We use __force to do different force. 1) pulling 2)pushing 3)gravity 4)muscular

Imagine a scenario in which balanced forces are exerted on a box. What does the motion of the box look like?

zepelin [54]

It would depend on the amount it would be still or crushed<span />

3 0

3 years ago

Let A be the last two digits, and let B be the last three digits, and the C be the sum of the last 4 digits of your 8-digit stud

UNO [17]

Answer:

66.053m/s

Explanation:

A = 47

B = 347

C = 19

Train moves at

(23 + A)m/s

= 23 + 47 = 60m/s

At (250.0+B) seconds

250.0+347 =

547 seconds

Distance d,

= 70 x 597

= 41790

It also moves at

(45.0 + c)

= 45 + 19

= 64m/s

Time = 800 + B

= 800 + 347

= 1147

Distance,

= 64 x 1147

= 73408m

Total distance,

= 73408 + 41790

= 115,198

Total time,

= 597 + 1147

= 1744

Average speed,

= Total distance / total time

= 115198/1174

= 66.053m/s

7 0

3 years ago

A snowboarder is sliding down a hill to the right. What are the 3 forces being used?

Softa [21]

Answer:

ANswers

Explanation:

Friction, knetic, and physics

8 0

3 years ago

The earth's magnetic moment is 8 x 1022 a-m2. If that moment were created by a loop of wire going around the earth (r = 6378 km)

Elena L [17]

Answer:

M = I A definition of magnetic moment - current * area

A = π R^2 = π * (6.4E6)^2 = 1.3E14 m^2

I = 8E22 A-m^2 / 1.3E14 m^2 = 6.2E8 amperes

I = 620,000,000 amps

4 0

2 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:

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.

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

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

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

4 years ago