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lys-0071 [83]
3 years ago
6

Most of the bright stars in our galaxy are located in the galactic

Physics
1 answer:
GenaCL600 [577]3 years ago
3 0

False. Most of the bright stars in our galaxy are located in the disk.

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The more twists per foot in a pair of wires, the more resistant the pair will be to ____.
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The more twist per foot in a pair of wires, the more resistant the pair will be to cross talk. A cross talk in network planning and design is a disturbance produced by electromagnetic interference beside a circuit or a cable pair. A telecommunication signal interrupts a signal in an adjacent circuit and can source the signals to turn out to be confused and cross over each other.  
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A pendulum is observed to complete 20 full cycles in 60 seconds. Determine the period and the frequency of the pendulum.
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i hope i have been useful buddy.

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3 years ago
Read 2 more answers
IM A plastic sphere floats in water with 50.0% of its volume submerged. This same sphere floats in glycerin with 40.0 \% of its
arsen [322]

Density of glycerin is 5/4 times water density.

Density of the sphere is 1/2 times water density.

<h3>How to find the density of sphere floats and glycerin?</h3>

Density is defined as the ratio of mass to volume, that is density is the amount of mass per unit of volume.

A substance's density exist defined as its mass per unit of volume. Density is most frequently represented by the symbol, however Latin letter D may also be utilized. If the mass or volume of an object is altered, the density of the thing may also change. Water, for example, has a specific density. An object will sink if its density exists greater than that of water; it will float if its density is lower.

Since a plastic sphere floats in water with 50% of its volume submerged.

So, the Density of the sphere = 1/2× water density

Since the same sphere floats in glycerin with 40% of its volume submerged.

So, the Density of the sphere = 4/100 × Glycerin density

= 2/5 × Glycerin density

Density of Glycerin = 5/2 of sphere density

Density of glycerin = 5/2 × 1/2 × water density

Density of glycerin = 5/4 × water density

To learn more about the volume of sphere floats refer to:

brainly.com/question/10827296

#SPJ4

6 0
1 year ago
a ball kicked with a velocity of 8m/s at an angle of 30 degree to horizontal. calculate the time of flight of the ball. (g=10ms^
posledela

Answer:

Approximately 0.8\; \rm s (assuming that air resistance is negligible.)

Explanation:

Let v_0 denote the initial velocity of this ball. Let \theta denote the angle of elevation of that velocity.

The initial velocity of this ball could be decomposed into two parts:

  • Initial vertical velocity: v_0(\text{vertical}) = v_0 \cdot \sin(\theta).
  • Initial horizontal velocity: v_0(\text{vertical}) = v_0 \cdot \cos(\theta).

If air resistance on this ball is negligible, v_0(\text{vertical}) alone would be sufficient for finding the time of flight of this ball.

Calculate v_0(\text{vertical}) given that v_0 = 8 \; \rm m \cdot s^{-1} and \theta = 30^\circ:

\begin{aligned}& v_0(\text{vertical}) \\ &= v_0 \cdot \sin(\theta) \\ &= \left(8 \; \rm m \cdot s^{-1} \right) \cdot \sin\left(30^{\circ}\right) \\ &= 4\;\rm m \cdot s^{-1} \end{aligned}.

Assume that air resistance on this ball is zero. Right before the ball hits the ground, the vertical velocity of this ball would be exactly the opposite of the value when the ball was launched.

Since v_0(\text{vertical}) = 4\; \rm m \cdot s^{-1}, the vertical velocity of this ball right before landing would be v_1(\text{vertical}) = -4\; \rm m \cdot s^{-1}.

Calculate the change to the vertical velocity of this ball:

\begin{aligned}& \Delta v(\text{vertical}) \\ & = v_1(\text{vertical}) - v_0(\text{vertical}) \\ &= -8\; \rm m \cdot s^{-1}\end{aligned}.

In other words, the vertical velocity of this ball should have change by 8\; \rm m \cdot s^{-1} during the entire flight (from the launch to the landing.)

The question states that the gravitational field strength on this ball is g = 10\; \rm m \cdot s^{-2}. In other words, the (vertical) downward gravitational pull on this ball could change the vertical velocity of the ball by 10\; \rm m\cdot s^{-1} each second. What fraction of a second would it take to change the vertical velocity of this ball by 8\; \rm m \cdot s^{-1}?

\begin{aligned}t &= \frac{\Delta v(\text{initial})}{g} \\ &= \frac{8\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-2}} = 0.8\; \rm s\end{aligned}.

In other words, it would take 0.8\; \rm s to change the velocity of this ball from the initial velocity at launch to the final velocity at landing. Therefore, the time of the flight of this ball would be 0.8\; \rm s\!.

5 0
3 years ago
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