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pishuonlain [190]
3 years ago
13

A stone is dropped from a bridge and hits the pavement below in two seconds. What is the velocity of the stone when it hits the

pavement?
A. 0.20 m/s
B. 19.6 m/s
C. 4.9 m/s
D. 9.8 m/s
Physics
2 answers:
Helen [10]3 years ago
4 0
We have: a = v/t
Here, t = 2 s  [ Given ]
a = 9.8 m/s²  [constant value for earth system ]

Substitute their values into the expression:
9.8 = v/2
v = 9.8 × 2
v = 19.6 m/s

In short, Your Answer would be Option B

Hope this helps!
Yakvenalex [24]3 years ago
4 0

i just did the quiz and so the only 2 options that made logical sense would be B (19.6 m/s) or D (9.8 m/s) the reason someone could get confused between these two is because 19.6 divided by 2 is 9.8 m/s so for each second it will go down 9.8 m/s but since it wants to know the velocity it would be 19.6 m/s

hope this helps

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Question 4 of 10
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Answer:

22.3 kg•m/s

Explanation:

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3 years ago
An elevator is accelerating upward at a rate of 3.6 m/s2. a block of mass 24 kg hangs by a low-mass rope from the ceiling, and a
kozerog [31]
The answer:

<span>When the elevator accelerates upward at a rate of 3.6 m/s², the value of the acceleration becomes

</span>A=g+3.6=13.4 m/s²
and by using the newton's law, F=mass x A, we have 
T1= (24 + 90 )x 13.4= 1527.6 N, where T1 is the <span>Tension in upper rope
</span> and 
T2= ( 90 )x 13.4= 1206N, where T2 is the Tension in lower rope

When the elevator accelerates downward at a rate of 3.6 m/s², the value of the acceleration becomes
A=9.8 - 3.6 = 6.2 m/s²

T1= (24 + 90 )x 6.2= 706.8 N, where T1 is the Tension in upper rope
 and 
T2= ( 90 )x 6.2= 558N, where T2 is the Tension in lower rope


5 0
3 years ago
A certain freely falling object, released from rest, requires 1.95 s to travel the last 23.5 m before it hits the ground. (a) Fi
Ratling [72]

Answer:

(a). The velocity of the object is -2.496 m/s.

(b).  The total distance of the object travels during the fall is 23.80 m.

Explanation:

Given that,

Time = 1.95 s

Distance = 23.5 m

(a). We need to calculate the velocity

Using equation of motion

s = ut+\dfrac{1}{2}gt^2

Put  the value into the formula

-23.5=u\times1.95+\dfrac{1}{2}\times(-9.8)\times(1.95)^2

u=\dfrac{-23.5+4.9\times(1.95)^2}{1.95}

u=-2.496\ m/s

(b). We need to calculate the total distance the object travels during the fall

Using equation of motion

v = u+gt

Put the value in the equation

-2.496=0-9.8\times t

t =\dfrac{2.496}{9.8}

t=0.254\ sec

The total time is

t'=t+1.95

t'=0.254+1.95

t'=2.204\ sec

We need to calculate the distance

Using equation of motion

s = ut+\dfrac{1}{2}gt^2

Put the value into the formula

s=0+\dfrac{1}{2}\times9.8\times(2.204)^2

s=23.80\ m

Hence, (a). The velocity of the object is -2.496 m/s.

(b).  The total distance of the object travels during the fall is 23.80 m.

4 0
3 years ago
Can we always see the same amount of the illuminated side of the Moon from Earth? Explain. Thanks so much for everyone's help! :
Nadusha1986 [10]

Answer:

Nope.

Explanation:

No. The Moon rotates on its own axis at the same rate that it orbits around Earth. That means we always see the same side of the Moon from our position on Earth. The side we don't see gets just as much light, so a more accurate name for that part of the Moon is the "far side."

5 0
3 years ago
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