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3 years ago

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A ball hangs on a string. The force of tension (Ft) in the string is 15n. The force due to gravity (Fg) pulls the ball down with

a force of 15 N. Show your work.

1 answer:

Tanya [424]3 years ago

7 0

Answer: 0N

Explanation: 15N-15N =0N

they are a balanced force,meaning they canceled each other out.

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What does Pascal's Law state? what are the implications of Pascal's law?

dybincka [34]

Answer:

Pascal's Law states that the pressure applied to a fluid in a closed container is transmitted equally to all points in the fluid and act in all directions of the container. ... Therefore, it can rightly be said that since the liquid does not flow, it definitely has equal pressure acting on it at all the points.

Explanation:

plzzzzzzz Mark my answer in brainlist

4 0

3 years ago

A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal fr

nikklg [1K]

Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

a= μ g ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g =ω²X

We know for spring mass system natural frequency given as

$\omega=\sqrt{\dfrac{k}{M+m}}$

By putting the values

$\omega=\sqrt{\dfrac{150}{5.67+1}}$

ω = 4.47 rad/s

μ g =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

$\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2$

$kX^2=(m+M)v^2$

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

7 0

3 years ago

Read 2 more answers

What are non-contact forces please include example in your answer

Alina [70]

Answer:

Gravitational force. Magnetic force. Electrostatics. Nuclear force.

Explanation:

Apple falling from a tree

raindrops falling from the sky

4 0

3 years ago

Read 2 more answers

Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0

3 years ago

Two subway stops are separated by 1210 m. If a subway train accelerates at 1.30 m/s2 from rest through the first half of the dis

solong [7]

Answer:

Part 1) Time of travel equals 61 seconds

Part 2) Maximum speed equals 39.66 m/s.

Explanation:

The final speed of the train when it completes half of it's journey is given by third equation of kinematics as

$v^{2}=u^2+2as$

where

'v' is the final speed

'u' is initial speed

'a' is acceleration of the body

's' is the distance covered

Applying the given values we get

$v^2=0+2\times 1.30\times \frac{1210}{2}\\\\v^{2}=1573\\\\\therefore v=39.66m/s$

Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as

$v=u+at\\\\v=0+1.30\times t\\\\\therefore t=\frac{39.66}{1.30}=30.51seconds$

Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance

Thus total time of journey equals $T=2\times 30.51\approx61seconds$

Part b)

the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals $39.66m/s$

4 0

3 years ago