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Ira Lisetskai [31]
3 years ago
9

Question: Which one of the following foods contains simple carbohydrates?

Physics
1 answer:
Katena32 [7]3 years ago
5 0

Answer:

Milk

Hope this helps,if not sorry

You might be interested in
A proton is initially at rest. After some time, a uniform electric field is turned on and the proton accelerates. The magnitude
marusya05 [52]

Answer:

a) 8.83*10⁵ m/s  b) 2.80*10⁶ m/s

Explanation:

a) Assuming no other forces acting on the proton, the acceleration on it is produced by the electric field.

By definition, the  force due to the electric field is as follows:

F = q*E = e*E (1)

where e is the elementary charge, the charge carried by only one proton, and is e = 1.6*10⁻¹⁹ C.

According to Newton's 2nd law, this force is at the same time, the product of the mass of the proton, times the acceleration a:

F = mp*a (2)

From (1) and(2), being left sides equal, right sides must be equal too:

F = e*E = mp*a

Solving for a:

a = \frac{e*E}{mp} =\frac{1.6e-19C*1.36e5N/C}{1.67e-27kg} =1.3e13 m/s2

⇒ a = 1.3*10¹³ m/s²

As we have the value of a (which is constant due to the field is uniform), the displacement x, and we know that the initial velocity is 0, in order to get the value of the speed, we can use the following kinematic equation:

vf^{2} -vo^{2} = 2*a*x

Replacing by v₀ = 0, a= 1.3*10¹³ m/s² and  x = 0.03 m, we can find vf as follows:

vf =\sqrt{2*(1.3e13 m/s2)*0.03m} = 8.83e5 m/s

⇒ vf = 8.83*10⁵ m/s

b) We can just repeat the equation from above, replacing x=0.03 m by x=0.3 m, as follows:

vf =\sqrt{2*(1.3e13 m/s2)*0.3m} = 2.80e6 m/s

⇒ vf = 2.80*10⁶ m/s

4 0
3 years ago
A 0.5 kg mass on a spring undergoes simple harmonic motion with a total mechanical energy of 12 J. If the oscillation amplitude
Darya [45]

Answer:

The frequency of the oscillation is 2.45 Hz.

Explanation:

Given;

mass of the spring, m = 0.5 kg

total mechanical energy of the spring, E = 12 J

Determine the spring constant, k as follows;

E = ¹/₂kA²

kA² = 2E

k = (2E) / (A²)

k = (2 x 12) / (0.45²)

k = 118.519 N/m

Determine the angular frequency, ω;

\omega = \sqrt{\frac{k}{m} } \\\\\omega =  \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s

Determine the frequency of the oscillation;

ω = 2πf

f = (ω) / (2π)

f = (15.396) / (2π)

f = 2.45 Hz

Therefore, the frequency of the oscillation is 2.45 Hz.

8 0
2 years ago
rs. Rushing fills a balloon with hydrogen gas to demonstrate its ability to burn. Which combination could she use to produce the
anyanavicka [17]

Answer:

Mg and HCl

Explanation:

Here, we want to get the combination that could be used in the production of the needed hydrogen

An important chemical property of inorganic acids is that when they react with metals, they give off hydrogen gas in conjunction with the formation of a salt

HCl is a mineral acid while Mg is a metallic substance

So the reaction between this metal and the mineral acids will give the needed hydrogen gas to be produced

4 0
2 years ago
Read 2 more answers
At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/
Nata [24]

Answer:

Explanation:

Given that,

At one instant,

Center of mass is at 2m

Xcm = 2m

And velocity =5•i m/s

One of the particle is at the origin

M1=? X1 =0

The other has a mass M2=0.1kg

And it is at rest at position X2= 8m

a. Center of mass is given as

Xcm = (M1•X1 + M2•X2) / (M1+M2)

2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

2(M1+0.1) = 0.8

2M1 + 0.2 =0.8

2M1 = 0.8-0.2

2M1 = 0.6

M1 = 0.6/2

M1 = 0.3kg

b. Total momentum, this is an inelastic collision and it momentum after collision is given as

P= (M1+M2)V

P = (0.3+0.1)×5•i

P = 0.4 × 5•i

P = 2 •i kgm/s

c. Velocity of particle at origin

Using conversation of momentum

Momentum before collision is equal to momentum after collision

P(before) = M1 • V1 + M2 • V2

We are told that M2 is initially at rest, then, V2=0

So, P(before) = 0.3V1

We already got P(after) = 2 •i kgm/s in part b of the question

Then,

P(before) = P(after)

0.3V1 = 2 •i

V1 = 2/0.3 •i

V1 = 6 ⅔ •i m/s

V1 = 6.667 •i m/s

4 0
3 years ago
A single conservative force acts on a 5.30-kg particle within a system due to its interaction with the rest of the system. The e
cupoosta [38]

Answer:

Given that

m = 5.3 kg

Fx = 2x + 4

We know that work done by force F given as

w= ∫ F. dx

a)

Given that x=1.08 m to x=6.5 m

Fx = 2x + 4

w= ∫ F. dx

w=\int_{1.08}^{6.5}(2x+4) .dx

w=\left [x^2+4x \right ]_{1.08}^{6.5}

w=(6.5^2-1.08^2)+4(6.5-1.08)\ J

w=62.7 J

b)

We know that potential energy given as

F=-\dfrac{dU}{dx}

∫ dU =  -∫F.dx           ( w= ∫ F. dx)

ΔU= -62.7 J

c)

We know that form work power energy theorem

Net work = Change in kinetic energy

W= KE₂ - KE₁

62.7 =KE₂ - (1/2)x 5.3 x 3²

KE₂ = 86.55 J

This is the kinetic energy at 6.5m

8 0
2 years ago
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