You can make sure there's no change in volume by keeping
your gas in a sealed jar with no leaks. Then you can play with
the temperature and the pressure all you want, and you'll know
that the volume is constant.
For 'ideal' gases,
(pressure) times (volume) is proportional to (temperature).
And if volume is constant, then
(pressure) is proportional to (temperature) .
So if you increase the temperature from 110K to 235K,
the pressure increases to (235/110) of where it started.
(400 kPa) x (235/110) = 854.55 kPa. (rounded)
Obviously, choice-b is the right one, but
I don't know where the .46 came from.
Answer:
Explanation:
I got everything but i. Don't know why but it's eluding me. So let's do everything but that.
a. PE = mgh so
PE = (2.5)(98)(14) and
PE = 340 J
b.
so
and
KE = 250 J
c. TE = KE + PE so
TE = 340 + 250 and
TE = 590 J
d. PE at 8.7 m:
PE = (2.5)(9.8)(8.7) and
PE = 210 J
e. The KE at the same height:
TE = KE + PE and
590 = KE + 210 so
KE = 380 J
f. The velocity at that height:
and
so
v = 17 m/s
g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:
590 = KE + PE and
PE = (2.5)(9.8)(11.6) so
PE = 280 then
590 = KE + 280 so
KE = 310 then
and
so
v = 16 m/s
h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:
and
26 = 0 + 9.8t and
26 = 9.8t so the time at 26 m/s is
t = 2.7 seconds. Now we use that in the equation for displacement:
Δx =
and filling in the time the object was at 26 m/s:
Δx = 0t +
so
Δx = 36 m
i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.
Explanation:
Simply, shake the boxes. The box that is difficult to shake will be the sand and the box that will be easier will be the feathers.
Answer:
A) 1.67 x 10 ⁻⁶ m/s
B)5.59 x
%
Explanation:
A)
Given:
d = 5.0 km,
mₐ = 2.5 x
kg
u₁ = 4.0 x 10⁴ m/s
= 5.98 x 10 ²⁴ kg
Solve using kinetic conserved energy
mₐ x u₁ +
x u₂ = uₓ x (mₐ +
)
(2.5 x
) (4.0 x 10⁴ )+ (5.98 x 10 ²⁴ )(0) = uₓ x (2.5 x
+ 5.98 x 10 ²⁴ )
uₓ = ( 2.5 x
x 4.0 x 10⁴ ) / (2.5 x
+ 5.98 x 10 ²⁴ )
uₓ = 1.67 x 10 ⁻⁶ m/s
B) Assuming earth radius as a R = 1.5 x 10 ¹¹ m
t = 365 days x 24 hr / 1 day x 60 minute / 1 hr x 60s / 1 minute = 31536000 s
t = 31536000 s
D = 2 π R = 2 π( 1.5 x 10 ¹¹ )
D = 9.4247 x 10 ¹¹ m
u₂ = D / t = 9.4247 x 10 ¹¹ / 31536000
u₂ = 29885.775 m/s
% = ( 1.67 x 10 ⁻⁶ m/s ) / (29885.775 m/s) x 100
% = 5.59 x
%
Answer:
We have the position vector given in terms of time t. r(t) = t^3*i + t^2*j
To find the velocity vector we have to differentiate r(t) with respect to time.
r'(t) = 3t^2*i + 2t*j
The vector representing acceleration is the derivative of the position vector
r''(t) = 6t*i + 2*j
When time t = 2.
The velocity vector is 3*2^2*i + 2*2*j
=> 12*i + 4*j
The speed is the absolute value of the velocity vector or sqrt(12^2 + 4^2) = sqrt (144 + 16) = sqrt 160
The acceleration vector is 6*2*i + 2*j
=> 12*i + 2*j
The required acceleration at t=2 is 12*i + 2*j and the speed is sqrt 160.
Explanation:
Can I have thx and brainliest?