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2 years ago

Study the diagram and calculate the effort required to balance the load

Physics

1 answer:

Lisa [10]2 years ago

6 0

Answer:

Solution given;

load =600N

effort=?

load distance: 0.6m

effort distance;2.6m

we have

load *load distance :effort *effort distance

0.6*600=effort *2.6

360/2.6=effort

effort:138.46N

Required effort is 138.5N.

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an iron rod of length 100m at 10 degree Celsius is used to measure a distance of 2km on a day when the temperature is 40 degree

german

Answer:

0.68 m

Explanation:

α = dL / L1*(dT)

dL = L1(dT) * α

Initial length, L1 = 100

Chang in Length =dL

α linear expansivity ; dL = change in length ; dT = change in temperature ; L1 = initial length

α of iron rod = 1.13 * 10^-5 k

dL = 100(40 - 10) * 1.13 * 10^-5

dL = 100(30) * 1.13 * 10^-5

dL = 3000 * 1.13 * 10^-5

dL = 3390 * 10^-5

dL = 0.0339 m

Error :

Distance measured = 2km = (2 * 1000) = 2000m

[Distance measured / (initial length + change in length)] × change in length

Error = (2000 / (100 + 0.0339)) * 0.0339

Error = (2000 / 100.0339) * 0.0339

Error = 19.993222 * 0.0339

Error = 0.6777702

Error = 0.68 m

4 0

2 years ago

Determine the values of m and n when the following mass of the Earth is written in scientific notation: 5,970,000,000,000,000,00

yuradex [85]

Explanation:

Mass of the Earth is equal to,

$m=5,970,000,000,000,000,000,000,000\ kg$

Any number can be written in the form of scientific notation as :

$N=m\times 10^n$

m is the real number

n is any integer

Mass of the earth can be written in the form of scientific notation as :

$m=5.97\times 10^{24}\ m$

Here,

m = 5.97

n = 24

Hence, this is the required solution.

7 0

2 years ago

if you apply a Force of F1 to area A1 on one side of a hydraulic jack, and the second side of the jack has an area that is twice

Firlakuza [10]

Answer:

$2F_{1}$

Explanation:

F₁ = Force on one side of the jack

A₁ = Area of cross-section of one side of the jack

F₂ = Force on second side of the jack

A₂ = Area of cross-section of second side of the jack = 2 A₁

Using pascal's law

$\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{A_{_{2}}}$

$\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{2A_{_{1}}}$

$F_{1}= \frac{F_{2}}{2}\\$

$F_{2}= 2F_{1}$

7 0

2 years ago

A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro

yanalaym [24]

Answer:

a) $v_{o} =16m/s$

b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

$x=v_{ox}t$

$x=v_{o}cos(60)t$

$v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}$ (1)

That would be our first equation

Now, we need to analyze its vertical motion

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2$

Knowing $v_{o}$ in our first equation (1)

$8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2$

$\frac{1}{2}(9.8)t^2=16tan(60)-8$

Solving for t

$t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s$

So, the ball takes to seconds to get to the other building. Now we can calculate its initial velocity

$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To find the magnitude of the ball just before it strikes the building we need to calculate its x and y components

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

So, the magnitude of the velocity is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s$

c) The direction of the ball is:

$\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º$

4 0

2 years ago

jenny's model train is set up on a circular track. There are six telephone poles evenly spaced around the track. It takes the en

d1i1m1o1n [39]

Answer:

T = 60 s

Explanation:

There are 6 poles on the track which are equally spaced

so the angular separation between the poles is given as

$\theta = \frac{2\pi}{6}$

$\theta = \frac{\pi}{3}$

so the angular speed of the train is given as

$\omega = \frac{\theta}{t}$

$\omega = \frac{\pi}{30} rad/s$

now we have time period of the train given as

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{\frac{\pi}{30}}$

$T = 60 s$

5 0

2 years ago

Study the diagram and calculate the effort required to balance the load​

Study the diagram and calculate the effort required to balance the load