \frac{1}{12}ML^2$

Mass of the rod = 15 kg

Length of the rod = 2.0 m

Moment of Inertia, I = $\frac{1}{12}15 (2)^2$

= 5 kgm²

Therefore, the moment of inertia is 5kgm².

Learn more about moment of inertia here:

brainly.com/question/14119750

#SPJ4

4 0

2 years ago

A car has a mass of 1520 kg. While traveling at 20 m⁄s, the driver applies the brakes to stop the car on a wet surface with a 0.

docker41 [41]

Answer:

(a)d₁ = 51.02 m

(b)d₂ =51.02m

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

Known data

m=1520 kg : mass of the car

μk= 0.4 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the car

We define the x-axis in the direction parallel to the movement of the car and the y-axis in the direction perpendicular to it.

W: Weight of the block : In vertical direction downward

FN : Normal force : In vertical direction upward

f : Friction force: In horizontal direction

Calculated of the W

W= m*g

W= 1520 kg* 9.8 m/s² = 14896 N

Calculated of the FN

We apply the formula (1)

∑Fy = m*ay ay = 0

FN - Wy = 0

FN = Wy

FN = 14896 N

Calculated of the f

f = μk* N= (0.4)* (14896 N )

f = 5958.4 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax , ax= a : acceleration of the block

- f = m*a

-5958.4 = (1520)*a

a = (-5958.4) / ((1520)

a = -3.92 m/s²

(a) displacement of the car (d₁)

Because the car moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d₁ Formula (2)

Where:

d:displacement (m)

v₀: initial speed (m/s)

vf: final speed (m/s)

Data:

v₀ = 20 m⁄s

vf = 0

a = --3.92 m/s²

We replace data in the formula (2) to calculate the distance along the ramp the block reaches before stopping (d₁)

vf²=v₀²+2*a*d ₁

0 = (20)²+2*(-3.92)*d ₁

2*(3.92)*d₁ = (20)²

d₁ = (20)² / (7.84)

d₁ = 51.02 m

(b) Different car

m₂ = 1.5 *1520 kg

μk₂= 0.4

W₂= m*g

W₂= (1.5) *1520 kg* 9.8 m/s² = (1.5)*14896 N

FN₂= (1.5)*14896 N

f= 0.4* (1.5)*14896 N

a = - f/m₂ = - 0.4* (1.5)*14896 N /(1.5) *1520

a = -3.92 m/s²

vf²=v₀²+2*a*d₂

vf=0 , v₀=20 m⁄s , a = -3.92 m/s²

d₂ = d₁ = 51.02m

6 0

4 years ago

If gas costs $3.65 per gallon at a local service station, how many cubic feet of gas can a customer buy with $40 ?

svet-max [94.6K]

Hi, thank you for posting your question here at Brainly.

For consistency you must convert gallons to cubic foot. The conversions are:

7.481 gal = 1 ft3

Then, $3.65 per gallon becomes $27.3 per ft3. If you buy $40 worth of gas, you get $40/$27.3 = 1.46 ft3 of gas.

If you can travel for 4 miles per ft3, then with the amount of gas you have, you can travel up to 5.85 miles.

(4 miles/ft3)*(1.46 ft3) = 5.85 miles

3 0

3 years ago

4- Protein foods?? pls answer these i really need these​

4- Protein foods?? pls answer these i really need these