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3 years ago

When determining the number of significant digits in a measurement,

Physics

2 answers:

Black_prince [1.1K]3 years ago

6 0

B) All nonzero digits are significant.

myrzilka [38]3 years ago

3 0

Answer:

all nonzero digits are significant.

Explanation:

Significant digits are total number of digits that are used to define the final results.

So here we have to define as per rules

1) All non zero numbers are significant

2) All zeros between two non zero numbers are significant

3) All trailing zeros in a number without decimal are not significant

4) All zeros in a number which is greater than 1 with decimal are significant

So as per above rules correct answer will be

all nonzero digits are significant.

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Water drips from the nozzle of a shower onto the floor 189 cm below. The drops fall at regular (equal) intervals of time, the fi

laiz [17]

Answer:

0.83999 m

0.20999 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s² = a

s = 189 cm

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1.89=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.89\times 2}{9.81}}\\\Rightarrow t=0.62074\ s$

When the time intervals are equal, if four drops are falling then we have 3 time intervals.

So, the time interval is

$t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.62074}{3}\\\Rightarrow t'=0.206913\ s$

For second drop time is given by

$t''=2t'\\\Rightarrow t''=2\times 0.2069133\\\Rightarrow t''=0.4138266\ s$

Distance from second drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.4138266^2\\\Rightarrow s=0.839993\ m$

Distance from second drop is 0.83999 m

Distance from third drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.206913^2\\\Rightarrow s=0.20999\ m$

Distance from third drop is 0.20999 m

6 0

4 years ago

An athlete is practicing shotput using a 16-pound shot, and he throws it 40 feet. He then uses a 12-pound shot and throws it 50

Andrews [41]

Answer:

Lowering of mass of the shot has high kinetic energy which leads to covering more distance of the shot.

There is a difference in these throws because the mass of the shot decreases which increases its velocity and as a result the shot covers more distance. First, the athlete throws the shot having 16-pound weight or mass which covers 40 feet distance.

Whereas in the second throw, he uses a shot having 12-pound it covers 50 feet distance because the mass

is lower in the second throw and the athlete can easily throw the shot which covers more distance so the second shot has more kinetic energy as well as more velocity which compels it to cover more distance so we can conclude that lowering of mass leads to covering more distance of the shot.

7 0

3 years ago

Two skaters, both of mass 75 kg, are on skates on a frictionless ice pond. One skater throws a 0.3-kg ball at 5 m/s to his frien

iren [92.7K]

To solve this problem it is necessary to apply the concepts related to the momentum. The moment can be basically defined as the product of the mass and velocity of an object. Mathematically it is expressed as

$P = mv$

Where,

m = mass

V = Velocity

From the statement it is noted that there is a conservation of Momentum but given in different directions, to which we could affirm that there is the momentum of throw and momentum of return

$P_T = P_t+P_r$