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abruzzese [7]
3 years ago
6

You have landed on an unknown planet, Newtonia, and want to know what objects weigh there. When you push a certain tool, startin

g from rest, on a frictionless horizontal surface with a 12.0-N force, the tool moves 16.0 m in the first 2.00 s. You next observe that if you release this tool from rest at 10.0 m above the ground, it takes 2.58 s to reach the ground. What does the tool weigh on Newtonia, and what does it weigh on Earth?
Physics
1 answer:
a_sh-v [17]3 years ago
8 0

Answer:

Explanation:

Given

Force applied F=12\ N

time taken t_1=2\ s

Displacement s=16\ m

using

s=ut+\frac{1}{2}at^2

u=initial velocity

s=displacement

t=time

16=0+\frac{1}{2}a(2)^2

a=8\ m/s^2

thus mass of body m=\frac{F}{a}=\frac{12}{8}=1.5\ kg

Next it is released from a height of h=10\ m

time taken to reach ground t_2=2.58\ s

using  s=ut+\frac{1}{2}at^2 in vertical direction

here acceleration is due to acceleration due to gravity(g') of the planet

as it at rest so u=0 here

10=0+\frac{1}{2}(g')(2.58)^2

g'=3.004\ m/s^2

Thus acceleration due to gravity on Newtonian is 3\ m/s^2

Weight of tool on Newtonia W'=mg'

W'=1.5\times 3.004=4.506\ N

Weight on Earth W=1.5\times 9.8=14.7\ N

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Calculate the wavelength of light that has a frequency of 5.2 x 1012 1/s.
Travka [436]

Answer:

Wavelength = 5.77 * 10^-5 meters.

Explanation:

Given the following data:

Frequency of light = 5.2 *10^12 Hz

We know that the Speed of light = 3.0 * 10^8 m/s

To find the wavelength of light;

Mathematically, wavelength is calculated using this formula;

Wavelength = \frac {speed}{frequency}

Substituting into the equation, we have;

Wavelength = \frac {3*10^{8}}{5.2 *10^{12}}

Wavelength = 5.77 * 10^-5 meters.

6 0
3 years ago
Two identical satellites orbit the earth in stable orbits. one satellite orbits with a speed v at a distance r from the center o
wariber [46]
The available options are: (found the complete text on internet)
A- at a distance less than r
B- at a distance equal to r
C- at a distance greater than r

Solution:
The correct answer is C) at a distance greater than r.

In fact, the gravitational attraction between the satellite and the Earth provides the centripetal force that keeps the satellite in circular orbit, so we can write
G \frac{Mm}{r^2}=m \frac{v^2}{r}
where the term on the left is the gravitational force, while the term on the right is the centripetal force, and where
G is the gravitational constant
M is the Earth mass
m is the satellite mass
r is the distance of the satellite from the Earth's center
v is the satellite speed

Re-arranging the equation, we get
r= \frac{GM}{v^2}
and we see from this formula that, if the second satellite has a speed less than the speed v of the first satellite, it means that the denominator of the fraction is smaller, and so r is larger for the second satellite.
7 0
3 years ago
A student walks to the right 25-m along the 800 hall in 15-s. They turn around and walk 15-m to the left in 8.0-s. Calculate the
Ludmilka [50]

Answer:

Explanation:

Total distance covered  = 25 + 15 = 40 m

Total time = 15 + 8 = 23 s

Average speed = total distance covered / total time

= 40 / 23

= 1.74 m / s

Total displacement = 25 - 15 = 10 m

Total time = 15 + 8 = 23 s

Average velocity = total displacement / total time

= 10 / 23

= .434 m / s to the right .

3 0
3 years ago
Select the correct answer.
chubhunter [2.5K]
I believe it’s self-referent encoding
3 0
3 years ago
A swan on a lake gets airborne by flapping its wings and running on top of the water. If the swan must reach a velocity of 6.40
Veseljchak [2.6K]

Answer:

53.895 m.

Explanation:

Using the equation of motion,

v² = u² + 2as .............. Equation 1

Where v = final velocity of the swan, u = initial velocity of the swan, a = acceleration of the swan, s = distance covered by the swan.

make s the subject of the equation,

s = (v² - u²)/2a----------- Equation 2

Given: v = 6.4 m/s, u = 0 m/s ( from rest)  a = 0.380 m/s².

Substitute into equation 2

s = (6.4²-0²)/(2×0.380)

s = 40.96/0.76

s = 53.895 m.

Hence the swan will travel 53.895 m before becoming airborne.

6 0
4 years ago
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