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Fynjy0 [20]
3 years ago
13

Which statement best summarizes the central idea of “Applications of Newton’s Laws”?

Physics
2 answers:
Alexxx [7]3 years ago
4 0

The answer is: Newton’s laws can explain the forces that occur between objects every day.

Gravity as a force which causes any two bodies to be attracted to each other.  

The gravitational effect increases as the size of an object increases. Earth s surrounded by its own gravitational field.  

The weight of an object is the amount of force acting on the object due to gravity. SI unit for weight is newton (N).  


Shkiper50 [21]3 years ago
3 0

 

Newton's three forces, normal, tension and friction, are present in a surprising number of physical situations

Newton's Laws, that describe the relationship between an obejct and the forces acting upon it, apply in almost every physical situation, from quantum mechanics to electricity.

The correct answer is:

Newton’s laws can explain the forces that occur between objects every day


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NISA [10]

Answer:

200 ms-1

Explanation:

v = u+ at

v = 50+ 5×30

v = 200 ms-1

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A cup falls off of a table from a height of 0.75 m. What is the impact speed of the cup?
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4 years ago
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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
4 years ago
What is the energy of a single photon of visible light with a wavelength of 500.0 nm? 1 nm=10^−9 m
Dmitrij [34]

Answer:

3.97×10^−19 J

From the planks equation

E=hv

V= c/ wave length

V= 3×10^8/500×10^-9

=6×10^14

E= hv

6.63×10^-34×6×10^14

= 3.97×10^−19 J

Explanation:

8 0
2 years ago
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