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3 years ago

A body moving in a circular path has _______ displacement.

Physics

1 answer:

PtichkaEL [24]3 years ago

6 0

Answer:

Displacement is the shortest distance travelled from the initial point to the final point. After half revolution the displacement is the length of the diameter which is equal to 2r. Hence the displacement is 2r.

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A child walks 5m east, then 3m north, then 1m east. What is the magnitude of the child's displacement? What is the direction of

kykrilka [37]

You should first break up all the distances and directions into vectors and then add all horizontal vectors and verticle vectors.
He goes, 5 + 1 m east = 6m East
3 m north
displacement = root(6^2 + 3^2) = 6.708m
direction = arcTan(verticle/horizontal) = arcTan(3/6) = 26.565 degrees

7 0

3 years ago

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.03 s la

Nookie1986 [14]

Answer:

$h=53.09m$ (2)

$v_{min}>5.05m/s$

$v_{max}$

Explanation:

<u>a)Kinematics equation for the first ball:</u>

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=8.9m/s$

The ball reaches the ground, y=0, at t=t1:

$0=h+v_{o}t_{1}-1/2*g*t_{1}^{2}$

$h=1/2*g*t_{1}^{2}-v_{o}t_{1}$ (1)

Kinematics equation for the second ball:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=0$ the ball is dropped

The ball reaches the ground, y=0, at t=t2:

$0=h-1/2*g*t_{2}^{2}$

$h=1/2*g*t_{2}^{2}$ (2)

the second ball is dropped a time of 1.03s later than the first ball:

t2=t1-1.03 (3)

We solve the equations (1) (2) (3):

$1/2*g*t_{1}^{2}-v_{o}t_{1}=1/2*g*t_{2}^{2}=1/2*g*(t_{1}-1.03)^{2}$

$g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)$

$-2v_{o}t_{1}=g*(-2.06*t_{1}+1.06)$

$2.06*gt_{1}-2v_{o}t_{1}=g*1.06$

$t_{1}=g*1.06/(2.06*g-2v_{o})$

vo=8.9m/s

$t_{1}=9.81*1.06/(2.06*9.81-2*8.9)=4.32s$

t2=t1-1.03 (3)

t2=3.29sg

$h=1/2*g*t_{2}^{2}=1/2*9.81*3.29^{2}=53.09m$ (2)

b) $t_{1}=g*1.06/(2.06*g-2v_{o})$

t1 must : t1>1.03 and t1>0

limit case: t1>1.03:

$1.03>9.81*1.06/(2.06*g-2v_{o})$

$1.03*(2.06*9.81-2v_{o})$

$20.8-2.06v_{o}$

$(20.8-10.4)/2.06$

$v_{min}>5.05m/s$

limit case: t1>0:

$g*1.06/(2.06*g-2v_{o})>0$

$2.06*g-2v_{o}>0$

$v_{o}$

$v_{max}$

8 0

4 years ago

g As a proton moves in the direction the electric field lines A) it is moving from low potential to high potential and gaining e

Yuki888 [10]

Answer:

Option D => it is moving from high potential to low potential and losing electric potential energy.

Explanation:

Consider a big circle, within the circle we have force, F. That force, F is known as the Electric Field and inside the region or field or space, charged particle or object will be able to exerts force on the other objects.

Electric Field can be represented mathematically by using the formula below; E = kQ/r^2.

So, let us answer the question with what we have considered above. It is worthy of note to know that electric Field moves from a region of higher potential to a region of lower potential. So, any option that says this is correct.

But, there is only one problem and that is the fact that the question asked us about the direction of the movement of proton. Since, proton is s a positive charge, it is going to lose electric potential energy. So, Option D is correct.

3 0

3 years ago

The horizontal surface on which the block (mass 2.0 kg) slides is frictionless. The speed of the block before it touches the spr

ch4aika [34]

Answer:3.67 m/s

Explanation:

mass of block(m)=2 kg

Velocity of block=6 m/s

spring constant(k)=2 KN/m

Spring compression x=15 cm

Conserving Energy

energy lost by block =Gain in potential energy in spring