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GREYUIT [131]
3 years ago
15

Which of these forces could speed up a baseball?

Physics
2 answers:
kkurt [141]3 years ago
5 0
B) Applied and gravitational forces
Slav-nsk [51]3 years ago
5 0

Answer:

b)

Explanation:

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g 1. A mass undergoing simple harmonic motion along the x-axis has a period of T = 0.5 s and an amplitude of 25 mm. Its position
kondor19780726 [428]

Answer:

Explanation:

a )

Amplitude A = 14 mm , angular frequency ω = 2π / T

= 2π / .5

ω = 4π rad /s

φ₀ = initial phase

Putting the given values in the equation

14 = 25 cos(ωt + φ₀ )

14/25 = cosφ₀

φ₀ = 56 degree

x(t) = 25cos(4πt + 56° )

b )

maximum velocity = ω A

=  4π  x 25

100 x 3.14 mm /s

= 314 mm /s

At x = 0 ( equilibrium position or middle point , this velocity is achieved. )

maximun acceleration = ω² A

= 16π² x A

= 16 x 3.14² x 25

= 3943.84 mm / s²

3.9 m / s²

It occurs at x = A or at extreme position.

7 0
3 years ago
A package of mass 5 kg sits on an airless asteroid of mass 7.6 × 1020 kg and radius 8.0 × 105 m. We want to launch the package i
Effectus [21]

Answer:

s =  1.7 m

Explanation:

from the question we are given the following:

Mass of package (m) = 5 kg

mass of the asteriod (M) = 7.6 x 10^{20} kg

radius = 8 x 10^5 m

velocity of package (v) = 170 m/s

spring constant (k) = 2.8 N/m

compression (s) = ?

Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore  

• Ei = Ef

• Ei = energy in the spring + gravitational potential energy of the system

• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}

• Ef = kinetic energy of the object

• Ef = \frac{1}{2}mv^{2}  

• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}  

• s = \sqrt{\frac{m}[k}(v^{2}+\frac{2GM}{r})}

s = \sqrt{\frac{5}[2.8 x 10^5}(170^{2}+\frac{2 x 6.67 x10^{-11} x 7.6 x 10^{20}}{8 x 10^5})}

s =  1.7 m

7 0
3 years ago
(1 point) A rectangular tank that is 3 feet long, 9 feet wide and 12 feet deep is filled with a heavy liquid that weighs 110 pou
Aloiza [94]

Answer:

Explanation:

Work in pumping water from the tank is given as

W = ∫ y dF. From a to b

Where dF is the differential weight of the thin layer of liquid in the tank, y is the height of the differential layer

a is the lower limit of the height

b is the upper limit of the height.

We know that, .

F = ρVg

Where F is the weight

ρ is the density of water

V is the volume of water in tank

g is the acceleration due to gravity

Then,

dF = ρg ( Ady)

We know that the density and the acceleration due to gravity is constant, also the base area of the tank is constant, only the height that changes.

Then,

ρg = 62.4 lbs/ft³

Area = L×B = 3 × 9 = 27ft²

dF = ρg ( Ady)

dF = 1684.8dy

The height reduces from 12ft to 0ft

Then,

W = ∫ y dF. From a to b

W = ∫ 1684.8y dy From 0 to 12

W = 1684.8y²/2 from 0 to 12

W = 842.4 [y²] from y = 0 to y = 12

W = 842.4 (12²-0²)

W = 121,305.6 lb-ft

3 0
3 years ago
We can see our image in the mirror but not in the book​
marishachu [46]

Answer:

if the image is on the book yes , no its can be both

Explanation:

4 0
2 years ago
Whats a peer review?
Goryan [66]
B is the answer to your question
8 0
3 years ago
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