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2 years ago

1. One of the timing devices Galileo used was his pulse. Drop a rubber ball from a height of

1 answer:

7 0

The number of pulse beats elapsed before the rubber ball hits the ground can be obtained when you carry out the experiment yourself. However, the pulse beat method of timing used by Galileo is not reliable because it varies from time to time.

Galileo was interested in studying how objects fall. His discovery was that all objects had the same acceleration irrespective of their mass. This observation was in direct contrast to Aristotle's assertion that the velocity of objects is proportional to their mass.

However, he used his pulse beats as timer during the experiment. This method is unreliable because the pulse beats of a person changes depending on the person's state of mind. A stop clock could have been a more reliable timer than pulse beats.

Learn more: brainly.com/question/7201885

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A plane traveled west for 4.0 hours and covered a distance of 4,400 kilometers. What was its velocity? 18,000 km/hr 1,800 km/hr,

Airida [17]

west 1100 km / hr ..

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2 years ago

Read 2 more answers

A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.

fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

$F=\dfrac{mv^2}{r}-mg$

$F=m(\dfrac{v^2}{r}-g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)$

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

$F=\dfrac{mv^2}{r}+mg$

$F=m(\dfrac{v^2}{r}+g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)$

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

$T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}$

$T=\dfrac{m}{r}(g(r-h)+v^2)$

Since, $v^2=u^2-2gh$

$T=\dfrac{m}{r}(u^2-3gh+gr)$

Hence, this is the required solution.

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2 years ago

Determine the specific volume of refrigerant-134a at 1 MPa and 50°C, using (a) the ideal-gas equation of state and (b) the gener

Andrej [43]

Answer:

( a ) The specific volume by ideal gas equation = 0.02632 $\frac{m^{3} }{kg}$

% Error = 20.75 %

(b) The value of specific volume From the generalized compressibility chart = 0.0142 $\frac{m^{3} }{kg}$

% Error = - 34.85 %

Explanation:

Pressure = 1 M pa

Temperature = 50 °c = 323 K

Gas constant ( R ) for refrigerant = 81.49 $\frac{J}{kg k}$

(a). From ideal gas equation P V = m R T ---------- (1)

⇒ $\frac{V}{m}$ = $\frac{R T}{P}$

⇒ Here $\frac{V}{m}$ = Specific volume = v

⇒ v = $\frac{R T}{P}$

Put all the values in the above formula we get

⇒ v = $\frac{323}{10^{6} }$ ×81.49

⇒ v = 0.02632 $\frac{m^{3} }{kg}$

This is the specific volume by ideal gas equation.

Actual value = 0.021796 $\frac{m^{3} }{kg}$

Error = 0.02632 - 0.021796 = 0.004524 $\frac{m^{3} }{kg}$

% Error = $\frac{0.004524}{0.021796}$ × 100

% Error = 20.75 %

(b). From the generalized compressibility chart the value of specific volume

$\frac{V}{m}$ = v = 0.0142 $\frac{m^{3} }{kg}$

The actual value = 0.021796 $\frac{m^{3} }{kg}$

Error = 0.0142 - 0.021796 = $\frac{m^{3} }{kg}$

% Error = $\frac{- 0.0076}{0.021796}$ × 100

% Error = - 34.85 %

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2 years ago

Lizzie is pushing Alex on a scooter. Lizzie is pushing with 75 N of force to the left, and Alex is helping with 20 N to the left

nordsb [41]

Answer:

The net force on the scooter is 95 N to the left.

Explanation:

Lizzie is pushing with 75 N of force to the left, so the force is a vector with 75N of magnitude and to the left.

Alex is helping with 20 N to the left, so the force is a vector with 20 N magnitude and to the left.

When we add vectors, vectors that point in the same direction add up and vectors that point in opposite directions are subtracted.

Hence, the net force is equal to:

F=75 N + 20 N=95 N

As both forces point to the left, the net force is also to the left.

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2 years ago

A cart is pushed to the right with a force of 15 N while being pulled to the left with a force of 20 N. The net force on the car

9966 [12]

The net force of the cart when it is pushed to the right with a force of 15N.

<u>Explanation:</u>

To find the force of net, which is calculated by the formula.

The Net Force= Addition of the force applied on the respective direction.

The Net Force here is given by

The Net Force = 15-20 (A force towards the right and a force towards left, two opposite so subtraction).

Hence

Thus the Net Force = -5(The force towards left, so it gets a negative value).

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2 years ago