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forsale [732]
2 years ago
13

50 POINTS!! BRAINLEST

Physics
2 answers:
likoan [24]2 years ago
8 0

Answer: pls mark me as a brainliest friend pls, it would be a great help for me friend

Explanation:

1. B

2. The graph that shows students data is graph b. It is constant so it’s velocity is constant.

Hope it’s useful friend…

Good luck and have a great day buddy…

Stay safe and happy…

Keep rocking !!

sergey [27]2 years ago
4 0

Answer:2 one

Explanation:

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What are some of the downsides of using hydroelectric power?.
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Has an Environmental Impact. Perhaps the largest disadvantage of hydroelectric energy is the impact it can have on the environment.
It Displaces People.
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4 0
2 years ago
A Honda Hawk motorcycle and its rider with a combined mass of 450 kg travels around a curve of radius 106 m with a speed of 18 m
umka21 [38]

Answer: coefficient of static friction

= 0.31

Explanation: Since they negotiate the curve without skidding, the frictional force (F1) equals the centripetal force (F2).

F1= uN

F2 = M*(v²/r)

M is the combined mass 450kg

V is the velocity 18m/s

r is the radius 106m

N is the normal reaction 4410N

u is the coefficient of static friction

Making u subject of the formula we have that,

u = {450*(18²/106)} /4410

=1375.47/4410

=0.31

NOTE: coefficient of friction is dimensionless. It as no Unit.

7 0
3 years ago
The drawing below shows two different types of pulley systems designed to lift a weight. In pulley system A, the end of the rope
nadya68 [22]
The answer would be in the chart or graph A is 1 B is 2
4 0
3 years ago
Read 2 more answers
(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i
murzikaleks [220]

Answer:

a) t = -\frac{ln(2)}{k}

b) See the proof below

A(t) = A_o 2^{-\frac{t}{T}}

c) t = 3T \frac{ln(2)}{ln(2)}= 3T

Explanation:

Part a

For this case we have the following differential equation:

\frac{dA}{dt}= kA

With the initial condition A(0) = A_o

We can rewrite the differential equation like this:

\frac{dA}{A} =k dt

And if we integrate both sides we got:

ln |A|= kt + c_1

Where c_1 is a constant. If we apply exponential for both sides we got:

A = e^{kt} e^c = C e^{kt}

Using the initial condition A(0) = A_o we got:

A_o = C

So then our solution for the differential equation is given by:

A(t) = A_o e^{kt}

For the half life we know that we need to find the value of t for where we have A(t) = \frac{1}{2} A_o if we use this condition we have:

\frac{1}{2} A_o = A_o e^{kt}

\frac{1}{2} = e^{kt}

Applying natural log we have this:

ln (\frac{1}{2}) = kt

And then the value of t would be:

t = \frac{ln (1/2)}{k}

And using the fact that ln(1/2) = -ln(2) we have this:

t = -\frac{ln(2)}{k}

Part b

For this case we need to show that the solution on part a can be written as:

A(t) = A_o 2^{-t/T}

For this case we have the following model:

A(t) = A_o e^{kt}

If we replace the value of k obtained from part a we got:

k = -\frac{ln(2)}{T}

A(t) = A_o e^{-\frac{ln(2)}{T} t}

And we can rewrite this expression like this:

A(t) = A_o e^{ln(2) (-\frac{t}{T})}

And we can cancel the exponential with the natural log and we have this:

A(t) = A_o 2^{-\frac{t}{T}}

Part c

For this case we want to find the value of t when we have remaining \frac{A_o}{8}

So we can use the following equation:

\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}

Simplifying we got:

\frac{1}{8} = 2^{-\frac{t}{T}}

We can apply natural log on both sides and we got:

ln(\frac{1}{8}) = -\frac{t}{T} ln(2)

And if we solve for t we got:

t = T \frac{ln(8)}{ln(2)}

We can rewrite this expression like this:

t = T \frac{ln(2^3)}{ln(2)}

Using properties of natural logs we got:

t = 3T \frac{ln(2)}{ln(2)}= 3T

8 0
3 years ago
The first confirmed detections of extrasolar planets occurred in ____________. The first confirmed detections of extrasolar plan
nirvana33 [79]

Answer:

1992 (Early 1990s)

Explanation:

First of all, i would like to define an extrasolar planet as a planet that orbits a start that is not our own.

The first confirmed detections of extrasolar planets occured in the early 1990s (specifically 1992, some say 1995). The name of the first extrasolar planet is widely believed to be called Dimidium or 51 Pegasi b.  

Extrasolar were searched by monitoring stars for slight dimming that might occur as unseen planets pass in front of them.

4 0
3 years ago
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