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3 years ago

1.Round off the following measurement to the three significant figures. 1.296 g.

Physics

1 answer:

Kobotan [32]3 years ago

5 0

Answer:

<h2>1) 1.30g</h2><h2>2) 3 sf</h2><h2>3) 5.66 grams</h2>

Explanation:

1) Significant figure means the amount of figure we are interested in. Note that 0 is insignificant. For example 30 is a one significant figure i. e the only figure of interest is 3.

Given the figure 1.296 g, as we can see, there are initially for figures in the value. This means we are to eliminate one to make it 3. To do that we will remove the 6 at the back by round it up to 1 and add it to the value of 29 preceding it to give 30. The figure will become 1.30 (to 3sf).

<em>Note that only values greater than 4 are rounded up to 1 and added to the preceding value</em>

2) Given the figure 2020, the number of significant figures present is 3 significant figure. Note that all zeros at the end of a value are considered insignificant hence the only significant value here is 202 (which is 3 sig fig.)

3) For the figure 5.658, the value to 3sf is 5.66 (Note that the value of 8 at the end is rounded up to 1 and added to 65 to make it 66). This is as explained in the first answer above.

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A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in the linear speed

$\Delta t$ is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

$v=8322 m/s$

Therefore, its linear speed is not changing, so the change in linear speed is zero:

$\Delta v=0$

And therefore, the tangential acceleration is zero as well:

$a_t=\frac{0}{\Delta t}=0 m/s^2$

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3 years ago

A car travels 36 km in 30 minutes. The speed of the car is

iogann1982 [59]

Answer:

72km/hr

Explanation:

Speed in Km is usually represented in hours. so if the car is in constant velocity, and if the car travels 36km in 30 min then it travels 72km in 1 hour.

so the speed of the car is 72km/hr

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3 years ago

Which statement expresses the relationship between the frequency of emitted electrons and the energy they give off?

Nadusha1986 [10]

I think it would be the first choice

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4 years ago

Read 2 more answers

How should the magnetic field lines be drawn for the magnets shown below?

sergeinik [125]

Option B is the correct answer that show how magnetic field lines should be drawn for the magnets shown in the figure.

<h3>What is Magnetic Line of Force ?</h3>

The Magnetic Line of Force of a magnet is defined as the line along which a free N - pole would tend to move if placed in the field of a line such that the tangent to it at any point gives the direction of the field at that point.

When the two unlike poles are placed to each other, there will be attraction. And when the two like poles are placed to each other, there will be repulsion. The reason is that the line of force tend to move from the north pole to the south pole.

From the given diagram, the two magnets are of the same south pole. They are of like pole and there will be repulsion between the two magnets.

Therefore, Option B is the correct answer that show how magnetic field lines should be drawn for the magnets shown in the figure.

Learn more about Magnetic Field Lines here: brainly.com/question/17011493

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2 years ago

Curtis, a student in our class, makes the following statement: The puck reached a slightly higher location on the ramp than I pr

Sindrei [870]

Answer and Explanation: No, the explanation is not plausible. The puck sliding on the ice is an example of the <u>Principle</u> <u>of</u> <u>Conservation</u> <u>of</u> <u>Energy</u>, which can be enunciated as "total energy of a system is constant. It can be changed or transferred but the total is always the same".

When a player hit the pluck, it starts to move, gaining kinetic energy (K). As it goes up a ramp, kinetic energy decreases and potential energy (P) increases until it reaches its maximum. When potential energy is maximum, kinetic energy is zero and vice-versa.

So, at the beginning of the movement the puck only has kinetic energy. At the end, it gains potential energy until its maximum.

The representation is as followed:

$K_{i}+P_{i}=K_{f}+P_{f}$

$K_{i}+0=0+P_{f}$

$\frac{1}{2}mv^{2} = mgh$

As we noticed, mass of the object can be cancelled from the equation, making height be:

$h=\frac{v^{2}}{2g}$

So, the height the puck reaches depends on velocity and acceleration due to gravity, not mass of the puck.

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3 years ago