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3 years ago

10

An object is thrown upwards with a velocity of 3 meters per second, such that the only force acting on it is gravity (accelerati

on is -9.8). What will the object's velocity be after 2 seconds?
A.) -16.6 meters per second (down)
B.) 16.6 meters per second (up)
C.) -22.6 meters per second (down)
D.) 22.6 meters per second (up)
E.) 3 meters per second (up)

2 answers:

3241004551 [841]3 years ago

3 0

Answer:

Correct answer: A.) V = - 16.6 m/s down

Explanation:

Given:

V₀ = 3 m/s initial velocity

t = 2 seconds

g = 9.8 m/s²

V(t) = V(2) = ?

The movement described is a vertical upward shot

For velocity at any time is valid the next formula

V = V₀ - g · t

V = 3 - (9.8 · 2) = 3 - 19.6 = - 16.6 m/s down

Under condition that it has a enough drop height with respect to the ejection point.

God is with you!!!

NNADVOKAT [17]3 years ago

3 0

The answer is A. DOWN

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Richard is driving home to visit his parents. 150 mi of the trip are on the interstate highway where the speed limit is 65 mph .

Serggg [28]

Answer:

t = 25.5 min

Explanation:

To know how many minutes does Richard save, you first calculate the time that Richard takes with both velocities v1 = 65mph and v2 = 80mph.

$t_1=\frac{x}{v_1}=\frac{150mi}{65mph}=2.30h\\\\t_2=\frac{x}{v_2}=\frac{150mi}{80mph}=1.875h$

Next, you calculate the difference between both times t1 and t2:

$\Delta t=t_1-t_2=2.30h-1.875h=0.425h$

This is the time that Richard saves when he drives with a speed of 80mph. Finally, you convert the result to minutes:

$0.425h*\frac{60min}{1h}=25.5min=25\ min\ \ 30 s$

hence, Richard saves 25.5 min (25 min and 30 s) when he drives with a speed of 80mph

3 0

2 years ago

Anti-reflective coatings on lenses use thin-film interference to eliminate the reflection of a particular color. Suppose a glass

Sidana [21]

Answer with Explanation:

We are given that

$n_g=1.55$

$n_f=1.35$

$\lambda=521 nm=521\times 10^{-9} m$

$1 nm=10^{-9} m$

a.We have to find the expression for the minimum thickness the film can have ,t.

Condition for destructive interference

$2nt=(m+\frac{1}{2})\lambda$

$t=\frac{(m+\frac{1}{2})\lambda}{2n}$

For minimum thickness m=0

Then, $t=\frac{\lambda}{4n}$

b.Substitute the values

$t=\frac{521\times 10^{-9}}{4\times 1.35}=96.5nm$

7 0

2 years ago

You are observing a spacecraft moving in a circular orbit of radius 100,000 km around a distant planet. You happen to be located

Natalija [7]

To solve this problem we will apply the concepts related to centripetal acceleration, which will be the same - by balance - to the force of gravity on the body. To find this acceleration we must first find the orbital velocity through the Doppler formulas for the given periodic signals. In this way:

$v_{o} = c (\frac{\lambda_{max}-\bar{\lambda}}{\bar{\lambda}}})$

Here,

$v_{o} =$ Orbital Velocity

$\lambda_{max} =$ Maximal Wavelength

$\bar{\lambda}} =$ Average Wavelength

c = Speed of light

Replacing with our values we have that,

$v_{o} = (3*10^5) (\frac{3.00036-3}{3})$

<em>Note that the average signal is 3.000000m</em>

$v_o = 36 km/s$

Now using the definition about centripetal acceleration we have,

$a_c = \frac{v^2}{r}$

Here,

v = Orbit Velocity

r = Radius of Orbit

Replacing with our values,

$a = \frac{(36km/s)^2}{100000km}$

$a= 0.01296km/s^2$

$a = 12.96m/s^2$

Applying Newton's equation for acceleration due to gravity,

$a =\frac{GM}{r^2}$

Here,

G = Universal gravitational constant

M = Mass of the planet

r = Orbit

The acceleration due to gravity is the same as the previous centripetal acceleration by equilibrium, then rearranging to find the mass we have,

$M = \frac{ar^2}{G}$

$M = \frac{(12.96)(100000000)^2}{ 6.67*10^{-11}}$

$M = 1.943028*10^{27}kg$

Therefore the mass of the planet is $1.943028*10^{27}kg$

7 0

2 years ago

An adult elephant has a mass of about 5.0 tons. An adult elephant shrew has a mass of about 50 grams. How far r from the center

Diano4ka-milaya [45]

Answer:

2023857702.507m

Explanation:

$f=\frac{GMm}{r^{2} }$

recall from newton's law of gravitation

G=gravitational constant

mshew=50g

melephant=5*10^3kg

rearth=radius of the earth 6400km or 6400000m

mearth= masss of the earth

Gm(shrew)m(earth)/r(earth)^2 = Gm(elephant)m(earth)/r^2

strike out the left hand side and right hand side variables

m(shrew)/r(earth)^2 = m(elephant)/r^2

r^2 = m(elephant).r(earth)^2 / m(shrew) .........make r^2 the subject of the equation

r^2= $(5*10^{3} *(6400000)^{2} )/.05$

r^2=40960000000000

r=2023857702.507m

4 0

2 years ago

Which medium caused the most bending of light as it passed through?

babunello [35]

I wasn't there observing the experiment while you and your class
performed it, so I don't really know how it was set up, or what
happened.

But I can tell you this: Light doesn't bend while passing through
any medium. It only bends at the boundary where one medium
meets a different one.

4 0

3 years ago