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4 years ago

12

Could someone please help me with this problem? I’m thinking that they both will have the same force because they have the same

Fgrav and equal masses. Thanks in advance!

1 answer:

Simora [160]4 years ago

7 0

You would be correct

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If the range of the projectile is 4.3 m, the time-of-flight is T = 0.829 s, and air resistance is negligible, determine the foll

ankoles [38]

Answer

given,

range of the projectile = 4.3 m

time of flight = T = 0.829 s

$v =\dfrac{d}{T}$

$v =\dfrac{4.3}{0.829}$

v = 5.19 m/s

vertical component of velocity of projectile

v_y = gt'

$v_y = 9.8 \times {\dfrac{T}{2}}$

$v_y = 9.8 \times {\dfrac{0.829}{2}}$

$v_y =4.06\ m/s$

a) Launch angle

$\theta = tan^{-1}(\dfrac{v_y}{v})$

$\theta = tan^{-1}(\dfrac{4.06}{5.19})$

θ = 38°

b) initial speed of projectile

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{5.19^2 + 4.06^2}$

v = 6.59 m/s

c) maximum height reached by the projectile

$y_{max}=v_{avg}t'$

$y_{max}=\dfrac{1}{2}v_y\dfrac{T}{2}$

$y_{max}=\dfrac{1}{2}\times(g\dfrac{T}{2})\times\dfrac{T}{2}$

$y_{max}=\dfrac{gT^2}{8}$

7 0

4 years ago

There is more land in the Northern Hemisphere than in the Southern Hemisphere. How might this difference affect CO2, concentrati

alexira [117]

Oxygen is a product of photosynthesis, therefor lover levels of C02 would be in the area with more water due to more phtosynthetic processes happening converting it into oxygen.

hope this helps, please mark brainliest. :)

5 0

3 years ago

Someone please hellpp!!! Will mark brainliest

____ [38]

<em>Answer:</em>

<em>r=x+y</em>

<em>sorry if its not correct you can delete if you want.</em>

6 0

3 years ago

Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

mrs_skeptik [129]

Answer:

surface charge density on each sphere is $440 \times 10^{-9}$ C

Explanation:

given data

radius of smaller sphere = 5 cm

radius of larger sphere is 12 cm

electric field at surface of larger sphere = 660 kV/m = 660 × 1000 v/m

solution

we apply here electric field formula that is express as

E = $(\frac{1}{4\pi\epsilon })\times (\frac{Q_{1} }{R^{2} } )$ .................1

put here value

660000 = $9 \times 10^9 \times \frac{Q1}{0.12^2}$

Q1 = 1056 × $10^{-9}$

and

here field inside a conductor is zero so that electric potential ( V ) is constant

$\frac{Q{1} }{R} = \frac{Q{2} }{r}$ ..................2

so Q2 will be

Q2 = $\frac{5}{12} \times 1056 \times 10^{-9}$

Q2 = $440 \times 10^{-9}$ C

6 0

3 years ago

Charged beads are placed at the corners of a square in the various configurations shown in

Akimi4 [234]

Answer:

The consecutive charge configuration has a more intense field than alternating

Explanation:

In each corner we place a different account there are only two different settings, see attached.

In the case of alternating charging (+ - + -) see diagram 1, the electric field in the center is canceled in pairs, resulting in a zero field

In the case of consecutive loads (+ + - -) in this case we have a result between the two charges, therefore the total field is

E = 2 k q / ra2 a cos 45

The consecutive charge configuration has a more intense field than alternating

8 0

3 years ago