If the sun exploded would the solar system collaspe?

A stone is thrown at an angle of 30 degrees above the horizontal from the top edge of a cliff with an initial speed of 12 m/s. A

natta225 [31]

v = initial velocity of launch of the stone = 12 m/s

θ = angle of the velocity from the horizontal = 30

Consider the motion of the stone along the vertical direction taking upward direction as positive and down direction as negative.

v₀ = initial velocity along vertical direction = v Sinθ = 12 Sin30 = 6 m/s

a = acceleration of the stone = - 9.8 m/s²

t = time of travel = 4.8 s

Y = vertical displacement of stone = vertical height of the cliff = ?

using the kinematics equation

Y = v₀ t + (0.5) a t²

inserting the values

Y = 6 (4.8) + (0.5) (- 9.8) (4.8)²

Y = - 84.1 m

hence the height of the cliff comes out to be 84.1 m

5 0

2 years ago

Fill in the blanks for the following:

storchak [24]

Answer:

a. 4.21 moles

b. 478.6 m/s

c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank

Explanation:

Volume of container = 100.0 L

Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

R = 0.08206 L-atm- $mol^{-1}$ $K^{-1}$

Therefore,

n = (1.01325 x 100)/(0.08206 x 293)

n = 101.325/24.04 = 4.21 moles

The equation for root mean square velocity is

Vrms = $\sqrt{\frac{3RT}{M} }$

R = 8.314 J/mol-K

where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol

Vrms = $\sqrt{\frac{3*8.314*293}{0.0319} }$ = 478.6 m/s

For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship

$\frac{Voxy}{Vnit}$ = $\sqrt{\frac{Mnit}{Moxy} }$

where

Voxy = root mean square velocity of oxygen = 478.6 m/s

Vnit = root mean square velocity of nitrogen = ?

Moxy = Molar mass of oxygen = 31.9 g/mol

Mnit = Molar mass of nitrogen = 14.00 g/mol

$\frac{478.6}{Vnit}$ = $\sqrt{\frac{14.0}{31.9} }$

$\frac{478.6}{Vnit}$ = 0.66

Vnit = 0.66 x 478.6 = 315.876 m/s

the root mean square velocity of the oxygen gas is

478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank

6 0

2 years ago

A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$