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2 years ago

The acceleration of the car with the data in the table above would be

Physics

1 answer:

neonofarm [45]2 years ago

6 0

The acceleration of the car would be 0.33 first and then it would be 0.17.

<u>Explanation:</u>

An applied force is a force that is applied to an object by an individual or another item. On the off chance that an individual is pushing a work area over the room, at that point there is an applied power following up on the article. The applied power is the power applied on the work area by the individual.

The net force applied to the object rises to the mass of the article increased by the measure of its acceleration. The net power following up on the soccer ball is equivalent to the mass of the soccer ball duplicated by its adjustment in speed each second (its acceleration).

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Explain how sound travels through the air to observer

Harlamova29_29 [7]

Sample:
As soon as the lightning is seen, sound travels in the air for several kilometers until it eventually reaches the observer.

3 0

3 years ago

What is the radius of a communications satellite’s orbital path that is in a uniform circular orbit around Earth and has a perio

pantera1 [17]

If the period of a satellite is T=24 h = 86400 s that means it is in geostationary orbit around Earth. That means that the force of gravity Fg and the centripetal force Fcp are equal:

Fg=Fcp

m*g=m*(v²/R),

where m is mass, v is the velocity of the satelite and R is the height of the satellite and g=G*(M/r²), where G=6.67*10^-11 m³ kg⁻¹ s⁻², M is the mass of the Earth and r is the distance from the satellite.

Masses cancel out and we have:

G*(M/r²)=v²/R, R=r so:

G*(M/r)=v²

r=G*(M/v²), since v=ωr it means v²=ω²r² and we plug it in,

r=G*(M/ω²r²),

r³=G*(M/ω²), ω=2π/T, it means ω²=4π²/T² and we plug that in:

r³=G*(M/(4π²/T²)), and finally we take the third root to get r:

r=∛{(G*M*T²)/(4π²)}=4.226*10^7 m= 42 260 km which is the height of a geostationary satellite.

3 0

2 years ago

Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

Artist 52 [7]

Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

b) $(x_4,y_4)=(-1.917,-1.75)m$

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m$

If we solve for a we got:

$(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0$

$a=-\frac{(5kg*2.3m)}{6kg}=-1.917m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m$

$(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0$

And solving for b we got:

$b=-\frac{(7kg*1.5m)}{6kg}=-1.75m$

So the coordinates for this new particle are:

$(x_4,y_4)=(-1.917,-1.75)m$

5 0

3 years ago

A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl

zysi [14]

Answer: $\alpha =9.302\ rad/s^2$

Explanation:

Given

mass of sphere $m=250\ gm$

diameter of sphere $d=4.30\ cm$

radius $r=\frac{4.30}{2}\ cm$

$f=0.0200\ N$

friction will provide resisting torque so

$f\times r=I\times \alpha$

where $I=\text{moment of Inertia}$

$f=\text{friction force}$

$\alpha =\text{angular acceleration}$

$I=\frac{2}{5}mr^2$

$0.02\times r=\frac{2}{5}mr^2\times \alpha$

$\alpha =\frac{5}{2r}\times f$

$\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02$

$\alpha =9.302\ rad/s^2$

(b)time taken to decrease its rotational speed by $21\ rad/s$

$t=\dfrac{\Delta \omega }{\alpha }$

$t=\dfrac{21}{9.302}$

$t=2.25\ s$

6 0

3 years ago

Explain the difference in reactivity of the elements based on their electron configurations.

victus00 [196]

Well a basic explanation is that some elements have enough electrons to be considered stable. These elements do not need to react with other elements to gain more electrons. Reactive elements are no where near stable; they respond with other elements in order to become stable. The more unstable, the harsher the reaction is.

5 0

3 years ago