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3 years ago

The requirements for a master’s degree in psychology generally include

Physics

1 answer:

erastova [34]3 years ago

6 0

It takes 2 years of graduate study in a subfeild, practicum experience, and a thesis.
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A 3.00-m rod is pivoted about its left end. A force of 7.80 N is applied perpendicular to the rod at a distance of 1.60 m from t

dalvyx [7]

Answer:

The net torque about the pivot is and the answer is 'c'

c. $T_{net}=8.58$

Explanation:

$T=F*d$

The torque is the force apply in a distance so it is the moment so depends on the way to be put it the signs so:

$T_1=F_1*d_1$

$T_1=7.8N*1.6m=12.48N*m$

$T_2=F_2*d_2$

$T_2=2.60N**cos(30)*3.0m$

$T_2= - 3.9 N*m$

Now to find the net Torque is the summation of both torques

$T_{net}=T_1+T_2$

$T_{net}=12.48N-3.9N=8.58N$

8 0

3 years ago

Which of the following would be a valid method to increase the buoyant force acting on an object?

shepuryov [24]

I think it’s B hope this helps.

4 0

2 years ago

Read 2 more answers

15 Points , Physics HW, can someone please show me step by step how to calculate percent difference for this problem. My numbers

timofeeve [1]

The percent difference between two numbers $x$ and $y$ is given by

$\dfrac{|y-x|}x \times 100\%$

The absolute value is there because we only care about the absolute percent difference, and not taking into account whether we go from $x$ to $y$ or vice versa. If we remove them, we have two possible interpretations of percent difference.

For example, the (absolute) percent difference between 3 and 6 is

$\dfrac{|6-3|}3 \times 100\% = 100\%$

In other words, we add 100% of 3 to 3 to end up with 6. This is the same as the percent difference going from 3 to 6. On the other hand, the percent difference going from 6 to 3 is

$\dfrac{3-6}3\times100\%=-50\%$

which is to say, we take away 50% of 6 away from 6 to end up with 3.

"Make comparisons to object measurements" tells us that the differences should be computed relative to "measurements for object". In other words, take $x$ from the left column and $y$ from the right column.

$\dfrac{|7.1-7.3|}{7.1} \times 100\% \approx 2.82\%$

$\dfrac{|4.8-5.0|}{4.8} \times 100\% \approx 4.17\%$

$\dfrac{|7.2-7.5|}{7.2} \times 100\% \approx 4.17\%$

3 0

2 years ago

Mumz [18]

No se ha da han dicho nada más de lo dicho y han ido de vuelta y han dicho nada más de que se pueda hacer el favor del niño

4 0

3 years ago

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Sphere 1 with radius R_1 has positive charge q, Sphere 2 with radius 4.50 R_1 is far from sphere 1 and initially uncharged. The

tia_tia [17]

Answer:

Explanation:

capacitance of sphere 2 will be 4.5 times sphere 1

a ) when spheres are in contact they will have same potential finally . So

V_1 / V_2 = 1

b )

Charge will be distributed in the ratio of their capacity

charge on sphere1 = q x 1 / ( 1 + 4.5 )

= q / 5.5

fraction = 1 / 5.5

c ) charge on sphere 2

= q x 4.5 / 5.5

fraction = 4.5 / 5.5

d ) surface charge density of sphere 1

= q /( 5.5 x A ) where A is surface area

surface charge density of sphere 2

= q x 4.5 /( 5.5 x 4.5² A ) where A is surface area

= q /( 5.5 x 4.5 A )

q_1/q_2 = 4.5

6 0

3 years ago