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4 years ago

What statements best define percent error? Check all that apply.

Physics

2 answers:

oksian1 [2.3K]4 years ago

5 0

3, 1 could be.

And there's always Heisenberg's uncertainty principle !!!

Maru [420]4 years ago

3 0

1,3,6 answers to the question

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A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache

Paladinen [302]

Answer:

$F_a=1470\ N$

Explanation:

<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

$\displaystyle F_a-F_{r1}-F_{r2}=m.a=0$

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

$\displaystyle F_a=F_{r1}+F_{r2}.....[1]$

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

$\displaystyle F_{r2}-T=0$

The friction forces are computed by

$\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g$

$\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g$

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

$\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g$

Simplifying

$\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)$

Plugging in the values

$\displaystyle F_{a}=0.25(9.8)[400+2(100)]$

$\boxed{F_a=1470\ N}$

8 0

3 years ago

Kepler’s third law can be used to derive the relation between the orbital period, P (measured in days), and the semimajor axis,

NikAS [45]

Kepler's 3rd law is given as
P² = kA³
where
P = period, days
A = semimajor axis, AU
k = constant

Given:
P = 687 days
A = 1.52 AU

Therefore
k = P²/A³ = 687²/1.52³ = 1.3439 x 10⁵ days²/AU³

Answer: 1.3439 x 10⁵ (days²/AU³)

8 0

3 years ago

Read 2 more answers

The mass of the skier including his equipment is 75 kg in the ski race, the total vertical change in height is 880m

Bogdan [553]

75 percent off of water and please water the light water and water water and then go back and please water pollution please 880m

8 0

3 years ago

A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p

wel

Answer:

Normal force = 8.75 N

Explanation:

given,

frictional force between the steel spatula and the Teflon frying pan=0.350 N

coefficient of friction between material =0.04

normal force = ?

using formula,

Frictional force = coefficient of friction × normal force

$normal\ force = \dfrac{Frictional\ force}{coefficient\ of\ friction}$

$normal\ force = \dfrac{0.350}{0.04}$

Normal force = 8.75 N

8 0

3 years ago

A ball rolls of buildings that is 100m high calculate the time that it takes for ball to hit the ground

LUCKY_DIMON [66]

Answer:

2as=v2-u2

2000=v2

V=44

V=u+at

44/10=t

T=4.4seconds

5 0

3 years ago