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3 years ago

Milk has a density of 1.035 g/ml and olive oil has a density of 0.850 g/ml. Which has more mass

Physics

2 answers:

professor190 [17]3 years ago

8 0

Answer:

Milk with 10.35g

Explanation:

Considering the density equation:

p=m/v

Where p is density, m the mass and w the volume we will consider 10ml of each substance and calculate their masses using their densities rearranging the equation:

m=pv

for the olive oil m=(0.850 g/ml )(10ml)= 8.50g

I hope you find this information useful and interesting! Good luck!

marysya [2.9K]3 years ago

6 0

Answer:

Milk.

Explanation:

Density is used to describe how much space an object or substance takes up in relation to the amount of matter in that object or substance (its mass).

Another way to put it is that, density of a substance is the amount of mass of that substance per unit volume. If an object is heavy and compact, it has a high density.

Density = mass/volume

So,

Milk = 1.035 g of milk per 1 ml

Olive oil = 0.85 g of olive oil in 1 ml.

Therefore, milk has a higher mass.

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The area vector of a square loop of 5 turns of a conductor each with side length of 0.2 m carrying a current of 2 A is antiparal

Anon25 [30]

Answer:

$M= 0.4 Am^2$

Explanation:

From the question we are told that:

Number of turns $N=5$ $N=5$

Conductor each with side length $L=0.2m$

Current $I=2A$

Magnetic field $B=50.0T$

Generally the equation for the total magnetic moment M is mathematically given by

$M = current * area$

$M= I * A$

$M = 2 * (5* 0.2*0.2)$

$M = 2 * 0.2$

$M= 0.4 Am^2$

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3 years ago

Assume that Parker Company will receive SF200,000 in 360 days. Assume the following interest rates:

Anna35 [415]

Answer:

b. $96,914

Explanation:

360-day borrowing rate = 5%

spot rate = 0.48

360-day deposit rate = 6%

Borrow at the rate of 5% to get

SF200,000/1.05 = $190,476.19

Convert at the spot rate of $0.48 to get

190,476.19*0.48 = $91,428.57

Invest at the interest rate of 6% to get

91,428.57/1.06 = 96,914.28

Therefore, Parker Company will receive $96,914 in 360 days.

7 0

3 years ago

SHOW WORK

Helga [31]

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

$T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\$

For Point b:

$E=\frac{1}{2} m a^2 w^2\\\\$

$=\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J$

For Point C:

$V_{max}= a w$

$= (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}$

For point D:

$X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm$

5 0

3 years ago

You kick a soccer ball of mass 0.41 kg. the ball leaves your foot with an initial speed of 23 m/s. (a what is the magnitude of t

svetoff [14.1K]

Impulse describes the change of momentum. Since we don't know the momentum of the soccer ball before the hit, this question is hard to answer. If you assume the momentum of the ball before the hit was p = 0, then the change in momentum is just Δp = Impulse = mv.

7 0

3 years ago

Read 2 more answers

A particle moves along line segments from the origin to the points (1, 0, 0), (1, 5, 1), (0, 5, 1), and back to the origin under

Kaylis [27]

Answer:

0 J

Explanation:

Since work done W = ∫F.dr and F(x, y, z)= z²i + 4xyj + 5y²k and dr = dxi + dyj + dzk

F.dr = (z²i + 4xyj + 5y²k).(dxi + dyj + dzk) = z²dx + 4xydy + 5y²dz

W = ∫F.dr = ∫z²dx + 4xydy + 5y²dz = z²x + 2xy² + 5y²z

We now evaluate the work done for the different regions

W₁ = work done from (0,0,0) to (1,0,0)

W₁ = {z²x + 2xy² + 5y²z}₀₀₀¹⁰⁰ = 0²(1) + 2(1)(0)² + 5(0)²(0) - [(0)²(0) + 2(0)(0)² + 5(0)²(0)] = 0 - 0 = 0 J

W₂ = work done from (1,0,0) to (1,5,1)

W₂ = {z²x + 2xy² + 5y²z}₁₀₀¹⁵¹ = (1)²(1) + 2(1)(5)² + 5(5)²(1) - [0²(1) + 2(1)(0)² + 5(0)²(0)] = 1 + 50 + 125 - 0 = 176 J

W₃ = work done from (1,5,1) to (0,5,1)

W₃ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁵¹ = 1²(0) + 2(0)(5)² + 5(5)²(1) - [(1)²(1) + 2(1)(5)² + 5(5)²(1)] = 125 - (1 + 50 + 125) = 125 - 176 = -51 J

W₄ = work done from (0,5,1) to (0,0,0)

W₄ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁰⁰ = (0)²(0) + 2(0)(0)² + 5(0)²(0) - [1²(0) + 2(0)(5)² + 5(5)²(1)] = 0 - 125 = -125 J

The total work done W is thus

W = W₁ + W₂ + W₃ + W₄

W = 0 J + 176 J - 51 J - 125 J

W = 176 J - 176 J

W = 0 J

The total work done equals 0 J

4 0

3 years ago