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3 years ago

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What’s the correct answer

1 answer:

White raven [17]3 years ago

8 0

Looks like you simply substitute the length of the femur

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What is the slope of (0,0)(4,200)

Kruka [31]

Answer:

.02

Explanation:

X-x/ Y-y= 4-0/200-0 =.02

5 0

3 years ago

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Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fil

laila [671]

Answer:

$W_n_e_t=7.648512 \approx 7.6J$

$K.E=0.8J$

$v=0.7844645406 \approx 0.78m/s$

Explanation:

From the question we are told that

Mass of pitcher $M= 2.6kg$

Force on pitcher $f=8.8N$

Distance traveled $48cm=>0.48m$

Coefficient of friction $\mu=0.28$

a)Generally frictional force is mathematically given by

$F=\mu N$

$F=0.28*2.6*9.8$

$F=7.1344N$

Generally work done on the pitcher is mathematically given as

$W_n_e_t=W_f+W_F$

$W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N$

$W_n_e_t=4.224-3.424512$

$W_n_e_t=0.799488\approx 0.8J$

b)Generally K.E can be given mathematically as

$K.E= W_n_e_t$

Therefore

$K.E=0.8J$

c)Generally the equation for kinetic energy is mathematically represented by

$K.E=1/2mv^2$

$0.8=1/2mv^2$

Velocity as subject

$v=\sqrt{\frac{K.E*2}{m} }$

$v=\sqrt{\frac{0.8*2}{2.6} }$

$v=0.7844645406 \approx 0.78m/s$

6 0

3 years ago

If two different theories describe experimental observations equally well, can one be said to be more valid than the other (assu

Andreas93 [3]

Answer:

No

Explanation:

In such situations we cannot determine which one is more valid as both serves the purpose well.

Two theories are carried out in different environment and circumstance keeping different parameters and one can opt for any number of ways to carry out that experiment but what matter at the end is the accuracy they bring.

Each of the theory is a new discovery and follows all the possible logical rules hence it is not possible to decide which one is more valid.

3 0

3 years ago

Calculate the orbital speed (in m/s) of a satellite that circles the Earth with a time period of 12.00 hours. The mass of the Ea

Hoochie [10]

Answer:

$v = 3869 m/s$

Explanation:

As we know that the orbital speed of the satellite is given as

$v = \sqrt{\frac{GM}{r}}$

also we know that

time period of the revolution is given as

$T = \frac{2\pi r}{v}$

now from above equation we know that

$T = \frac{2\pi (\frac{GM}{v^2})}{v}$

$T = \frac{2\pi GM}{v^3}$

so we will have

$v = (\frac{2\pi GM}{T})^{1/3}$

now plug in all data in this equation

$v = (\frac{2\pi (6.67 \times 10^{-11})(5.97 \times 10^{24})}{12 \times 3600})^{1/3}$

$v = 3869 m/s$

3 0

3 years ago

ShEs jUsT a FrIenD (Answer if you understand)

GarryVolchara [31]

I hate it so muchhhhhhhh

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3 years ago

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