PLZ HELP GUYS PLZ!!!!! The answer is not motion, particles.

A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule

ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

$p_i = 0$

The final total momentum is instead:

$p_f = m_a v_a + m_c v_c$

where

$m_a = 125 kg$ is the mass of the astronaut

$v_a = 2.50 m/s$ is the velocity of the astronaut

$m_c = 1900 kg$ is the mass of the capsule

$v_c$ is the velocity of the capsule

Since the total momentum must be conserved, we have

$p_i = p_f = 0$

so

$m_a v_a + m_c v_c=0$

Solving the equation for $v_c$ , we find

$v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s$

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

$F \Delta t = \Delta p$

The change in momentum of the astronaut is

$\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s$

And the duration of the push is

$\Delta t = 0.600 s$

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

$F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N$

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

$K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J$

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

$K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J$

3 0

3 years ago

A liquid of density 1200kg/m³ is filled in a beaker upto the depth of 50 cm. calculate the pressure at bottom of the breaker.(g=

sweet-ann [11.9K]

Density=1200kg/m^3=p
Height=50cm=0.5m=h
Acceleration due to Gravity=9.8m/s^2=g

$\boxed{\sf Pressure=pgh}$

$\\ \qquad\quad\sf{:}\dashrightarrow Pressure=1200(0.5)(9.8)$

$\\ \qquad\quad\sf{:}\dashrightarrow Pressure=600(9.8)$

$\\ \qquad\quad\sf{:}\dashrightarrow Pressure=5880Pa$

6 0

3 years ago

You run<br> completely around a 400m track in<br> 80s. What was your average velocity?

dalvyx [7]

Answer:

V=?

S=400m

t=80s

V=S/t

V=400/80

V=5m/sec

6 0

3 years ago

If a hockey stick exerts a force of 40 N on a 0.5-kg puck, what will the

antoniya [11.8K]

Answer:

B. 80 m/s²

Explanation:

F = ma

a = F/m = (40 N)/(0.5 kg) = 80 m/s²

3 0

3 years ago

An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 14.0 m if her initial speed is 2.20 m/

pochemuha

We have the equation of motion

$v^2 = u^2+2as$ , where v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.

In this case initial velocity = 2.20 m/s, final velocity = 0 m/s, displacement = 14 m

On substitution we will get 0 = $2.20^2+2*a*14$

On solving we can find the acceleration value as -0.173 $m/s^2$

So free fall acceleration = 0.173 $m/s^2$

3 0

3 years ago