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3 years ago

6

select all that apply the force of gravity a. changes slightly with the location on the earth b. decreases with height above sea

level c. is unaffected by altitude

1 answer:

Reptile [31]3 years ago

7 0

The answer to your question is C

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when a man throws a stone in the direction of maximum range the stone rises to a height of 20m . what is the range and how long

jeyben [28]

Answer:

50m

Explanation:im just smart thank me later

3 0

3 years ago

A vertical spring (ignore its mass), whose spring constant is 1070 N/m, is attached to a table and is compressed 0.100 m.

harina [27]

I can not solve the problem if I do not have the mass.

3 0

3 years ago

A parallel-plate air capacitor with a capacitance of 260 pF has a charge of magnitude 0.155 Î¼C on each plate. The plates have a

butalik [34]

Answer:

The potential difference between the plates is 596.2 volts.

Explanation:

Given that,

Capacitance $C=260\ pF$

Charge $q=0.155\ \mu\ C$

Separation of plates = 0.313 mm

We need to calculate the potential difference between the plates

Using formula of potential difference

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

Where, Q = charge

C = capacitance

Put the value into the formula

$V=\dfrac{0.155\times10^{-6}}{260\times10^{-12}}$

$V=596.2\ volts$

Hence,The potential difference between the plates is 596.2 volts.

7 0

3 years ago

The force of gravity on a 1 kg object on the Earth's surface is approximately 9.8 N. For the same object in low-earth orbit arou

mojhsa [17]

Answer:

g = 8.61 m/s²

Explanation:

distance of the International Space Station form earth is 200 Km

mass of the object = 1 Kg

acceleration due to gravity on earth = 9.8 m/s²

mass of earth = 5.972 x 10²⁴ Kg

acceleration due to gravity = ?

r = 6400 + 200 = 6800 Km = 6.8 x 10⁶ n

using formula

$g = \dfrac{GM}{r^2}$

$g = \dfrac{6.67\times 10^{-11}\times 5.972\times 10^24}{(6.8\times 10^6)^2}$

g = 8.61 m/s²

3 0

3 years ago

Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t

salantis [7]

Explanation:

Mass of bumper cars, $m_1=m_2=110\ kg$

Initial speed of car A, $u_1=8\ m/s$

Initial speed of car Z, $u_2=-10\ m/s$

Final speed of car A after the collision, $v_1=-10\ m/s$

We need to find the velocity of car Z after the collision. Let it is equal to $v_2$ . Using the conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$110\times 8+110\times (-10)=110\times (-10)+110v_2$

$v_2=\dfrac{-1320}{110}\ m/s$

$v_2=-12\ m/s$

So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.

5 0

3 years ago