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3 years ago

An object traveling in a straight line accelerates. What will definitely happen due to the acceleration?

2 answers:

7 0

"Acceleration" means any change in the speed or direction of motion ... speeding up, slowing down, or turning. So . . .

<span>-- </span><span>The distance traveled in a certain time may increase or decrease.

-- The displacement covered in a certain time may increase or decrease.

-- The speed of the object may increase or decrease.

-- The velocity of the object (speed/direction) will change.
</span>
Hope this helps..

marshall27 [118]3 years ago

7 0

Well 2 things could happen 1 it could just keep going 2 well basically acceleration does not depend on time so it could lift off

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Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at

dedylja [7]

Answer:

$50.91 \mu C$

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for $q_{1}$ and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by $q_{1}$ on q. Then we can know the magnitude of the force exerted by $q_{2}$ about q, finally this will allow us to know the magnitude of $q_{2}$

$q_{1}$ exerts a force on q in +y direction, and $q_{2}$ exerts a force on q in -y direction.

$F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\$

The net force on q is:

$F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}$

Rewriting for $q_{2}$ :

$q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C$

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3 years ago

The light coming out of a concave lens:

torisob [31]

The light coming out of a concave lens will never meet.

So, the answer is A. will never meet.

Happy Studying! ^^

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3 years ago

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Why it is difficult to run fast in sand

balu736 [363]

Running on sand requires 1.6 times more energy spent than running on hard surface, so the force applied by our foot on sand is less.

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3 years ago

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WILL UPVOTE!!!Physics help please!!

liubo4ka [24]

Speed v = initial speed u + acceleration a x time t
v=u+at = 2 + 4*3 = 14 m/s

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3 years ago

A flat horizontal surface with an area of 518 cm^2 is inside a uniform electric field of 1.33 x 10^4 N/C. The angle between the

GarryVolchara [31]

Answer: 576.48 N*m^2/C

Explanation: In order to calculate the electric flux through the any surface we have to take into account the scalar product between the electric field vector and the normal vector to the surface.

So we have:

ФE= E*A= 1.33 * 10^4*0.0518* cos (33.2°)= 576.48 N*m^2/C

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3 years ago