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Helen [10]
3 years ago
8

Ddxxxxxxxdxxxxxxxvvgvvgggghhhh

Physics
2 answers:
Luba_88 [7]3 years ago
8 0

Answer:

I like your blue hair

Explanation:

thanks for points

balandron [24]3 years ago
7 0

Answer:

huh? thanks for points

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How do you find acceleration due to gravity with time and height given?
Feliz [49]

Here, height is given which will be the distance for a freely falling object.

The velocity will be

v=\text{ }\frac{h}{t}

and the acceleration will be

a=\frac{v}{t}

In this way, the formula works.

3 0
11 months ago
1. How would you expect an instantaneous acceleration vs. time graph to look for a cart moving with a constant
r-ruslan [8.4K]

Answer:

a=0   v = v₀ + a t

a=0    line is horizontal

Explanation:

1, In a graph of acceleration vs. time, we have lines, when the line is horizontal it is zero, when the line has a positive slope the increasing accelerations and when the slope is negative the decreasing acceleration

2, speed and relationship of a car is given by

        v = v₀ + a t

where vo is the initial velocity, a is the acceleration and tel time

in this case I will calcograph velocity vs. time the constant acceleration is a straight line.

In general from the graph we can find the initial velocity with the cut at that x and the acceleration of the car with the slope

6 0
3 years ago
A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above t
elena55 [62]

Answer:

The first part can be solved via conservation of energy.

mgh = mg2R + K\\K = mg(h-2R)

For the second part,

the free body diagram of the car should be as follows:

- weight in the downwards direction

- normal force of the track to the car in the downwards direction

The total force should be equal to the centripetal force by Newton's Second Law.

F = ma = \frac{mv^2}{R}\\mg + N = \frac{mv^2}{R}

where N = 0 because we are looking for the case where the car loses contact.

mg = \frac{mv^2}{R}\\v^2 = gR\\v = \sqrt{gR}

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

mgh = mg2R + \frac{1}{2}mv^2\\mgh = mg2R + \frac{1}{2}m(gR)\\gh = g2R + \frac{1}{2}gR\\h = 2R + \frac{R}{2}\\h = \frac{5R}{2}

Explanation:

The point that might confuse you in this question is the direction of the normal force at the top of the loop.

We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.

7 0
3 years ago
What is used to create the sequential order of elements?
algol [13]
Atomic Number
or
Number of Protons

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5 0
2 years ago
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Question- Gases the oceans exchange with the atmosphere include..
lutik1710 [3]
I am pretty sure it’s B (carbon dioxide)
5 0
2 years ago
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