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2 years ago

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Help!!

2 answers:

scZoUnD [109]2 years ago

7 0

If Fg=mg=ma and, Fg(planetX)=1/5Fg(earth)

then the time would be 5x of the time as gravity is acceleration. So 3.9s*5=19.5s

As the force of gravity is less, then the acceleration of masses is also less, therefore it will take more time for the object to fall by the factor of the force of gravity difference

Neko [114]2 years ago

4 0

D= 1/2 g T^2

T = √(2D/g)

If g changes to g/5, then T increases by the factor √5 .

If it was originally 3.9 sec, then the new T is 8.72 sec.

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The _______ is responsible for determining the frequency of vibration of the air column in the tube within a wind instrument. A.

Vladimir [108]

"A is correct answer." The effective length of the tube is responsible for determining the frequency of vibration of the air column in the tube within a wind instrument. "Hope this helps!" "Have a great day!" "Thank you for posting your question!"

8 0

3 years ago

The kinetic energy of an object with a mass of 6.8 kg and a velocity of 5.0 m/s is J. (Report the answer to two significant figu

abruzzese [7]

Answer:

$\boxed{\sf Kinetic \ energy \ (KE) = 85 \ J}$

Given:

Mass (m) = 6.8 kg

Speed (v) = 5.0 m/s

To Find:

Kinetic energy (KE)

Explanation:

Formula:

$\boxed{ \bold{\sf KE = \frac{1}{2} m {v}^{2} }}$

Substituting values of m & v in the equation:

$\sf \implies KE = \frac{1}{2} \times 6.8 \times {5}^{2}$

$\sf \implies KE = \frac{1}{ \cancel{2}} \times \cancel{2} \times 3.4 \times 25$

$\sf \implies KE =3.4 \times 25$

$\sf \implies KE = 85 \: J$

8 0

3 years ago

Read 2 more answers

To measure the acceleration due to gravity on a distant planet, an astronaut hangs a 0.070-kg ball from the end of a wire. The w

mote1985 [20]

Answer:

g = 1.64m/s²

Explanation:

1.5m in 0.078s

V = 15 / 0.078

= 19.23m/s

Tension = mg

μ = 3.10 × 10⁻⁴

T = V²μ

mg = V²μ

g = V²μ / m

g = ((19.23)²(3.10 × 10⁻⁴)) / (0.070)

g = 1.64m/s²

7 0

3 years ago

A uniform rod of length L is pivoted at L/4 from one end. It is pulled to one side through a very small angle and allowed to osc

ludmilkaskok [199]

Answer:

T= 4.24sec

Explanation:

We are going to use the formula below to calculate.

$T=2\pi \sqrt{\frac{L}{g} }$

Where T is period

L is length of rod

g is acceleration due to gravity = $9.8m/s^{2}$

From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this $L_{O}$

$L_{O} = 3/4 * 5.95m$

= 4.4625m

thus $T=2\pi \sqrt{\frac{L_{O} }{g} }$

$T=2\pi \sqrt{\frac{4.4625 }{9.8} }$

T= 4.24sec

8 0

3 years ago

Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker poi

Nataly [62]

Answer

given,

mass of ball, m = 57.5 g = 0.0575 kg

velocity of ball northward,v = 26.7 m/s

mass of racket, M = 331 g = 0.331 Kg

velocity of the ball after collision,v' = 29.5 m/s

a) momentum of ball before collision

P₁ = m v

P₁ = 0.0575 x 26.7

P₁ = 1.535 kg.m/s

b) momentum of ball after collision

P₂ = m v'

P₂ = 0.0575 x (-29.5)

P₂ = -1.696 kg.m/s

c) change in momentum

Δ P = P₂ - P₁

Δ P = -1.696 -1.535

Δ P = -3.231 kg.m/s

d) using conservation of momentum

initial speed of racket = 0 m/s

M u + m v = Mu' + m v

M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)

0.331 u' = 3.232

u' = 9.76 m/s

change in velocity of the racket is equal to 9.76 m/s

5 0

3 years ago