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grandymaker [24]
2 years ago
13

A hotel elevator ascends 200m with maximum speed of 5m/s. Its acceleration and deceleration both have a magnitude of 1.0m/s2. Pa

rt APart complete How far does the elevator move while accelerating to full speed from rest? Express your answer with the appropriate units. 12.5 m Previous Answers Correct Part B How long does it take to make the complete trip from bottom to top? Express your answer with the appropriate units.
Physics
1 answer:
ValentinkaMS [17]2 years ago
4 0

Answer:

45 s .

Explanation:

The accelerator will first accelerate , then move with uniform velocity and at last it will decelerate to rest .

displacement s = ?

acceleration a = 1 m /s²

Final speed v = 5 m/s

initial speed u = 0

v² = u² + 2as

5² = 0 + 2 x 1 x s

s = 12.5  m

B)  Let time of acceleration or deceleration be t

v = u + a t

5 = 0 + 1 t

t = 5 s

Similarly displacement during deceleration = 12.5 m

Total distance during uniform motion = 200 - ( 12.5 + 12.5 ) =  175 m .

velocity of uniform motion = 5 m /s

time during which there was uniform velocity = 175 / 5 = 35 s

Total time = 5 + 35 + 5 = 45 s .

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I think its a tbh bc it seems to be the best answer out of a b c and d
6 0
3 years ago
d. The force is doubled and the object’s mass is halved? 18. ||| A man pulling an empty wagon causes it to accelerate at 1.4 m/s
Harrizon [31]

Answer:

a' = 0.35 m/s^2

Explanation:

Let say the empty wagon has mass "M"

now by newton's II Law we will have

F = Ma

now it is given that empty wagon is pulled with acceleration 1.4 m/s/s

now we will have

F = 1.4 M

now a child of mass three times the mass of wagon is sitting on the empty wagon

so here we have

F = (M + 3M) a

1.4 M = 4M a'

so we have

a' = 0.35 m/s^2

8 0
3 years ago
projectile motion of a particle of mass M with charge Q is projected with an initial speed V in a driection opposite to a unifor
SIZIF [17.4K]

Answer:

Range, R = MV²/2QE

Explanation:

The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.

Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses  all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.

So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops

Therefore {tex}R = MV²/2QE{/tex}

5 0
3 years ago
. A 2.0-kg block is on a perfectly smooth ramp that makes an angle of 30° with the horizontal. (a) What is the block’s accelerat
AfilCa [17]

Answer:

a) a = 4.9 m / s²,  N = 16.97 N   and b)   F = 9.8 N

Explanation:

a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry

        sin 30 = Wx / W

        cos 30 = Wy / W

        Wx = W sin30

        Wy = W cos 30

Let's write the equations on each axis

X axis

        Wx = ma

Y Axis  

       N- Wy = 0

       N = Wy = mg cos 30

       N = 2.0 9.8 cos 30

       N = 16.97 N

We calculate the acceleration

       a = Wx / m

       a = mg sin 30 / m

       a = g sin 30

       a =9.8 sin 30

       a = 4.9 m / s²

b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component

      F -Wx = 0

      F = Wx

      F = m g sin 30

      F = 2.0 9.8 sin 30

      F = 9.8 N

5 0
3 years ago
One half of a balanced chemical equation is shown.
Zarrin [17]

Answer:

A: 3 Mg, 2 P, 14 O, 12 H

Explanation:

I took the test on Edgenuity

6 0
2 years ago
Read 2 more answers
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