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2 years ago

if we ignore air resistance the mass of an object does not affect the rate at which it accelerate why?

Physics

1 answer:

quester [9]2 years ago

5 0

Answer:

See explanation

Explanation:

The acceleration due to gravity on an object is independent of the mass of the object. This is so because, the acceleration due to gravity depends only on the radius of the earth and the mass of the earth.

As a result of this, all objects are accelerated to the same extent and should reach the ground at the same time when released from a height as long as other forces other than gravity are not at work.

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What kind of weather would an occuluded front likely bring

abruzzese [7]

An occluded front forms when a cold front catches up with a warm front. So it would bring Sun and warmth.

3 0

2 years ago

A mover loads a crate onto a truck bed 1.6m from the street using a ramp that is 4.6m long. What is a mechanical advantage?

Ne4ueva [31]

Answer:

Mechanical advantage = 2.875

Explanation:

Given:

A diagram is shown below for the above scenario.

Length of ramp (Effort arm) = 4.6 m

Height of truck bed ( Resistance length) = 1.6 m

Mechanical advantage (MA) is the ratio of effort arm and resistance length.

So, mechanical advantage is given as,

$MA=\frac{\textrm{Effort arm}}{\textrm{Resistance length}}= \frac{4.6}{1.6}=2.875$

6 0

2 years ago

A system of pulleys is used to raise a load of bricks that weighs 1,700 newtons. The force applied to the pulley is 340 newtons.

Law Incorporation [45]

By definition, we have that the mechanical advantage is given by the following equation:
$MA = \frac{W}{T} 
$
Where,
W: is the load
T: is the tension
Substituting the values in the given equation we have:
$MA = \frac{1700}{340}$
$MA = 5
$
Therefore, the mechanical advantage is equal to 5.
Answer:
The mechanical advantage of this machine is:
MA = 5

4 0

3 years ago

Read 2 more answers

Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

2 years ago

A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s

gulaghasi [49]

(a) 0.96 m/s

The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:

$T=2\cdot 2.5 s=5.0 s$

The frequency of the waves is the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{5.0 s}=0.20 Hz$

The wavelength instead is just the distance between two consecutive crests, so

$\lambda=4.8 m$

And the wave speed is given by:

$v=\lambda f=(4.8 m)(0.20 Hz)=0.96 m/s$

(b) 0.265 m

The total distance between the highest point of the wave and its lowest point is

d = 0.53 m

The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is

$A=\frac{d}{2}=\frac{0.53 m}{2}=0.265 m$

(c) Amplitude: 0.15 m, wave speed: same as before

In this case, the amplitude of the wave would be lower. In fact,

d = 0.30 m

So the amplitude would be

$A=\frac{d}{2}=\frac{0.30 m}{2}=0.15 m$

Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.

8 0

3 years ago

if we ignore air resistance the mass of an object does not affect the rate at which it accelerate why?​

if we ignore air resistance the mass of an object does not affect the rate at which it accelerate why?