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3 years ago

A dragster is traveling at a constant rate of 45 m/s. How fast is it going when it has traveled 325 m?

Physics

1 answer:

Sergeeva-Olga [200]3 years ago

6 0

The speed of the dragster after travelling 325 m is still 45 m/s

Explanation:

In this problem, we are told that the dragster is travelling at constant rate, therefore its speed is constant and it is

v = 45 m/s

Therefore, its speed will remain constant during the entire motion: therefore, after having travelled for 325 m, its speed will still be 45 m/s.

We can also calculate the time taken for the dragster to cover the 325 m. In fact, the speed in a uniform motion is given by

$v=\frac{d}{t}$

where:

v is the speed

d is the distance covered

t is the time elapsed

In this case, we have:

v = 45 m/s

d = 325 m

Solving for t, we find:

$t=\frac{d}{v}=\frac{325}{45}=7.2 s$

Learn more about speed:

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Which parts of Dalton’s atomic theory had to be revised

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3 years ago

Astronomers suspect that a galaxy’s type can be affected both by the conditions in the protogalactic cloud from which it forms (

Katen [24]

Answer:

The items here are describing either a condition in a later interacton or a protogalactic cloud. The results matching with spiral and elliptical galaxy are:

For spiral galaxy are options 6,3,2 and 5.

and for elliptical galaxy are options 4 and 1.

Explanation:

Here it is given that astrnomers suspect that types of galaxy can be affected both by the conditions which occurs due to protogalactic cloud and then from it forms the initial conditions and then by the later interactions with the other galaxies.

so, both types of galaxies are matched with their respective items given:

A. Spiral galaxy:

2. A galaxy collision results tostripping of gas.

3. The protogalactic cloud rotates in a very slow motion.

5. The density of protogalactic cloud is very high.

6. when the protogalactic cloud shrinks cloud forms very rapidly.

B. Elliptical galaxy:

1. The protogalactic cloud has high angular momentum.

4. Most of the protogalactic gases settles down into a disk.

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3 years ago

A felt-covered beanbag is fired into a empty wooden crate that sits on a concrete floor and is open on one side. The beanba

Phantasy [73]

Answer:

31.42383 m/s

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

$\mu$ = Coefficient of kinetic friction = 0.48

s = Displacement = 0.935 m

$m_1$ = Mass of bean bag = 0.354 kg

$m_2$ = Mass of empty crate = 3.77 kg

$v_1$ = Speed of the bean bag

$v_2$ = Speed of the crate

Acceleration

$a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g$

$a=--9.81\times 0.48=4.7088\ m/s^2$

From equation of motion

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.7088\times 0.935+0^2}\\\Rightarrow v=2.96739\ m/s$

In this system the momentum is conserved

$m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_1=\frac{(m_1+m_2)v_2}{m_1}\\\Rightarrow u=\frac{(0.354+3.77)\times 2.69739}{0.354}\\\Rightarrow u=31.42383\ m/s$

The speed of the bean bag is 31.42383 m/s

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3 years ago

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

If atoms [Blank] energy during a change of state, they are pulled together by attractive forces and become more organized.

Rainbow [258]

If atoms lose , they become more organized.
When energy is removed from a given amount of atoms, they enter a more organized state because their attractive forces overcome their repulsive forces, which causes them to come closer. This is observed when a gas is condensed to a liquid, and then also when a liquid freezes into a solid. This moving closer also limits the motion of the particles.

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3 years ago