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Ilia_Sergeevich [38]
3 years ago
5

To overcome an object's inertia, it must be acted upon by __________. A. gravity B. energy C. force D. acceleration

Physics
2 answers:
astra-53 [7]3 years ago
7 0
In order to overcome an object’s inertia (resistance to change), it must be acted upon by an unbalanced force, so the answer to the problem is letter C.
Jlenok [28]3 years ago
4 0
According to Newton's first law, the law of inertia, an object's velocity will not change unless it is acted upon by an outside force. This means that objects at rest tend to stay at rest, and objects in motion tend to stay in motion.
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a college student produces about 100 kcal of heat per hour on the average what is the rate of energy production and joules
Bond [772]

Given:

Amount of heat produced = 100 kcal per hour

Let's find the rate of energy production in joules.

We know that:

1 calorie = 4.184 Joules

1 kcal = 4.184 Joules

To find the rate of energy production in Joules, we have:

\begin{gathered} Rate=100\ast4.184 \\  \\ \text{Rate}=418.4\text{ KJ/hour} \end{gathered}

Therefore, the rate of energy production in joules is 418.4 kJ/h which is equivalent to 418400 Joules

ANSWER:

418.4 kJ/h

6 0
1 year ago
Lymph ducts are more similar to arteries than veins.<br> A. <br> True<br> B. <br> False
Strike441 [17]

Answer:

False

Explanation:

5 0
4 years ago
Oxygen (O) is a gas found in the 16th column of the periodic table. What statement is true about oxygen and the other elements i
Wittaler [7]

The answer is C atleast thats what i'm gonna put but i'll come correct it if i'm wrong

7 0
3 years ago
Problem 4 A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true ano
Nikitich [7]

Answer:

Part a: The eccentricity is 1.086.

Part b: The altitude at closest approach is 5088 km

Part c: The velocity at perigee is 8.516 km/s

Part d: The turn angle is 134.08 while the aiming radius is 5641.28 km

Explanation:

<h2>Part a </h2>

Specific energy is given by

\epsilon=\frac{v^2}{2}-\frac{\mu}{r}

Here

  • ε is the specific energy
  • v is the velocity which is given as 2.23 km/s
  • μ is the gravitational constant whose value is 398600
  • r is the distance between earth and the meteorite which is 402,000 km

                         \epsilon=\frac{v^2}{2}-\frac{\mu}{r}\\\epsilon=\frac{2.2^2}{2}-\frac{398600}{402,000}\\\epsilon=1.495 km^2/s^2

Value of specific energy is also given as

\epsilon=\frac{\mu}{2a}\\a=\frac{\mu}{2\epsilon}\\a=\frac{398600}{2\times 1.495}\\a=13319 km

Orbit formula is given as

r=a(\frac{e^2-1}{1+ecos \theta})\\ae^2-recos\theta-(a+r)=0

Putting values in this equation and solving for e via the quadratic formula gives

ae^2-recos\theta-(a+r)=0\\(133319)e^2-(402000)(cos 150) e-(133319+402000)=0\\133319e^2+348142.21 e-535319=0\\\\e=\frac{-348142.21 \pm \sqrt{348142.21^2-4(133319)(535319)}}{2 (133319)}\\\\e=1.086 \, or \, -3.69

As the value of eccentricity cannot be negative so the eccentricity is 1.086.

<h2>Part b</h2>

The radius of trajectory at perigee is given as

r_p=a(e-1)\\

Substituting values gives

r_p=133319 (1.086-1)\\r_p=11465.4 km

Now for estimation of altitude z above earth is given as

z=r_p-R_E\\z=11465.4-6378\\z=5087.434\\z\approx 5088 km

So the altitude at closest approach is 5088 km

<h2>Part c</h2>

radius of perigee is also given as

r_p=\frac{h^2}{\mu}\frac{1}{1+e}

Rearranging this equation gives

h=\sqrt{r_p\mu(1+e)}\\h=\sqrt{11465.4 \times 3986000 \times (1+1.086)}\\h=97638.489 km^2/s

Now the velocity at perigee is given as

v_p=\frac{h}{r_p}\\v_p=\frac{97638.489}{11465.4}\\v_p=8.516 km/s\\

So the velocity at perigee is 8.516 km/s

<h2>Part d</h2>

Turn angle is given as

\delta =2 sin^{-1} (\frac{1}{e})

Substituting value in the equation gives

\delta =2 sin^{-1} (\frac{1}{e})\\\delta =2 sin^{-1} (\frac{1}{1.086})\\\delta =134.08

Aiming radius is given as

\Delta =a \sqrt{e^2-1}

Substituting value in the equation gives

\Delta =a \sqrt{e^2-1}\\\Delta =13319 \sqrt{1.086^2-1}\\\Delta=5641.28 km

So the turn angle is 134.08 while the aiming radius is 5641.28 km

3 0
3 years ago
Why is external fertilization not possible in organisms that live on land
Softa [21]


because the sperm and the egg will dry out

4 0
3 years ago
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