Answer:
The work done on the suitcase is, W = 1691 J
Explanation:
Given data,
The force on the suitcase is, F = 89 N
The distance Russell dragged the suitcase, S = 19 m
The work done on the suitcase by Russell is equal to the work done on the suitcase to overcome the friction
The work done on the suitcase by Russell is given by the formula
W = F · S
Substituting the given values,
W = 89 N x 19 m
W = 1691 J
Hence, the work done on the suitcase is, W = 1691 J
Assuming north as positive direction, the initial and final velocities of the ball are:
(with negative sign since it is due south)
the time taken is
, so the average acceleration of the ball is given by
And the positive sign tells us the direction of the acceleration is north.
My answers would be as follows:
<span>You step on the scale and notice that you have lost five pounds. Which of the following has changed?
B) Only your weight changed since mass is conserved. It cannot be destroyed or created.
How could you increase the mass of a wooden block?
D) None of the above. Mass cannot be created so the mass will stay the same it will only be the weight you can increase depending on the acceleration.
If you went to the moon, how would your mass change relative to your mass on Earth?
A) It would be the same on the moon as on Earth. Mass will be the same. It is weight that will change since the gravitational acceleration in the moon is different as that to the earth.
</span>
Answer:
Explanation:
The speed of the astronaut can be found with the help of law of conservation of momentum .
mv = MV , M is mass of astronaut , m is mass of object thrown , v is velocity of object thrown and V is velocity of astronaut.
Putting the values
77.5 x V = .94 x 12
V = .14554 m /s
This will be the uniform velocity of astronaut.
Distance to be covered = 37.3 m
time taken = distance / velocity
= 37.3 / .14554
= 256.28 s
= 4.27 minutes.
<span>THIS IS A GAS PHASE REACTION AND WE ARE GIVE PARTIAL PRESSURES . I WRITE IN TERMS OF P RATHER THAN CONCENTRATION :
lnPso2cl12=-kt+lnPso2cl1
initial partial pressure Pso2cl12 the rate constant k and the time t
lnPso2cl12=(4.5*10-2*s-1)*65*s+ln (375)
so lnPso2cl12=3.002
we take the base e antilog:
lnPso2cl12=e3.002
Pso2cl12=20 torr
we use the integrated first order rate
lnPso2cl12=3.002=k*t+ lnPso2cl12=3.002
we use the same rate constant and initial pressure
k=4.5*10-2*s-1
Pso2cl12=375
Pso2cl12=1* so2cl12
Pso2cl12=37.5 torr
subtract in Pso2cl12 grom both side
lnPso2cl12- lnPso2cl12=-kt
ln(x)-ln(y)=ln (x/y)
ln (Pso2cl12/Pso2cl20)=-kt
we get t
-1/k*ln(Pso2cl12/Pso2cl20)=t
t=51 s</span>
