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Archy [21]
3 years ago
14

Andy has been consuming a set number of calories and recently added an additional 100 calories in the form of a daily snack. Alt

hough he is not trying to lose weight, he has lost one pound per week over the last two months. He wants to increase his level of activity and exercise, and he wants to return to his weight from two months ago. Which strategy should he follow to prevent becoming underweight?
Increase his activity and exercise levels and add more calories to gradually gain weight
Increase his activity and exercise levels and eat the same number of calories to gradually gain weight
Increase his snacks to two or three per day until he reaches his target weight
Increase his calories consumed and do less activity and exercise until he reaches his target weight
Physics
1 answer:
I am Lyosha [343]3 years ago
7 0

Answer:

Increase his activity and exercise levels and add more calories to gradually gain weight

Explanation:

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It requires 2,500 joules to raise a certain amount of water (c = 4.186 J/g C) from 20 C to 60 C
Ksivusya [100]

The mass of the water is 14.9 g

Explanation:

When a certain amount of a susbtance is heated, the temperature of the substance increases according to the equation

Q=mC_s \Delta T

where

Q is the amount of energy supplied to the substance

m is the mass of the substance

C_s is its specific heat capacity

\Delta T is the change in temperature

In this problem, we have:

Q = 2500 J of energy supplied to the water

C_s = 4.186 J/gC is the specific heat capacity of water

\Delta T=60 C - 20 C = 40^{\circ}C is the change in temperature of the water

Therefore we can solve for m to find the mass of the water:

m=\frac{Q}{C_s \Delta T}=\frac{2500}{(4.186)(40)}=14.9 g

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

7 0
3 years ago
12. A flat circular coil of wire having 200 turns and diameter 6.0 cm carries a current of 7.0 A. It is placed in a magnetic fie
IgorLugansk [536]

Answer:

The magnitude of the magnetic torque on the coil is 1.98 A.m²

Explanation:

Magnitude of magnetic torque in a flat circular coil is given as;

τ = NIASinθ

where;

N is the number of turns of the coil

I is the current in the coil

A is the area of the coil

θ is the angle of inclination of the coil and magnetic field

Given'

Number of turns, N = 200

Current, I = 7.0 A

Angle of inclination, θ = 30°

Diameter, d = 6 cm = 0.06 m

A = πd²/4 = π(0.06)²/4  = 0.002828 m²

τ = NIASinθ

τ = 200 x 7 x 0.002828 x Sin30

τ = 1.98 A.m²

Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²

4 0
3 years ago
FOR BRAINLIEST!<br><br> Someone draw me a Rube Goldberg Machine.
tatyana61 [14]
This ain’t mine but here is someone else’s

8 0
2 years ago
Dry air is primarily composed of nitrogen. In a classroom demonstration, a physics instructor pours 3.6 L of liquid nitrogen int
neonofarm [45]

Answer:

The  value is  V_n  =  2.2498 \  m^3

Explanation:

From the question we are told that

   The volume of  liquid nitrogen is  V_n  =  3.6 \  L=  3.6 *10^{-3} \ m^3

   The  density of  nitrogen at gaseous form   is  \rho_n =  1.2929 \  kg/m^3  =  The dry air at sea level

   

Generally the density of nitrogen at liquid form is  

         \rho _l = 808 \  kg/m^3

And this is mathematically represented as

      \rho_l  =  \frac{m}{V_l }

=>   m  =  \rho_l  *  V_l

Now the density of  gaseous nitrogen is

       \rho_n  =  \frac{m}{V_n }

=>   m  =  \rho_n  *  V_n

Given that the mass is constant

       \rho_n  *  V_n  =   \rho_l  *  V_l

        1.2929*  V_n  =   808  *  3.6*10^{-3}

=>   V_n  =  2.2498 \  m^3

       

3 0
3 years ago
It's nighttime, and you've dropped your goggles into a 3.2-mm-deep swimming pool. If you hold a laser pointer 1.1 mm above the e
Savatey [412]

Answer: 5.30m

Explanation:

depth of pool = 3.2 m

i = 67.75°

Using snell's law, we have,

n₁ × sin(i) = n₂ × 2 × sin(r)

n₁ = 1, n₂ =1.33, r= 44.09°

Hence,

Distance of Google from edge if pool is:

2.2 + d×tan(r) = 2.2 + (3.2 × tan(44.09°) =5.30m

5 0
3 years ago
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