Metal sphere A is hung from the ceiling by a long, thin string and given a positive charge. An identical sphere B is suspended n

The answer is letter d, 8.3.

Here’s a solution for the given problem:

We have:

B = 6i - 8j 

Let A be unknown; we'll denote A as = mi + nj 

The resultant A+B lies along the x-axis (which implies A+B = Ki + 0j, where K is yet to be determined,

and we also know the magnitude of A+B is equivalent to the magnitude of A,

therefore, mag(A+B)=K=sqrt(m^2+n^2), or K^2 = m^2+n^2. 

Using vector addition, A+B becomes (m+6)i + (n-8)j.

Since we know A+B = Ki + 0j, we can establish that: 

m + 6 = K 

n - 8 = 0, which gives n=8. 

Thus, K^2=m^2+n^2 means (m+6)^2 = m^2 +8^2 

= m^2 + 12m + 36 = m^2 + 64 

which gives us 12m = 28 

m = 2.33333... 

Consequently, the magnitude of A is sqrt[(2.333...)^2 + 8^2] = 8.3333.

Answer:

The acceleration of the platform is - 1.8 m/s²

Explanation:

The net force on a body causes that body to accelerate in the direction of the resultant force applied.

Setting up the force equilibrium for the configuration:

ma = 800 - mg

100a = 800 - 100×9.8

100a = - 180

100a = - 180

a = - 1.8 m/s²

This indicates that the body is falling downward.

Answer:

b) TA = TB = TC

Explanation:

• When the blocks are brought into contact and isolated from the environment, they will exchange heat until they achieve thermal equilibrium.
• During this exchange, the hotter body will lose heat, which will be gained by the cooler body.
• The equilibrium state will be established once this equation is satisfied:

• Substituting the initial temperatures T₀A = 300º C and T₀B = 400ºC, while simplifying for equal block masses mA = mB, enables us to solve for the final temperature, Tfin:

• At equilibrium, when both blocks combine, they will yield a uniform final temperature of 350ºC.
• When block C, also at this temperature, makes contact, all three blocks will simultaneously reflect this final temperature of 350 ºC.
• Therefore, option b) is correct.

Answer:

F=126339.5N

Explanation:

To compute the force required to escape, a free-body diagram for the hatch must be drawn. We will equate the downward and upward forces, thus applying the following equation:

Fw=W+Fi+F

where

Fw=   force or weight exerted by the water column above the submarine.

To calculate Fw, we can use:

Fw=h. γ. A

h=height

γ= specific weight of seawater = 10074N / m ^ 3

A=Area

Fw=28x10074x0.7=197467N

w represents the hatch weight = 200N

Fi denotes the internal pressure force in the submarine, which is 1 atm = 101325Pa. We can calculate this force using:

Fi=PA=101325x0.7=70927.5N

Finally, the force needed to open the hatch is determined by the original equation:

Fw=W+Fi+F

F=Fw-W+Fi

F=197467N-200N-70927.5N

F=126339.5N