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trasher [3.6K]
3 years ago
10

Is it possible for an object to be going 100 miles per hour but have a velocity of zero?

Physics
1 answer:
Lesechka [4]3 years ago
5 0
It is possible for on object to be going at 100 miles per hour, but still have a velocity. This is because the object going at 100 miles per hour has speed, which is a scalar quantity, which is defined by only magnitude, but the velocity of the object can be 0, since velocity is a vector quantity which is defined by both magnitude and direction.

Since this object only has magnitude and no direction (which is not given), then the velocity can be 0
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With the simplified model of the eye, what corrective lens (specified by focal length as measured in air) would be needed to ena
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Answer:

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Explanation:

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3 years ago
A very humble bumble bee is flying horizontally due North at a constant speed of 3.11 m/s. At the current location of the bumble
Reil [10]

To solve this problem we will apply the concepts of the Magnetic Force. This expression will be expressed in both the vector and the scalar ways. Through this second we can directly use the presented values and replace them to obtain the value of the magnitude. Mathematically this can be described as,

\vec{F_B} = q(\vec{V}\times \vec{B})

F_B = q|v||B| sin\theta

Here,

q = Charge

v = Velocity

B = Magnetic field

\theta = \text{Angle between } \vec{B} \text{ and } \vec{V}

Our values are given as,

\theta = 35.7\°

q = 22.5*10^{-9}C

B = 1.05*10^{-5}T

v = 3.11m/s

Replacing,

F_B = (22.5*10^{-9}C)(3.11 \times 1.05*10^{-5}) sin(35.1\°)

F_B = 4.224*10^{-13}N

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3 0
3 years ago
The component of the ball's velocity whose magnitude is most affected by the collisions is
bazaltina [42]
<span>The component most affected by the collisions is vertical. The ball's vertical will either decrease or increase due to the collision. If the velocity is high during the collsion the ball's vertical will likely be higher and if the ball's velocity is low the vertical will be as well.</span>
8 0
3 years ago
HELP!!! 30 POINTS+BRAINLIEST!!!!!QUICK!!
tia_tia [17]

Answer:

A:7.2ms^{-1}

B:14.25ms^{-1}

C:1.45sec

D:10.3m

E:2.9sec

F:20.88m

Explanation:

Let v be the velocity and \alpha be the angle between the velocity and ground.

Question A:

Horizontal component of velocity is given by vcos(\alpha ).

So,horizontal component is 16\times cos(63)=16\times 0.45=7.2ms^{-1}

Question B:

Vertical component of velocity is given by vsin(\alpha ).

So,vertical component is 16\times sin(63)=16\times 0.89=14.25ms^{-1}

Question C:

Time required is given by \frac{\text{vertical component of velocity}}{g}}=\frac{14.25}{9.8}=1.45 seconds

Question D:

Maximum height is given by \frac{\text{vertical component of velocity}^{2}}{2g}}=\frac{203.06}{19.6}=10.3m

Question E:

Time of flight is twice the time required to reach maximum height=2\times 1.45=2.9 seconds.

Question F:

The distance between the player and ball after landing is called range and is given by \text{horizontal component of velocity}\times \text{time of flight}=7.2\times 2.9=20.88m

8 0
3 years ago
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