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3 years ago

11

assuming birdman flies at height of 72m, how fast should he fly to hit bucket at 63m from start of field. gravity is -9.8m/s^2 n

o air resistance drops thing at start of field. what is his speed/velocity????????

1 answer:

algol [13]3 years ago

7 0

Answer:

.....Birdman?

Explanation:

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A ball is bouncing to a height that is 1/3 of the previous height after every bounce. What kind of sequence is this? How could y

tamaranim1 [39]

Answer and Explanation:

The ball is bouncing to a height of 1/3 of its previous height this is a type of geometric sequence the total distance can be found by the sum of geometric sequence

For example let the initial height is 243 fit

After one bounce it will reach 243/3 =81 feet

After second bounce 81/3=27 feet

After third bounce 27/3 =9 feet

After fourth bounce 9/3 =3 feet

So a sequence is formed that is 243,81,27,9,3..........

Here $r=\frac{3}{9}=\frac{1}{3}$

Sum of infinite GP = $\frac{a}{1-r}$

From this formula we can find the total distance traveled by the ball

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5. How does a jack make changing a tire easier?

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Explanation:

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2 years ago

Find the volume of a rectangular prism that is 8m long, 4m wide, and 300cm high

konstantin123 [22]

L=8m
B=4m
H=300cm=3m

$\\ \bull\tt\longmapsto Volume=LBH$

$\\ \bull\tt\longmapsto Volume=8(4)(3)$

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2 years ago

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What is meant by the motion of a force

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2 years ago

In the diagram, q1 = -6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? pls help

Alexxx [7]

Answer:

Below

Explanation:

First draw the vectors that represent both electric fields.

E1 is the elictric field created by q1, E2 is the one created by q2.

● q1 is negative so E1 will point from P.

● q2 is positive so E2 will point out of P

(Picture below)

■■■■■■■■■■■■■■■■■■■■■■■■■■

The resulting electric field is equal to the sum of the two fields since both vectors are colinear.

Let E be the total field.

● E = E1 + E2

The formula of the electric field intensity is:

● E = K ×(q/d^2)

-K is Coulomb's constant

-d is the distance between the charge and the object ( here P)

-q is the charge

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E1 = K × (q1/d1^2)

The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)

Coulombs constant is 9×10^9 m^2/C^2

● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]

● E1 = -359.43 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E2 = K ×(q2/d^2)

The distance between q2 and P is 0.25 m.

● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]

● E2 = 463.68 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E = E1 + E2

● E = -359.43+463.68

● E = 105.25 N/C

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2 years ago

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