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SCORPION-xisa [38]
3 years ago
14

The calorie count of a serving of food can be computed based on its composition of​ carbohydrate, fat, and protein. The calorie

count C for a serving of food can be computed using the formula Upper C equals 4 h plus 9 f plus 4 p​, where h is the number of grams of carbohydrate contained in the​ serving, f is the number of grams of fat contained in the​ serving, and p is the number of grams of protein contained in the serving. A serving of raisins contains 148 calories and 29 grams of carbohydrate. If raisins are a​ fat-free food, how much protein is provided by this serving of​ raisins?
Physics
1 answer:
Tresset [83]3 years ago
3 0

Answer:

The raisins provide 8 grams of protein.

Explanation:

First wrote the information as a equation:

C=4h+9f+4p

Where: C is calories, h carbohidrates, f fats and p protein

The information given for the raisin, 148 calories, 0 fat and 29 carbohidrates. That into a equation.

148=4(29)+9(0)+4p

We clear p from the equation

14\frac{148-4(29)}{4} =p=8 gr

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Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

\omega_f = \omega_i + \alpha t

where

\omega_i = 0.300 rev/s is the initial angular velocity

\alpha = 0.895 rev/s^2 is the angular acceleration

Substituting t = 0.200 s, we find

\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s

Let's now convert it into rad/s:

\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

r=\frac{0.800}{2}=0.400 m

And so now we can find the tangential speed at t = 0.200 s:

v=\omega_f r =(3.01)(0.400)=1.2 m/s

2) 2.25 m/s^2

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

a_t = \alpha r

where \alpha is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

a_t = (5.62)(0.400)=2.25 m/s^2

3) 3.6 m/s^2

The centripetal acceleration is given by

a=\frac{v^2}{r}

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

a=\frac{1.2^2}{0.400}=3.6 m/s^2

7 0
3 years ago
Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth’s center. Satellite A is to o
Vera_Pavlovna [14]

Answer:

Explanation:

Orbital radius of satellite A , Ra = 6370 + 6370 = 12740 km

Orbital radius of satellite B , Rb = 6370 + 19110 = 25480 km

Orbital potential energy of a satellite = - GMm / r where G is gravitational constant , M is mass of the earth and m is mass of the satellite

Orbital potential energy of a satellite A = - GMm / Ra

Orbital potential energy of a satellite B = - GMm / Rb

PE of satellite B /PE of satellite A

=  Ra / Rb

= 12740 / 25480

= 1 / 2

b ) Kinetic energy of a satellite is half the potential energy with positive value , so ratio of their kinetic energy will also be same

KE of satellite B /KE of satellite A

= 1 / 2

c ) Total energy will be as follows

Total energy = - PE + KE

- P E + PE/2

= - PE /2

Total energy of satellite B / Total energy of A

= 1 / 2

Satellite B will have greater total energy because its negative value is less.

5 0
3 years ago
Silicon (chemical symbol Si) is located in Group 14, Period 3. Which is silicon
VARVARA [1.3K]

Answer:

I dont know lol sorry ddhd

3 0
2 years ago
Which property of a star can help us deduce its surface temperature?
horrorfan [7]

I think I has to be C

8 0
3 years ago
Read 2 more answers
A ball is rolled of a 2.14m table at 8m/s. Find the time of flight for the ball and the horizontal range
Mama L [17]

Answer:

0.661 s, 5.29 m

Explanation:

In the y direction:

Δy = 2.14 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(2.14 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.661 s

In the x direction:

v₀ = 8 m/s

a = 0 m/s²

t = 0.661 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (8 m/s) (0.661 s) + ½ (0 m/s²) (0.661 s)²

Δx = 5.29 m

Round as needed.

3 0
3 years ago
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