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3 years ago

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An astronaut drops a rock on the surface of an asteroid.The rock is released from rest at a height of 0.86 m above the ground, a

nd hits the ground 1.37 s later. Part A) What is the acceleration due to gravity on this asteroid?

1 answer:

SCORPION-xisa [38]3 years ago

5 0

Answer:

$a_y=0.92m/s^2$

Explanation:

To solve this problem we use the formula for accelerated motion:

$y=y_0+v_{y0}t+\frac{a_yt^2}{2}$

We will take the initial position as our reference ( $y_0=0m$ ) and the downward direction as positive. Since the rock departs from rest we have:

$y=\frac{a_yt^2}{2}$

Which means our acceleration would be:

$a_y=\frac{2y}{t^2}$

Using our values:

$a_y=\frac{2(0.86m)}{(1.37s)^2}=0.92m/s^2$

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Two magnets are arranged so that they pull together. Which conclusion can be reached? A.Two north poles are together. B. Two sou

Virty [35]

Answer:

Answer is D. a north pole is near a south pole

Explanation:

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6 0

2 years ago

Read 2 more answers

A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp

NISA [10]

A) 750 m

First of all, let's find the wavelength of the microwave. We have

$f=12GHz=12\cdot 10^9 Hz$ is the frequency

$c=3.0\cdot 10^8 m/s$ is the speed of light

So the wavelength of the beam is

$\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m$

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

$y=\frac{m\lambda D}{a}$

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

$y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m$

and so, the diameter is

$d=2y = 750 m$

B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

$r=\frac{750 m}{2}=375 m$

So the area is

$A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2$

And since the power is

$P=100 kW = 1\cdot 10^5 W$

The average intensity is

$I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2$

4 0

3 years ago

A golf ball is struck with a five iron on level ground. it lands 100.0 m away 4.60 s later. what was the magnitude and direction

finlep [7]

consider the motion in x-direction

$v_{ox}$ = initial velocity in x-direction = ?

X = horizontal distance traveled = 100 m

$a_{x}$ = acceleration along x-direction = 0 m/s²

t = time of travel = 4.60 sec

Using the equation

X = $v_{ox}$ t + (0.5) $a_{x}$ t²

100 = $v_{ox}$ (4.60)

$v_{ox}$ = 21.7 m/s

consider the motion along y-direction

$v_{oy}$ = initial velocity in y-direction = ?

Y = vertical displacement = 0 m

$a_{y}$ = acceleration along x-direction = - 9.8 m/s²

t = time of travel = 4.60 sec

Using the equation

Y = $v_{oy}$ t + (0.5) $a_{y}$ t²

0 = $v_{oy}$ (4.60) + (0.5) (- 9.8) (4.60)²

$v_{oy}$ = 22.54 m/s

initial velocity is given as

$v_{o}$ = sqrt(( $v_{ox}$ )² + ( $v_{oy}$ )²)

$v_{o}$ = sqrt((21.7)² + (22.54)²) = 31.3 m/s

direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg

6 0

3 years ago

sest yourse 1. A pencil lies on the dashboard of a car. a) i) What happens to the pencil when the car suddenly stops? suddenly a

julia-pushkina [17]

Answer:

1. the pencil would have the momentum and would keep going until it hits the windshield. 2. when the car suddenly accelerates, the pencil would be inert and it would move toward the back of the car until a constant speed from the car is reached.

8 0

2 years ago

The newly formed xenon nucleus is left in an excited state. Thus, when it decays to a state of lower energy a gamma ray is emitt

nevsk [136]

Answer:3.87*10^-4

Explanation:

What is the decrease in mass, delta mass Xe , of the xenon nucleus as a result of this deca

We have been given the wavelength of the gamma ray, find the frequency using c = freq*wavelength.

C=f*lambda

3*10^8=f*3.44*10^-12

F=0.87*10^20 hz

Then with the frequency, find the energy emitted using equation

E=hf E = freq*Plank's constant

E=.87*10^20*6.62*10^-34

E=575.94*10^(-16)

With this energy, convert into MeV from joules.

With the energy in MeV, use E=mc^2 using c^2 = 931.5 MeV/u.

Plugging and computing all necessary numbers gives you

3.87*10^-4 u.

6 0

3 years ago