I need help for a review sheet

You weigh 716 newtons on Earth. You

Otrada [13]

Answer:

537 N

Explanation:

The force due to gravity of a planet is:

F = GMm / r²

where G is the universal gravitational constant

M is the mass of the planet

m is the mass of the object

and r is the distance between the object and the center of the planet

On Earth, you weigh 716 N, so:

716 N = GMm / r²

On planet X:

F = G (3M) m / (2r)²

F = 3/4 GMm / r²

F = 3/4 (716 N)

F = 537 N

8 0

3 years ago

A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the

vladimir2022 [97]

Answer:

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k

The Pressure-Volume relation is PV = C

T = C for isothermal process

Calculating for the work done in isothermal process

W = P₁V₁ $ln[\frac{P_{1} }{P_{2} }]$

= mRT₁ $ln[\frac{P_{1} }{P_{2} }]$ [∵pV = mRT]

= (5) (0.287) (272.039) $ln[\frac{2.0}{1.0}]$

= 270.588 kJ

Since the process is isothermal, Internal energy change is zero

ΔU = $mc_{v}(T_{2} - T_{1} ) = 0$

From 1st law of thermodynamics

Q = ΔU + W

= 0 + 270.588

= 270.588 kJ

4 0

3 years ago

The rocket's acceleration has components $a_{x}(t)= \alpha t^{2}$ and $a_{y}(t)= \beta - \gamma t$, where \(\alpha = 2.50 {\

lbvjy [14]

it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]

5 0

3 years ago

Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

ddd [48]

Answer:

Al's mass is 102.92 kg

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

$F_{net} = 0$

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

$F_{net} = \frac{dp_{horizontal}}{dt} = 0$

$p_{horizontal_i }= p_{horizontal_f}$

where the suffix i and f means initial and final respectively.

The initial momentum will be:

$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

But, as they are at rest, initially

$p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0$

$p_{horizontal_i} = 0$

So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

5 0

3 years ago

A baby carriage is sitting at the top of a hill that is 21m high. The carriage with the baby weighs 12 n. What's the potential e

Strike441 [17]

A baby carriage is sitting at the top of a hill that is 21m high. The carriage with the baby weighs 12N. The carriage has... energy. Calculate it = 252J

7 0

3 years ago