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4 years ago

Galileo is heliocentric or geocentric

Physics

2 answers:

Ymorist [56]4 years ago

8 0

Answer:

Heliocentric

Explanation:

Just took the test

algol [13]4 years ago

6 0

Galileo is geocentric, just like all the rest of us.

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The block A and attached rod have a combined mass of 50 kg and are confined to move along the guide under the action of the 796-

Luba_88 [7]

Answer:

The bending moment is 459.16 N.m

Explanation:

From the given information;

Let's assume that the angle is 66°

Then, the free body diagram is draw and attached in the file below.

Now, the calculation of the acceleration from the first part of the free body diagram is:

$\sum F_x = ma_x \\ \\ 796 - 50(9.81) sin 66=50a \\ \\ 796 - 448.094 = 50 a \\ \\ a = \dfrac{347.906}{50} \\ \\ a = 6.96 \ m/s^2$

Bending moment M:

From the second part of the diagram:

$\sum M_B = mad \\ \\ M - (15 \times 9.81) (1.5) = (25 \times 6.96)(1.5 sin 66) \\ \\ M - 220.725 = 238.435 \\ \\ M = 238.435 + 220.725 \\ \\ \mathbf{M = 459.16 \ N.m}$

6 0

3 years ago

NEED HELP ASAP!! Will give Brainliest.

olga2289 [7]

a. 25 joules of work is done by the object.

Explanation:

According to the work-energy theorem, the work done ON an object is equal to the change in kinetic energy of the object, therefore:

$W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

$K_f = \frac{1}{2}mv^2$ is the final kinetic energy of the car, with

m = 2 kg being the mass of the car

v = 0 is the final speed of the car (brought to rest)

$K_i = \frac{1}{2}mu^2$ is the initial kinetic energy of the car, with

u = 5 m/s being the initial speed of the car

Substituting numbers, we find:

$W=\frac{1}{2}(2)(0)^2 - \frac{1}{2}(2)(5)^2=-25 J$

The negative sign means that the work done ON the car is negative: this means therefore that the work has been actually done BY the car. In fact, the car has lost its kinetic energy: this means that it has done work on the surrounding, converting its kinetic energy into other forms of energy.

Learn more about work and kinetic energy:

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brainly.com/question/6536722

#LearnwithBrainly

5 0

4 years ago

The work function for silver is 4.73 eV. (a) Convert the value of the work function from electron volts to joules.

pogonyaev

Answer:

$W=7.56\times 10^{-19}\ J$

Explanation:

Given that,

The work function for silver is 4.73 eV.

We need to find the value of the work function from electron volts to joules.

We know that,

$1\ eV=1.6\times 10^{-19}\ J$

For 4.73 eV,

$4.73\ eV=1.6\times 10^{-19}\times 4.73\\\\=7.56\times 10^{-19}\ J$

So, the work function for silver is $7.56\times 10^{-19}\ J$ .

6 0

3 years ago

Which does a reference point provide? Select two options. a position from which to measure future distance a set of standard uni

In-s [12.5K]

Answer:

hope this helps you :)

Explanation:

An object is in motion if its position changes relative to another object. To decide if you are moving, you can use your chair as a reference point. A reference point is a place or object used for comparison to determine if something is in motion.

4 0

4 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago