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salantis [7]
3 years ago
8

A book is dropped from a window. It takes 5 seconds to reach the ground. What is its velocity after 2 seconds? What’s the veloci

ty at the time when it hits the ground?
Physics
1 answer:
ira [324]3 years ago
3 0

Answer:

Explanation:

Initial velocity is 0. In the equation v = v0+at where v0 is the initial velocity of 0, we only have to fill in -9.8 for a and 2 for t to get the velocity after 2 seconds -19.6 m/s; after 5 seconds, when it hits the ground, a = -9.8 and t = 5 to give a velocity of -49 m/s. Gravity pulls down everything at the same rate, it doesn't matter whether we drop a feather or an elephant from the window!

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For an object to slow down and stop due to friction while sliding across the floor which of the following would be true.
EastWind [94]

Answer:10-4

Explanation:

4 0
2 years ago
Question 9 please. How do you find instantaneous velocity and how do I sketch a velocity vs time graph from a position vs time o
pickupchik [31]

How do you find instantaneous velocity

Select a point on a distance-time curve graph. Draw a tangent to the curve at that point. Tangent -> hypotenuse of right angled triangle. Opp/adjacent in graph units is vel at that point -> in distance and/or time

3 0
3 years ago
As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, afte
Klio2033 [76]

Answer:

 r^2 = \frac{ 2 k  \ Ze^2}{ 2m}

Explanation:

For this exercise we must use the principle of conservation of energy

starting point. The proton very far from the nucleus

          Em₀ = K = ½ m v²

final point. The point where the proton is stopped (v = 0)

          Em_f = U = q V

where the potential is

          V = k Ze / r²

Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching

Energy is conserved

          Em₀ = Em_f

           ½ m v² = e (k \frac{Ze}{r^2})

          r^2 = \frac{ 2 k  \ Ze^2}{ 2m}

with this expression we can find the closest approach distance (r)

3 0
3 years ago
Jupiter's satellite Europa orbits Jupiter with a period of 3.55 d at an orbital radius of 6.71 108 m (assume the orbit to be cir
yawa3891 [41]

Answer:

(a)

M = 1.898 x 10^27 kg

(b)

v = 13.74 km/s

(c) E = 0.28 N/kg

Explanation:

Time period, T = 3.55 days = 3.55 x 24 x 3600 second = 306720 s

Radius, r = 6.71 x 10^8 m

G = 6.67 x 10^-11 Nm^2/kg^2

(a) T=2\pi \sqrt{\frac{r^{3}}{GM}}

M=\frac{4\pi ^{2}r^{3}}{GT^{2}}

M=\frac{4\times3.14^{2}\times 6.71^{3}\times 10^{24}}{6.67\times 10^{-11}\times 306720^{2}}

M = 1.898 x 10^27 kg

(b) Let v be the orbital velocity

v=\frac{2\pi r}{T}

v=\frac{2\times 3.14\times 6.71\times 10^{8}}{306720}

v = 13739.5 m/s

v = 13.74 km/s

(b) The gravitational field E is given by

E = \frac{GM}{r^{2}}

E = \frac{6.67\times10^{-11}\times 1.898\times 10^{27}}{6.71^{2}\times 10^{16}}

E = 0.28 N/kg

6 0
2 years ago
When measuring accelerations with three different graphs (x-t, v-t, a-t) which gives the most reliable results and why?
kipiarov [429]

Answer:

the best graph to find the acceleration is v-t since calculating the slope averages the different experimental errors.

Explanation:

The different graphics depending on time give various information, let's examine what we can get from some

Graph of x -t. from this graph we can obtain the speed through the slope, but the acceleration is not directly obtainable

v-t chart. We can get the acceleration not through the slope and the distance traveled by the area under the curve. Obtaining acceleration is very accurate since it is an average that avoids possible errors in measurements. This is the best graph to find the acceleration

Graph of a-t In this graph the acceleration is a point on the Y axis, it gives some errors because it depends strongly on the possible experimental errors.

In conclusion, the best graph to find the acceleration is v-t since calculating the slope averages the different experimental errors.

5 0
3 years ago
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