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3 years ago

How do i calculate the net force of the following question.Plz answer correctly

1 answer:

5 0

<h2>Explanation:</h2><h2>force = mass×acceleration</h2>

mass=30 force=50

(a) 50÷30 

<h2>because we don't have acceleration which is known as m/s2 that stands for meter per seconds square . Hope this help pls follow thank you</h2>

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In part one of this experiment, a 0.20 kg mass hangs vertically from a spring and an elongation below the support point of the s

statuscvo [17]

To solve this problem it is necessary to apply the concepts related to Hooke's Law as well as Newton's second law.

By definition we know that Newton's second law is defined as

$F = ma$

m = mass

a = Acceleration

By Hooke's law force is described as

$F = k\Delta x$

Here,

k = Gravitational constant

x = Displacement

To develop this problem it is necessary to consider the two cases that give us concerning the elongation of the body.

The force to keep in balance must be preserved, so the force by the weight stipulated in Newton's second law and the force by Hooke's elongation are equal, so

$k\Delta x = mg$

So for state 1 we have that with 0.2kg there is an elongation of 9.5cm

$k (9.5-l)=0.2*g$

$k (9.5-l)=0.2*9.8$

For state 2 we have that with 1Kg there is an elongation of 12cm

$k (12-l)= 1*g$

$k (12-l)= 1*9.8$

We have two equations with two unknowns therefore solving for both,

$k = 3.136N/cm$

$l = 8.877cm$

In this way converting the units,

$k = 3.136N/cm(\frac{100cm}{1m})$

$k = 313.6N/m$

Therefore the spring constant is 313.6N/m

3 0

3 years ago

An incompressible fluid flows steadily through two pipes of diameter 0.15 m and 0.2 m which combine to discharge in a pipe of 0.

QveST [7]

Answer:

Average velocity = 1.835 m/s

Explanation:

Detailed explanation and calculation is shown in the image below

3 0

3 years ago

What two aspects of a force do scientists measure?

malfutka [58]

_Award brainliest if helped!
Force is a vector, A. Magnitude and Direction.

5 0

3 years ago

Josh starts his sled at the top of a 2.9-m-high hill that has a constant slope of 25∘. After reaching the bottom, he slides acro

algol13

Answer:

S=48.29 m

Explanation:

Given that the height of the hill h = 2.9 m

Coefficient of kinetic friction between his sled and the snow μ = 0.08

Let u be the speed of the skier at the bottom of the hill.

By applying conservation of energy at the top and bottom of the inclined plane we get.

Potential Energy=kinetic Energy

mgh = (1/2) mu²

u² = 2gh

u²=2(9.81)(2.9)

=56.89

u=7.54 m/s

a = - f / m

a = - μ*m*g / m

a = - μg

From equation of motion

v²- u² = 2 -μ g S

v=0 m/s

-(7.54)²=-2(0.06)(9.81)S

S=48.29 m

3 0

3 years ago

Yellow light of wavelength 590 nm passes through a diffraction grating and makes an interference pattern on a screen 80 cm away.

riadik2000 [5.3K]

Answer:B

Explanation:

Given

Wavelength of light $\lambda =590\ nm$

Screen distance $L=80\ cm$

First fringe is at a distance $y_1=1.9\ cm$

No of lines per mm is given by N

$N=\frac{1}{d}$

where d=slit width

From N-slits Experiment

$\sin \theta _m=\frac{m\lambda }{d}$

$d=\frac{m\lambda }{\sin \theta _m}-----1$

Position of bright fringe is given by

$y=\tan \theta _m\cdot L$

$\tan \theta _m=\frac{y}{L}$

$\theta _m=\tan^{-1}(\frac{y}{L})$

Put the value of $\theta _m$ in eq. 1

$d=\frac{m\lambda }{\sin (\tan^{-1}(\frac{y}{L}))}$

Therefore $N=d^{-1}$

$N=\frac{\sin (\tan^{-1}(\frac{y}{L}))}{m\lambda }$

for $m=1$

$N=\frac{\sin (\tan^{-1}(\frac{1.9\times 10^{-2}}{0.8}))}{1\times 590\times 10^{-9}}$

$N=40243\ line/m$

$N=40\ line/mm$

4 0

4 years ago