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3 years ago

How is acceleration impacted when mass remains the same but the amount of force is decreased?

1 answer:

3 0

The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

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A bullet of mass m 1 is fired into a rod of length L and mass m 2 which is pivoted on one end and rests on a frictionless horizo

Reptile [31]

Answer:

The angular speed of the rod is $w=(\frac{9m_{1}v_{i} }{4m_{2}L } )(1-\frac{1}{\sqrt{2} } )$

Explanation:

The final kinetic energy is:

$\frac{1}{2} mv_{f}^{2} =\frac{1}{2}(\frac{1}{2} mv_{i}^{2} )$

Clearing vf:

$v_{f} =\frac{1}{\sqrt{2} } v_{i}$

The conservation of angular momentum before and after collision is:

$m_{1} v_{i} (\frac{3L}{4} )=Iw+m_{1}v_{f} (\frac{3L}{4} )$

Clearing w:

$w=(\frac{9m_{1}v_{i} }{4m_{2}L } )(1-\frac{1}{\sqrt{2} } )$

3 0

4 years ago

Eye at the lowest radiated power of 1,2 x10 x (- 17) W. Determine how many photons of light with a wavelength of 500nm fall on t

mars1129 [50]

Answer:

$\frac{n}{t} = 30\ photons/s$

Explanation:

The radiated power can be given in terms of the wavelength as follows:

$Rasiated\ Power = \frac{nE}{t} = \frac{nhc}{\lambda t}$

where,

Radiated Power = 1.2 x 10⁻¹⁷ W

n = no. of photons = ?

h = plank's constant = 6.625 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength = 500 nm = 5 x 10⁻⁷ m

t = time

Therefore,

$1.2\ x\ 10^{-17}\ W = \frac{n(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{(5\ x\ 10^{-7}\ m)(t) }\\\\\frac{n}{t} = \frac{1.2\ x\ 10^{-17}\ W}{3.975\ x\ 10^{-19}\ J}\\\\\frac{n}{t} = 30\ photons/s$

8 0

3 years ago

A 12,500 N alien UFO is hovering about the surface of Earth. At time , its position can be given as () = ((0.24 m/s^3)^3 + 25 m)

stepladder [879]

a) $F=(3675i-4543k)N$

b) 5843 N

Explanation:

The position of the UFO at time t is given by the vector:

$r(t)=(0.24t^3+25)i+(4.2t)j+(-0.43t^3+0.8t^2)k$

Therefore it has 3 components:

$r_x=0.24t^3+25\\r_y=4.2t\\r_z=-0.43t^3+0.8t^2$

We start by finding the velocity of the UFO, which is given by the derivative of the position:

$v_x=r'_x=\frac{d}{dt}(0.24t^3+25)=3\cdot 0.24t^2=0.72t^2\\v_y=r'_y=\frac{d}{dt}(4.2t)=4.2\\v_x=r'_z=\frac{d}{dt}(-0.43t^3+0.8t^2)=-1.29t^2+1.6t$

And then, by differentiating again, we find the acceleration:

$a_x=v'_x=\frac{d}{dt}(0.72t^2)=1.44t\\a_y=v'_y=\frac{d}{dt}(4.2)=0\\a_z=v'_z=\frac{d}{dt}(-1.29t^2+1.6t)=-2.58t+1.6$

The weight of the UFO is W = 12,500 N, so its mass is:

$m=\frac{W}{g}=\frac{12500}{9.8}=1276 kg$

Therefore, the components of the force on the UFO are given by Newton's second law:

$F=ma$

So, Substituting t = 2 s, we find:

$F_x=ma_x=(1276)(1.44t)=(1276)(1.44)(2)=3675 N\\F_y=ma_y=0\\F_z=ma_z=(1276)(-2.58t+1.6)=(1276)(-2.58(2)+1.6)=-4543 N$

So the net force on the UFO at t = 2 s is

$F=(3675i-4543k)N$

The magnitude of a 3-dimensional vector is given by

$|v|=\sqrt{v_x^2+v_y^2+v_z^2}$

where

$v_x,v_y,v_z$ are the three components of the vector

In this problem, the three components of the net force are:

$F_x=3675 N\\F_y=0\\F_z=-4543 N$

Therefore, substituting into the equation, we find the magnitude of the net force:

$|F|=\sqrt{3675^2+0^2+(-4543)^2}=5843 N$

7 0

3 years ago

Heat in the atmosphere always moves which way?

jeka94

Answer:

Explanation:

heat rises. That powers convection which is in the link

6 0

3 years ago

Read 2 more answers

We showed that the length of the pendulum of period 2.000 seconds on the Earth’s surface was 0.99396 meters. What period would t

Kitty [74]

To solve this problem it is necessary to apply the concepts related to the Period based on gravity and length.

Mathematically this concept can be expressed as

$T= 2\pi \sqrt{\frac{l}{g}}$

Where,

l = Length

g = Gravitational acceleration

First we will find the period that with the characteristics presented can be given on Mars and then we can find the length of the pendulum at the desired time.

The period on Mars with the given length of 0.99396m and the gravity of the moon (approximately $1.62m / s ^ 2)$ will be