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soldi70 [24.7K]
2 years ago
7

Two ice skaters, Lilly and John, face each other while stationary and push against each others hands. John's mass is twice the m

ass of Lilly. How do their speeds compare after the push offs?
Physics
1 answer:
VladimirAG [237]2 years ago
7 0

Answer:

lily's speed would be twice john's speed

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A person's height will increase from birth until about age 25, and it may decrease starting at about age 70. This is an example
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A submarine completed a 450 km training with an average speed of 50 km/h. For the first 180 km, it travelled at an average speed
Kryger [21]

Answer:

45km/hr

Explanation:

Total distance=450km

Total speed=50km/hr

Total time= distance/speed

=450/50

=9hrs

distance a=180km

speed a=60km/hr

Time a=180/60

=3hrs

Distance b=450-180=270km

Speed b=?

Time b=270/speed b

Total time=time a + time b

9=3+(270/speed b)

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2 years ago
What are the magnitude and direction of the force the pitcher exerts on the ball? (enter your magnitude to at least one decimal
murzikaleks [220]
Details are missing in the question. Complete text of the problem:

"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.

(a) Through what distance does it move before its release? (m)
(b) What are the magnitude and direction of the force the pitcher exerts on the ball? (Enter your magnitude to at least one decimal place.)"


Solution

(a) The pitcher accelerates the baseball from rest to a final velocity of v_f = 16.5 m/s, so \Delta v=16.5 m/s, in a time interval of \Delta t = 181 ms=0.181 s. The acceleration of the ball in the horizontal direction (x-axis) is therefore

a_x =  \frac{\Delta v}{\Delta t}= \frac{16.5 m/s}{0.181 s}=91.2 m/s^2

And the distance covered by the ball during this time interval, before it is released, is:

S= \frac{1}{2} a_x (\Delta t)^2 = \frac{1}{2} (91.2 m/s^2)(0.181 s)^2=1.49 m

(b) For this part we need to consider also the weight of the ball, which is W=mg=2.28 N

From this, we find its mass: m= \frac{W}{g}= \frac{2.28 N}{9.81 m/s^2}=0.23 Kg

Now we can calculate the magnitude of the force the pitcher exerts on the ball. On the x-axis, we have

F_x = m a_x = (0.23 kg)(91.2 m/s^2)=20.98 N

We also know that the ball is moving straight horizontally. This means that the vertical component of the force exerted by the pitcher must counterbalance the weight of the ball (acting downward), in order to have a net force of zero along the y-axis, and so:

F_y=W=mg=2.28 N (upward)

So, the magnitude of the force is

F= \sqrt{F_x^2+F_y^2}=  \sqrt{(20.98N)^2+(2.28N)^2}=21.2 N

To find the direction, we should find the angle of F with respect to the horizontal. This is given by

\tan \alpha =  \frac{F_y}{F_x}= \frac{2.28 N}{20.98 N}=0.11

From which we find \alpha=6.2^{\circ}

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A rigid, 2.50 L bottle contains 0.458 mol He. The pressure of the gas inside the bottle is 1.83 atm. If 0.713 mol Ar is added to
stellarik [79]
<span>These are inert gases, so we can assume they don't react with one another. Because the two gases are also subject to all the same conditions, we can pretend there's only "one" gas, of which we have 0.458+0.713=1.171 moles total. Now we can use PV=nRT to solve for what we want.

The initial temperature and the change in temperature. You can find the initial temperature easily using PV=nRT and the information provided in the question (before Ar is added) and solving for T.

You can use PV=nRT again after Ar is added to solve for T, which will give you the final temperature. The difference between the initial and final temperatures is the change. When you're solving just be careful with the units!
 
SIDE NOTE: If you want to solve for change in temperature right away, you can do it in one step. Rearrange both PV=nRT equations to solve for T, then subtract the first (initial, i) from the second (final, f):

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