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2 years ago

12

Give three examples from your life of magnetic force.

2 answers:

rewona [7]2 years ago

7 0

Magnets,popsocket, and a car

Margaret [11]2 years ago

4 0

Headphones, refrigerator magnets, and compasses

Hope that was helpful.

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Which kind of mirror can produce real images?<br><br><br> A)concave<br><br> B)convex<br><br> C)flat

Anton [14]

Answer:

C

Explanation:

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3 years ago

What effort force will be required to lift the 20 N object using the pulley above?

Umnica [9.8K]

Answer:

I think it is 5N.

Explanation:

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2 years ago

Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.20 m/s

Zielflug [23.3K]

Answer:

there will be collision

Explanation:

$v_{s}$ = speed of sue = 34 m/s

$v_{v}$ = speed of van = 5.20 m/s

$v_{sv}$ = speed of sue relative to van = $v_{s} - v_{v}$ = 34 - 5.20 = 28.8 m/s

$d_{s}$ = stopping distance after brakes are applied

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$v_{f}$ = final speed of sue = 0 m/s

$a$ = acceleration = - 1.80 m/s²

Using the kinematics equation

$v_{f}^{2} = v_{o}^{2} + 2 a d_{s}$

$0^{2} = 28.8^{2} + 2 (1.80) d_{s}$

$d_{s} = 230.4$ m

Since $d_{s} < D$

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3 years ago

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$

Solving for <em>M</em>, we get

$M = \dfrac{4\pi^2 r^3}{GT^2}$

Putting this expression back into Eqn(1), we get

$R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}$

$\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}$

$\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}$

$\:\:\:\:= 2.6×10^6\:\text{m}$

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2 years ago

did anyone do the, kinetic energy lab report in physics?? plzzz help a girl outttttttt i need answerss????ill give you as much b

Tema [17]

Answer:

yes

Explanation:

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2 years ago