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dangina [55]
4 years ago
7

A projectile is fired vertically from Earth's surface with an initial speed v0. Neglecting air drag, how far above the surface o

f Earth will it go? (Use any variable or symbol stated above along with the following as necessary: ME for the mass of the Earth, rE for the radius of the Earth, and G for the gravitational constant.)
Physics
1 answer:
lidiya [134]4 years ago
3 0
This may helpv^2=u^2+2as. v=0 at top of flight. a=acceleration of gravity(vo^2)/2a=s.
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if pete (mass =90.00kg) weighs himself and finds that he weighs 30.0 pounds, how far away from the earth is he?
Svet_ta [14]

Here we know that mass of the person is 90 kg

His weight is given as 30 lbf

so here we can convert it into Newton as we know that

1 lbf = 4.45 N

Now from above conversion

30 lbf = 30 \times 4.45 = 133.45 N

now we can use this to find the gravity at this height

mg = 133.45

90\times g = 133.45

g = 1.49 m/s^2

now we know that with height gravity varies as

g' = g(\frac{R}{R+H})^2

1.49 = 9.8(\frac{6.37\times 10^6}{6.37\times 10^6 + h})^2

h = 1.0 \times 10^7 m

so above is the height from the surface of earth

4 0
3 years ago
Write a hypothesis about the use of an object’s physical characteristics to determine its density. Use the format "if . . . then
Ad libitum [116K]

Answer:

If an object has a high density then the molicules making up that object are closly packed togeather. Because of this, objects with a higher density will have more mass than objects of the same size that have a lesser density.

7 0
3 years ago
(a) Calculate the distance that light travels in 0.5 seconds, and express the answer in units of the circumference of the earth.
Elan Coil [88]

Answer:

a)  3.74 circumference of earth

b)1.67× 10⁻⁸ sec

Explanation:

a) given,

time =  0.5 second

distance = speed  × time

              = 3 x 10⁸ × 0.5  

              = 1.5 × 10⁸ m

circumference of earth = 40075 km

                                       = 4.01 × 10⁷m

distance with respect to circumference=\dfrac{1.5 \times 10^8}{4.01 \times 10^7}

                                                                 = 3.74 circumference of earth

b) distance = 5 m

time =\dfrac{distance}{speed}

        =\dfrac{5}{3\times 10^8}

        = 1.67× 10⁻⁸ sec

6 0
4 years ago
When a point charge of +q is placed on one corner of a square, an electric field strength of 2 N/C is observed at the center of
melamori03 [73]

Electric field due to a single point charge at the center of the square is given as E = 2 N/C

Now we place three identical charges on rest of the corners

So the electric field at the center will be due to all four charges will be of same magnitude

but the direction of all four charges electric field be like they all cancel the field of its opposite charge

So here due to vector property of the charges we can say that the electric field of all four charges will cancel each other and net field due to all four charges at center will be ZERO

so correct answer will be

<em>b. The magnitude of the net electric field at the center of the square is 0 N/C.</em>

3 0
3 years ago
Read 2 more answers
9.For the following questions, use a periodic table and your atomic calculations to find the unknown information about each isot
myrzilka [38]
43 because the atomic number tells us how many protons there are in an atom.
3 0
3 years ago
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