Differentiate amihan and habagat

If F1 is the magnitude of the force exerted

aleksandrvk [35]

Answer: 3. F1 = F2

Explanation:

According to <u>Newton's law of Gravitation</u>, the force $F$ exerted <u>between two bodies</u> or objects of masses $M$ and $m$ and separated by a distance $r$ is equal to the product of their masses divided by the square of the distance:

$F=G\frac{Mm}{r^2}$ (1)

Where $G$ is the gravitational constant

Now, in the especific case of the Earth and the satellite, where the Earth has a mass $M$ and satellite a mass $m$ , being both separated a distance $r$ , the force exerted by the Earth on the satellite is:

$F1=G\frac{Mm}{r^2}$ (2)

And the force exerted by the satellite on the Earth is:

$F2=G\frac{Mm}{r^2}$ (3)

As we can see equations (2) and (3) are equal, hence the magnitude of the gravitational force is the same for both:

$F1=F2$

3 0

3 years ago

At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial

elena55 [62]

Answer : The partial pressure of $SO_3$ is, 67.009 atm

Solution : Given,

Partial pressure of $SO_2$ at equilibrium = 30.6 atm

Partial pressure of $O_2$ at equilibrium = 13.9 atm

Equilibrium constant = $K_p=0.345$

The given balanced equilibrium reaction is,

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}$

Now put all the values of partial pressure, we get

$0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}$

$p_{SO_3}=67.009atm$

Therefore, the partial pressure of $SO_3$ is, 67.009 atm

6 0

3 years ago

Please help on this one ?

Marianna [84]

D has the largest range

3 0

3 years ago

If τ=r×F then F.τ is equal

Mademuasel [1]

Yes that is correct.

4 0

3 years ago

a ballistic pendulum is used to measure the speed of high-speed projectiles. A 6 g bullet A is fired into a 1 kg wood block B su

Galina-37 [17]

Answer:

(a) v-bullet = 399.04 m/s

(b) I = 2.38 kg m/s

(c) T = 2.59 N

Explanation:

(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.

The potential energy is given by:

$U=(M+m)gh$ (1)

U: potential energy

M: mass of the wood block = 1 kg

m: mass of the bullet = 6g = 6.0*10^-3 kg

g: gravitational constant = 9.8m/s^2

h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

$h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m$

The potential energy is:

$U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J$

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

$U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}$

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

$Mv_1+mv_2=(M+m)v$ (2)

v1: initial velocity of the wood block = 0m/s

v2: initial speed of the bullet

v: speed of bullet and block = 2.38m/s

You solve the equation (2) for v2:

$M(0)+mv_2=(M+m)v$

$v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}$

The speed of the bullet before the impact with the wood block is 399.04 m/s

(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:

$I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}$

The impulse is 2.38 kgm/s

(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:

$T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N$

The force on the cord after the impact is 2.59N

4 0

3 years ago