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2 years ago

A ball was positioned in the middle of a smooth ramp and allowed to roll

Physics

1 answer:

netineya [11]2 years ago

6 0

Answer:

it is quicker

Explanation:

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PLEEEEASE HELP ME PLEASE WILL GIVE BRAINLIEST

Drupady [299]

C. 7.8 hz I think! Hope this helped ☺️

8 0

3 years ago

Develop a hypothesis for why one of the two types of soup should indeed be rolling down faster than the other. This hypothesis s

Eva8 [605]

Answer:

Assume two identical cans filled with two types of soup having same mass are rolling down on an inclined plane in same conditions. In terms of inertia different types of soup will indicate different viscosity. The higher viscosity fillings indicates more part of the soup mass is rotating together with the can’s body. This means that for the can with lower viscosity soup has a lower moment of inertia and the can with higher viscosity has higher moment of inertia while the same gravity makes them to roll.

incline angle = θ ; can's mass = m ; Radius of the can's = R , Angular acceleration for Can 1 = α1 ; Angular acceleration for Can 2 = α2

T1 = Inertia of Can with high viscosity soup

T2 = Inertia of Can with low viscosity soup

M1 rolling moment of Can 1

M2 rolling moment of Can 2

equation is given by

T1*α1 = M1 - (a)

T2*α2 = M2 - (b)

M1 = M2 = m*g*R*sin(θ). (c)

as assumed T1 > T2

from the three equation (a), (b) & (c)

the α2 > α1

Angular acceleration of Can 2 is higher than Can 1. Already stated that Can 1 has more viscous soup as compared to Can 2.

7 0

3 years ago

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

4 years ago

What is the most likely reason that Mendeleev placed tellurium before iodine?

Ilya [14]

Mendeleev watched that tellurium has compound properties like different components in its gathering, and he didn't realize that neutrons cause the more noteworthy nuclear mass. Mendeleev expressed that he anticipated that tellurium would have a lower nuclear mass than iodine.

8 0

3 years ago

A car slows down uniformly from a speed of 30.0 m/s to rest in 7.20 s

Ede4ka [16]

When acceleration is constant, the average velocity is given by

$\bar v=\dfrac{v+v_0}2$