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Luba_88 [7]
3 years ago
5

When jogging outside you accidently bump into a curb. Your feet stop but your body continues to move forward and you end up on t

he ground. Which law of motion is being described in this scenario?
Law of Universal Gravitation

Newton’s First Law of Motion

Law of Conservation of Energy

Newton’s Second Law of Motion
Physics
2 answers:
gayaneshka [121]3 years ago
5 0
Newtons first law of motion
object in motion stays in motion, object at rest stays at rest unless acted upon an unbalanced force
VARVARA [1.3K]3 years ago
3 0

Answer:

Newton’s First Law of Motion

Explanation:

When jogging outside you accidentally bump into a curb. Your feet stop but your body continues to move forward and you end up on the ground. Newton's first law of motion is described in this scenario. This is also known as law of inertia.

According to this law, an object at rest will remain at rest and an object in motion will remain in motion until and unless and external force acts on it. In this case, feet stops and body of the person continues to move forward.

Hence, the correct option is (b) " Newton’s First Law of Motion  ".

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PilotLPTM [1.2K]
<h3><u>Answer and explanation;</u></h3>

Reflection;

  • Bounces back from an opaque material
  • Bounces back from the boundary of two mediums
  • Seen in mirrors

Refraction;

  • Passes through a prism  
  • Moves from one medium to another  
  • Seen in lenses

Reflection refers to the bouncing of waves such as light waves. It involves a change in direction of waves when they bounce off a barrier.

Refraction is the bending of waves when they move from one medium to another. It involves a change in the direction of waves as they pass from one medium to another.

8 0
3 years ago
Read 2 more answers
An electron moves along the z-axis with vz=4.5×10^7m/s. As it passes the origin, what are the strength and direction of the magn
Hitman42 [59]
This answer of the strength and directions of the magnetic field is a
5 0
3 years ago
Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes
pantera1 [17]

Answer:

The  magnitude of F1 is

|F1|=358.74 \ N

The magnitude of F2 is

|F2|=179.37\ N

And the direction of F2 is

\alpha = 44.214^o

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.  

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.  

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

\displaystyle -2F\ sin21^o+F\ cos\alpha =0

The sum of the vertical components of F1 and F2 must equal the total force Ft

\displaystyle -2F\ cos21^o+F\ sin\alpha =460

Solving for \alpha in the first equation

\displaystyle cos\alpha =\frac{2F\ sin21^o}{F}=2sin21^o

\displaystyle cos\alpha =0.717=>\alpha =44.214^o

\displaystyle F(2cos21^o+sin\alpha)=460

\displaystyle F=\frac{460}{2cos21^o+sin\alpha}

\displaystyle F=\frac{460}{2cos21^o+sin44.214^o}

\displaystyle F=179.37\ N

The  magnitude of F1 is

|F1|=2*F=358.74 \ N

The magnitude of F2 is

|F2|=179.37\ N

And the direction of F2 is

\alpha = 44.214^o

4 0
3 years ago
When two waves undergo complete, constructive interference, which of the following increases ?
Lostsunrise [7]
A speed cause it would slow down meeting eachother
4 0
3 years ago
A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.
MakcuM [25]

Answer:

b. 9.5°C

Explanation:

m_i = Mass of ice = 50 g

T_i = Initial temperature of water and Aluminum = 30°C

L_f = Latent heat of fusion = 3.33\times 10^5\ J/kg^{\circ}C

m_w = Mass of water = 200 g

c_w = Specific heat of water = 4186 J/kg⋅°C

m_{Al} = Mass of Aluminum = 80 g

c_{Al} = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C

The final equilibrium temperature is 9.50022°C

4 0
3 years ago
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