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professor190 [17]
3 years ago
10

Please help me with #20!!!

Physics
1 answer:
stiv31 [10]3 years ago
7 0
Hello!   for the solution of this answer you must go to the topic of tension
and in tension you go to modified atwood maschine and put the values in 
(2m1m2)g/m1+m2         you will get your answer

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Name one reason why you should pour milk in before cereal.
krok68 [10]

Answer:

splashing

Explanation:

if you put in the cereal after the milk it will splash everywhere, causing a waste of milk, and a loss of time.

5 0
3 years ago
It takes you 8.3 min to walk with an average velocity of 1.6 m/s to the north from the bus stop to the museum entrance. How far
Anarel [89]

solution:

1.6 m/s = 96 m/min (in other words, 1.6 m/s x 60 s/min)  

96 m/min x 8.3 min = 796.8 m

s=ut +\frac{1}{2}at^2\\there is no accleration mentioned so,\\s= uv\\8.3\times60=498(s)\\510\times1.6=816(m)


3 0
3 years ago
Electric energy can be transferred into other types of energy we often experiences transformation of energy in our every day liv
Damm [24]

An amplifier.

Electrical energy provided to an amplifier is converted into sound energy as it is "fed" or provided to the speaker portion of an amplifier.

7 0
2 years ago
Which statement provides a complete scientific discription of an object in motion?
Alekssandra [29.7K]

Question: which statement provides a complete scientific discription of an object in motion?

Answer: the marble moved 30 cm north in 6 seconds.

Explanation: motion is a change in position of an object over time an object's motion cannot change unless it is acted upon by a force

question answered by

(jacemorris04)

7 0
2 years ago
Read 2 more answers
A car traveling on a flat (unbanked) circular track accelerates uniformly from rest with a tangential acceleration of 1.7 m/s^2.
Komok [63]

Answer:

The coefficient of static friction between the car and the track

u=0.572

Explanation:

We don't know the mass of the car or any other information so the acceleration is the reason to solve the friction coefficient

∑F=F_{f}=m*a_{t}

As we know

F_{f}=u*F_{N}=u*m*g

Also the center ward direction forces

F_{fc}=m*a_{c}

a_{c}=\frac{v_{t}^2}{r}

F_{fc}=m*\frac{v_{t}^2}{r}

But now vt relation with the tangential acceleration

v_{t}=2*a_{t}*\frac{\pi }{r}

replacing

F_{fc}=m*a_{t}*\frac{2\pi*r}{2r}

F_{fc}=m*a_{t}*\pi

So magnitude of the force can get by

F_{f}=\sqrt{(m*a_{t}*\pi)^{2}+(m*a_{t})^{2}}

Get the factor to simplify

F_{f}=a_{t}*m*\sqrt{(1+\pi^2)}

u*m*g=m*a_{t}*(\sqrt{1+\pi^2})

Solve to u'

u=\frac{a_{t}}{g}*(\sqrt{1+\pi^2})

u=\frac{17\frac{m}{s^2} }{9.8\frac{m}{s^2}}*(\sqrt{1+\pi^2})=0.572

5 0
3 years ago
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