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4 years ago

How does mass affect the distance a car can travel?

1 answer:

7 0

Well yes it does the heavier the object being launched it wont go as far as the lighter one.
Gravity is pulling it down and the heavier the mass the farther it will travel down the ramp. The heavier it is the more gravity has to pull down

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Cart 1 of mass m is traveling with speed 2vo in the +x-direction when it has an elastic collision with cart 2 of

iogann1982 [59]

Answer:

Explanation:

Momentum conservation

$m2v_0+2mv_0=mv_1+2mv_2 \quad (1/m) \quad 4v_0=v_1+2v_2\\$

Kinetic energy conservation

$\displaystyle \frac{1}{2}m(2v_0)^2+\frac{1}{2}2mv_0^2=\frac{1}{2}mv_1^2+\frac{1}{2}2mv_2^2 \quad (1/m) \quad 6v_0^2=v_1^2+2v_2^2$

Solve the system

6 0

4 years ago

A moss begins to grow on a rock in an area that was previously uninhabited by life. This is an example of a .

Kitty [74]

Answer: primary ecological succession

Explanation:

Primary ecological succession: This occurs when a certain or some organisms dominate an area that was uninhabited of life. This may happen after a natural disaster that leaves the land barren. The presence of these organisms on the barren land will help the soil occupied to gain nutrients which will eventually allow some other plants to grow on it. The growth of plants will later invite insects and insects may invite animals.

8 0

3 years ago

Read 2 more answers

ich statement best describes the fundamental difference between spiral and elliptical galaxies? Spiral galaxies are more massive

Ahat [919]

Answer:

The oldest stars in spiral galaxies are younger than the oldest stars in elliptical galaxies.

Explanation:

Spiral galaxies have a greater star formation than elliptical galaxies, since the latter contain less gas and dust. Therefore, the rate of creation of new stars is lower in elliptical galaxies. These galaxies are dominated by old stars, with a long evolution. While the spiral galaxies have stars with a faster evolution.

8 0

3 years ago

Read 2 more answers

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

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