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RSB [31]
3 years ago
12

Una banda transportadora se utiliza para cargar canastos en un avión de carga, si µs = 0.5. ¿Cuál es el ángulo máximo de elevaci

ón para evitar el deslizamiento? Respuesta: 26.6
Physics
1 answer:
Sonja [21]3 years ago
7 0
No se creo es 45.7 de nada
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Where is the center of gravity if the 9.00 kg mass of the barbell itself is taken into account?
Pavel [41]

The problem is missing some parts but nevertheless here is the answer:

 

Given:

1.7 m long barbell with 23 kg weight on the left and 34 in the right end.

 

Solution:

 Calculating from the left side:

Center of gravity = ΣM/ΣW = [23*0 + 9*(1.7/2) + 34*1.7]/[23+9+34] 

Center of gravity = .992 m from the left end

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What scientific breakthrough did Charles Darwin conclusion lead to? ANSSER ASAP PLZ
leva [86]
His breakthrough was the Theory of Evolution. 
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The temperature of the soda
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Answer:

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An electron that has a velocity with x component 2.4 x 106 m/s and y component 3.6 x 106 m/s moves through a uniform magnetic fi
likoan [24]

Answer:

(a) 7.315 x 10^(-14) N

(b) - 7.315 x 10^(-14) N

Explanation:

As you referred at the final remark, the electron and proton undergo a magnetic force of same magnitude but opposite direction. Using the definition of magnetic force,  a cross product must be done. One technique is either calculate the magnitude of the velocity and magnetic field and multiplying by sin (90°), but it is necessary to assure both vectors are perpendicular between each other ( which is not the case) or do directly the cross product dealing with a determinant (which is the most convenient approach), thus,

(a) The electron has a velocity defined as:  \overrightarrow{v}=(2.4x10^{6} i + 3.6x10^{6} j) \frac{[m]}{[s]}\\\\

In respect to the magnetic field; \overrightarrow{B}=(0.027 i - 0.15 j) [T]

The magnetic force can be written as;

\overrightarrow{F} = q(\overrightarrow{v} x \overrightarrow{B})\\ \\\\\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]

Bear in mind q =-1.6021x10^{-19} [C]  

thus,

\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= -1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=7.3152x10^{-14} [N]

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, k^{2}=k\cdot k = 1

(b) Considering the proton charge has the same magnitude as electron does, but the sign is positive, thus

\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= 1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(-7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (-7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=-7.3152x10^{-14} [N]

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, k^{2}=k\cdot k = 1

Final remarks: The cross product was performed in R3 due to the geometrical conditions of the problem.  

6 0
3 years ago
Sketch a position-time graph for a bear starting
Dmitrij [34]

Explanation:

hopefully that makes sense. the position doesn't change over the 5 seconds, meaning it's stopped but time still continues. then when the slope is negative this shows the bear's position becoming negative (backing up, changing direction).

3 0
3 years ago
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