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2 years ago

If you have 10 ounces of meat, what percentage of the calories comes from fat if the meat is 90% lean?

Physics

1 answer:

Lady bird [3.3K]2 years ago

4 0

Answer:

There are 602 calories in 10 ounces of Ground Beef (85% Lean / 15% Fat). serving sizes of Ground Beef

Explanation:

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A 1400 kg car traveling at 17.0 m/s to the south collides with a 4700 kg truck that is at rest. The car and truck stick together

STatiana [176]

Answer:

Final velocity = 7.677 m/s

KE before crash = 202300 J

KE after crash = 182,702.62 J

Explanation:

We are given;

m1 = 1400 kg

m2 = 4700 kg

u1 = 17 m/s

u2 = 0 m/s

Using formula for inelastic collision, we have;

m1•u1 + m2•u2 = (m1 + m2)v

Where v is final velocity after collision.

Plugging in the relevant values;

(1400 × 17) + (4700 × 0) = (1400 + 1700)v

23800 = 3100v

v = 23800/3100

v = 7.677 m/s

Kinetic energy before crash = ½ × 1400 × 17² = 202300 J

Kinetic energy after crash = ½(1400 + 1700) × 7.677² = 182,702.62 J

8 0

3 years ago

How heavy is an object that displaces 400N of water in a pool?

julia-pushkina [17]

All we can say is that the object's volume is about 41 liters. That's the same as the volume of water displaced.

We can't say anything about the object's weight. There is no direct connection between the weight of the object and the weight of the water it displaces.

7 0

3 years ago

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

6 0

4 years ago

What is the change in thermal energy E if the coefficent of kinetic friction between the box and floor is .4 , the distance the

Jet001 [13]

This question can be solved using the concept of friction energy.

The thermal energy change is b "258.4 J".

The change in thermal energy will be equal to the friction energy produced during the motion of the box.

$Change\ In\ Thermal\ Energy = E = Friction\ Energy\\\\E = \mu fd$

where,

μ = coefficient of kinetic friction = 0.4

f = force applied = 38 N

d = distance traveled by the box = 17 m

Therefore,

$E = (0.4)(38\ N)(17\ m)$

Learn more about friction energy here:

brainly.com/question/1343045?referrer=searchResults

7 0

2 years ago

A proton is moved from a position where the electric potential is 125 V to a position where the electric potential is 275 V. The

kakasveta [241]

We determine the electric potential energy of the proton by multiplying the net electric potential to the charge of the proton. The net electric potential is the difference of the final state to the that of the initial state. So, it would be 275 - 125 = 150 V.

electric potential energy = 150 (<span>1.602 × 10-19) = 2.4x10^-17 J</span>

7 0

3 years ago