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3 years ago

An object weighs 10N on earth .what is the objects weight on a planet one tenth the earths mass and one half its radius?

1 answer:

3 0

We know, weight = mass * gravity
10 = m * 9.8
m = 10/9.8 = 1.02 Kg

Now, Let, the gravity of that planet = g'
g' = m/r² [m,r = mass & radius of that planet ]
g' = M/10 / (1/2R)² [M, R = mass & radius of Earth ]
g' = 4M / 10R²
g' = 2/5 * M/R²
g' = 2/5 * g
g' = 2/5 * 9.8
g' = 3.92

Weight on that planet = planet's gravity * mass
W' = 3.92 * 1.02
W' = 4 N

In short, Your Answer would be 4 Newtons

Hope this helps!

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What is the period of oscillation of a mass 40kg on a spring with constant k=10N/m?

iVinArrow [24]

The period of oscillation of the system is 12.56 s

Explanation:

The period of oscillation of a spring-mass system is given by

$T=2\pi \sqrt{\frac{m}{k}}$

where

m is the mass attached to the spring

k is the spring constant

For the system in this problem, we have

m = 40 kg

k = 10 N/m

Substituting into the equation, we find

$T=2\pi \sqrt{\frac{40}{10}}=12.56 s$

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7 0

3 years ago

1. A car is traveling at 36 m/s hits a tree and comes to rest in 0.05 seconds.

QveST [7]

Answer: a. -720m/s^2

b. Yes, airbags will deploy

Explanation:

The formula for acceleration is:

= (Final velocity - Initial velocity)/Time

Final velocity = 0m/s

Initial velocity = 36m/s

Time taken = 0.05s

= (Final velocity - Initial velocity)/Time

= (0 - 36)/0.05

= -36/0.05

= -720m/s^2.

Since it's negative, it shows that there was a deceleration.

2. Yes the airbag will deploy since the acceleration gotten is more than -600 m/s^2.

3 0

3 years ago

A yo-yo of mass M has an axle of radius b and a spool of radius R. Its moment of inertia can be taken to be MR2/2 and the thickn

kow [346]

Answer:

The tension in the cord is $T=\frac{MR^{2}g }{2b^{2}+R^{2} }$

Explanation:

Given:

M = mass

b = radius

R = spool of radius

The equation is:

$bT=(\frac{MR^{2} }{2} )(\frac{a}{b} )\\T=\frac{MR^{2}a }{2b^{2} }$ (eq. 1)

The sum of forces in y:

∑Fy = Mg - T = Ma

$Mg=(M+\frac{MR^{2} }{2b^{2} } )a\\a=\frac{2b^{2}g }{2b^{2}+R^{2} }$

Replacing in eq. 1

$T=\frac{MR^{2} }{2b^{2} } (\frac{2b^{2}g }{2b^{2} +R^{2} } )\\T=\frac{MR^{2}g }{2b^{2}+R^{2} }$

3 0

2 years ago

Read 2 more answers

A massless string connects a 10.00 kg mass to a 13.00 kg cart which is resting on a frictionless horizontal surface. The mass ha

ch4aika [34]

The cart's acceleration to the right after the mass is released is determined as 7.54 m/s².

<h3>Acceleration of the cart</h3>

The acceleration of the cart is determined from the net force acting on the mass-cart system.

Upward force = Downward force

ma = mg

13a = 10(9.8)

13a = 98

a = 98/13

a = 7.54 m/s²

Thus, the cart's acceleration to the right after the mass is released is determined as 7.54 m/s².

Learn more about acceleration here: brainly.com/question/14344386

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6 0

2 years ago

Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th

Kipish [7]

Answer:

Probability of tunneling is $10^{- 1.17\times 10^{32}}$

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

Max velocity of the tennis ball, $v_{m} = 200\ mph$ = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

$KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J$

Kinetic energy of the tennis ball, KE' = $\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s$

Now, the distance the ball can penetrate to is given by:

$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

$\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js$

Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

$P(t) = e^{\frac{- 2t}{\eta}}$

$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

6 0

3 years ago