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Inga [223]
3 years ago
8

Which of the following would be most likely to contribute to molecules

Physics
1 answer:
lana66690 [7]3 years ago
8 0

Answer:

B. Containing charged regions

Explanation:

The term i.e. intermolecular forces would be used to explain the attraction forces. Here the interaction would be done between molecules etc that acts between the acts & the other types of particles i.e. neighboring like atoms or ions

So in the given case, the option b would be contributed to the molecules that have intermolecular forces

hence, the option b is correct

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A 2000 kg roller coaster is at the top of a loop with a radius of 24 m. If its speed is 18 m/s at this point, what force does it
levacccp [35]

Answer:

46620\ \text{N}

Explanation:

m = Mass of roller coaster = 2000 kg

r = Radius of loop = 24 m

v = Velocity of roller coaster = 18 m/s

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Normal force at the point will be

N-mg=\dfrac{mv^2}{r}\\\Rightarrow N=\dfrac{mv^2}{r}+mg\\\Rightarrow N=\dfrac{2000\times 18^2}{24}+2000\times 9.81\\\Rightarrow N=46620\ \text{N}

The force exerted on the track is 46620\ \text{N}.

8 0
3 years ago
Define second class lever​
Aleks04 [339]

Answer:

Please find detailed explanation of second class levers below

Explanation:

Levers are one of the classes of machine that possesses three levels namely: first class, second class and third claas. A second class lever is the level of levers in which the load (L) is in between the pivot (F) and the effort (E).

Examples of second class levers include; wheelbarrow, a bottle opener etc. In the bottle opener for example, the bottle lid (load) is in between the pivot of the opener and the hand opening it (effort).

6 0
3 years ago
A single mass (m1 = 3.5 kg) hangs from a spring in a motionless elevator. The spring constant is k = 278 N/m. 1)What is the dist
artcher [175]
<h2>Answer:</h2>

0.126m

<h2>Explanation:</h2>

According to Hooke's law, the force (F) acting on a spring to cause an extension or compression (e) is given by;

F = k x e            -------------------(i)

Where;

k = the spring's constant.

From the question, the force acting on the spring is the weight(W) of the mass. i.e

F = W               -----------------------(ii)

<em>But;</em>

W = m x g;

where;

m = mass of the object

g = acceleration due to gravity [usually taken as 10m/s²]

<em>From equation (ii), it implies that;</em>

F = W = m x g

<em>Now substitute F = m x g into equation(i) as follows;</em>

F = k x e

m x g = k x e      ------------------(iii)

<em>From the question;</em>

m = m1 = 3.5kg

k = 278N/m

<em>Substitute these values into equation (iii) as follows;</em>

3.5 x 10 = 278 x e

35 = 278e

<em>Now solve for e;</em>

e = 35/278

e = 0.126m

Therefore, the distance the spring is stretched from its unstretched length (which is the same as the extension of the spring) is 0.126m

3 0
4 years ago
A research group at Dartmouth College has developed a Head Impact Telemetry (HIT) System that can be used to collect data about
Olin [163]

Answer:

6.05 cm

Explanation:

The given equation is

2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)

The initial head velocity V₀ₓ =11 m/s

The final head velocity  Vₓ is 0

The accelerationis given by =1000 m/s²

the stopping distance = x-x₀=?

So we can wind the stopping distance by following formula

2 (-1000)(x-x₀)=[0^{2} -11^{2}]

x-x₀=6.05*10^{-2} m

       =6.05 cm

3 0
3 years ago
.A box falls to the ground from a delivery truck traveling at 30 m/s. After hitting the road, it slides 45 m to
Romashka-Z-Leto [24]

Answer:

t = 3 seconds

Explanation:

Given that,

Initial speed, u = 30 m/s

Final speed, v = 0

It slides 45 m to rest.it take the box to come to rest

We need to find how long it take the box to come to rest.

Let a be the acceleration and t is time.

v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(30)^2-u^2}{2(45)}\\\\=10\ m/s^2

Now finding time.

t=\dfrac{v-u}{a}\\\\t=\dfrac{30-0}{10}\\\\t=3\ s

So, the required time is 3 seconds.

8 0
3 years ago
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