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lorasvet [3.4K]
2 years ago
6

While Bob is demonstrating the gravitational force on falling objects to his class, he drops an 1.0 lb bag of feathers from the

top of the science building. Determine the distance the bag has traveled after falling for 1.5 seconds assuming it has reach free fall and given the gravitational acceleration of 9.8 m/sec2.
A) 7.4 m
B) 11 m
C) 15 m
D) 22 m
Physics
2 answers:
asambeis [7]2 years ago
4 0

The answer here is A) 7.4 m.

____ [38]2 years ago
3 0

As per the question Bob drops the bag full with feathers from the top of the building.

The mass of the bag(m)= 1.0 lb

Let the air resistance is neglected.As the bag is under free fall ,hence the only force that acts on the bag is the force of gravity which is in vertical downward direction.

Here the acceleration produced on bag due to the free fall will be nothing else except the acceleration due to gravity i.e g =9.8 m/s^2


Here we are asked to calculate the distance travelled by the bag at the instant 1.5 s

Hence time t= 1.5 s

From equation of kinematics we know that -

                S=ut + 0.5at^2     [ here S is the distance travelled]

For motion under free fall initial velocity (u)=0.

Hence   S= 0×1.5+{0.5×(-9.8)×(1.5)^2}

           ⇒ -S =0-11.025 m

            ⇒ S= 11.025 m

                   =11 m

Here the negative sign is taken only due to the vertical downward motion of the body .we may take is positive depending on our frame of reference .


Hence the correct option is B.

               

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Answer:

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Explanation:

The electric field can be calculated as shown below:

E = k*|q|/r^2

Where:

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For 0.100 m of the axon, the value of q is:

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Therefore:

E = (8.98755*10^9)*(8.972*10^-12)/0.05^2 = 32.255 N/C

A positive point charge always produce an electric field that is directed away from the field while a negative point charge produces an electric field that is directed toward the field

3 0
2 years ago
An ocean liner is cruising at 10 meters/second and is about to approach a stationary ferryboat. A parcel is released from the oc
Afina-wow [57]
The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.

First, we determine how long the parcel will fall using:

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5.5 = (0)(t) + 1/2 (9.81)(t)²
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Now, we may use this time to determine the horizontal distance covered by the parcel by using:
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The horizontal velocity of the parcel will be equal to the horizontal velocity of the cruise liner.

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4 0
2 years ago
A model train engine was moving at a constant speed on a straight horizontal track. As the engine moved​ along, a marble was fir
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v = 4.9 m/s.

The marble will reach maximum height after 0.5 seconds, at which the

height will be y(0.5) = (4.9)*(0.5) - (4.9)*(0.5)^2 = (4.9)*(0.25) = 1.225 m.

Now,  the marble has a vertical velocity component of 4.9 m/s and a horizontal velocity component

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This horizontal velocity component of the marble is the same as the

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3 0
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Answer:

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A jogger runs 10.0 blocks do east, 5.0 blocks due South, and another two. Zero blocks do east. Assume all blocks are equal size,
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d3 = 2 blocks East

now we can say

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d = \sqrt{12^2 + 5^2}

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<em>so magnitude of net displacement will be equal to 13 blocks</em>

6 0
1 year ago
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