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3 years ago

6

Look at the densities of the jovian planets given in figure 1. which of the following statements best describes the pattern of j

ovian planet densities?

2 answers:

barxatty [35]3 years ago

4 0

more than 5.5 times the density of water

is the answer.

avanturin [10]3 years ago

3 0

Given the densities of the jovian planets, the densities is just like random numbers, so the densities of the jovian planets have no concrete pattern due to its randomness. pattern can be either linear or geometric, where linear you see a common difference among data, while geometric you observed a common ratio

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3 years ago

Read 2 more answers

The 5-kg block A has an initial speed of 5 m/s as it slides down the smooth ramp, after which it collides with the stationary bl

Akimi4 [234]

Answer:

The coupled velocity of both the blocks is 1.92 m/s.

Explanation:

Given that,

Mass of block A, $m_1=5\ kg$

Initial speed of block A, $u_1=5\ m/s$

Mass of block B, $m_2=8\ kg$

Initial speed of block B, $u_2=0$

It is mentioned that if the two blocks couple together after collision. We need to find the common velocity immediately after collision. We know that due to coupling, it becomes the case of inelastic collision. Using the conservation of linear momentum. Let V is the coupled velocity of both the blocks. So,

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{5\times 5+0}{(5+8)}\\\\V=1.92\ m/s$

So, the coupled velocity of both the blocks is 1.92 m/s. Hence, this is the required solution.

8 0

3 years ago

Two 100 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Bat -10 m/s when th

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b) -10 m/s

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3 years ago

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3 years ago

Help with this physics task pls

cupoosta [38]

Answer:

Answers can be seen below

Explanation:

First we must explain the essential when we clear equations, and that is that if the term we need to clear is accompanied by other terms that are being added up, then those terms go to the other side of the equation to subtract if those terms are subtracting, then they go to the other side to add, if those terms are found multiplying then they go to the other side of the equation to divide and if those other terms are found dividing then they go to the other side of the equation to multiply.

(Primero debemos explicar lo esencial cuando despejamos ecuaciones, y es que si el término que necesitamos despejar va acompañado de otros términos que se están sumando, entonces esos términos van al otro lado de la ecuación para restar si esos términos están restando, luego van al otro lado para sumar, si esos términos se encuentran multiplicando luego van al otro lado de la ecuación a dividir, y si esos términos se encuentran dividiendo, pasan al otro lado de la ecuación a multiplicar.)

1 )

$t=\frac{v}{a}$ ; $d=s*(t-t_{0} )$

2)

$k=\frac{2*U}{x^{2} }$ ; $T_{2}=\frac{P_{2}*V_{2}*T_{1} }{P_{1}*V_{1} } \\$

3)

$L=\frac{F}{\pi*r*P}$ ; $d=\frac{w}{F*cos(o)}$

4)

$t^{2}=\frac{2*x}{g} ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} } \\$

5)

$h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}$

6)

$h=\frac{m}{(1/2)*\pi *r^{2} } ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2} }$

7)

$b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}$

8)

$a=\frac{v-v_{o} }{t} ; u=\frac{m_{1}+m_{2} }{M}$

9)

$v_{o}=\frac{x-\frac{1}{2}*a*t^{2} }{t} ; F=\frac{W+uNd}{d*cos(o)}$

10)

$h=\frac{E-\frac{1}{2}*m*v^{2} }{mg} ; v_{2} ^{2} = \frac{Dk-\frac{1}{2} m*v_{1}^{2} }{\frac{1}{2}m }$

11)

$N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2} }{\frac{1}{2}*k }$

12)

$x=x_{o} +\frac{v^{2-v_{o}^{2} } }{2a} ; m=\frac{P*A-F_{1}-F_{2} }{g}$

13)

$x_{o} = x-\frac{F}{k} ; u=\frac{cos(o)-\frac{a}{g} }{sin(o)}$

14)

$t=\frac{d}{v} +t_{o}$ ; $t_{o} = t-(\frac{v-v_{o} }{a} )$

15)

$F_{2}=\frac{W-F_{1} *d}{d}+F_{3} ; v_{2}^{2}=v_{1}^{2}+\frac{2*Dk}{m}$

16)

$y_{1}=y-\frac{u}{mg} ; x^{2} = \frac{2W}{k}+x_{o} ^{2}$

7 0

3 years ago