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nadezda [96]
4 years ago
9

Three events are observed at a baseball game: I. A baseball is thrown by a pitcher. It starts from rest and is traveling at 38 m

/s as it flies toward the catcher. II. A baseball is traveling at 38 m/s when it enters the catcher's glove and stops. III. A baseball is traveling at 38 m/s when it hits a wall and bounces away from the wall at -38 m/s. IV. The change in the momentum of the baseball has the largest magnitude in which case(s)
Physics
1 answer:
SashulF [63]4 years ago
8 0

Answer:

III. A baseball is traveling at 38 m/s when it hits a wall and bounces away from the wall at -38 m/s

Explanation:

The change in momentum of the baseball is a function of its change in velocity. And can be shown mathematically as;

∆p = m (∆v)

Where ∆p change in momentum

m mass of baseball

∆v change in velocity

Which implies that the higher the change in velocity of the baseball the higher the change in momentum.

I) ∆v = 38 - 0 = 38m/s

II) ∆v = 0 - 38 = -38m/s

III) ∆v = -38-38 = -76m/s

Therefore the third event have the highest change in velocity and thus have the highest change in momentum.

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Tartaric acid is present in spinach true or false ​
sertanlavr [38]

Answer:

<h2>false</h2>

Explanation:

because Oxalic acid is present in spinach

<h2>MARK ME AS BRAINLIST</h2>
7 0
3 years ago
A car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver n
satela [25.4K]

Answer:

a) t1 = v0/a0

b) t2 = v0/a0

c) v0^2/a0

Explanation:

A)

How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0

Vf = 0

Vf = v0 - a0*t

0 = v0 - a0*t

a0*t = v0

t1 = v0/a0

B)

How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

at this point

U = 0

v0 = u + a0*t

v0 = 0 + a0*t

v0 = a0*t

t2 = v0/a0

C)

The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.

t1 = t2 = t

Distance covered by the train = v0 (2t) = 2v0t

and we know t = v0/a0

so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0

now distance covered by car before coming to full stop

Vf2 = v0^2- 2a0s1

2a0s1 = v0^2

s1 = v0^2 / 2a0

After the full stop;

V0^2 = 2a0s2

s2 = v0^2/2a0

Snet = 2v0^2 /2a0 = v0^2/a0

Now the separation between train and car

= (2v0^2)/a0 - v0^2/a0

= v0^2/a0

8 0
4 years ago
At t = 0, one toy car is set rolling on a straight track with initial position 13.5 cm, initial velocity -4.2 cm/s, and constant
Blizzard [7]
(a) We must first look at the formulas of the velocities of each toy car. v1 =
-4.2 + 2.60t. v2 = 5.20. When the two cars have equal speed, then 
v1 = v2
-4.2 + 2.60t = 5.20
2.60t = 9.40
t = 3.62 s

(b) Their speed would then be 5.20 m/s. The toy car does not change speed since it doest not have any acceleration.
(c) The two cars will pass each other when their positions are equal.
x1 = 13.5 - 4.2t + 0.5*2.60t^2
x2 = 8.5 + 5.20t
x1 = x2
13.5 - 4.2t + 1.30t^2 = 8.5 + 5.20t
1.30t^2 - 9.40t + 5.0 = 0
t = 6.65s or t = 0.58 s

--------------------

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

3 0
3 years ago
A copper sphere was moving at 25 m/s when it hit another object. This caused all of the KE to be converted into thermal energy f
Verdich [7]

Answer:

\Delta T = 0.81 ^oC

Explanation:

As we know by energy conservation

All its kinetic energy will convert into thermal energy to raise its temperature

\frac{1}{2}mv^2 = ms\Delta T

now divide both sides by mass of the object

\frac{1}{2}v^2 = s\Delta T

so change in temperature is given as

\Delta T = \frac{v^2}{2s}

\Delta T = \frac{25^2}{2\times 387}

\Delta T = 0.81^oC

3 0
3 years ago
An attacker at the base of a castle wall 3.95 m high throws a rock straight up with speed 5.00 m/s from a height of 1.60 m above
s344n2d4d5 [400]

Answer:

a) No

b) the rock must have a minimum initial speed of 6.79m/s for it to reach the top of the building.

Explanation:

Given:

Height of the wall = 3.95m

Initial height = 1.60m

Initial speed = 5.00m/s

distance between the initial height and wall top = 3.95 - 1.60 = 2.35m

Using the formula;

v^2 = u^2 + 2as ....1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance travelled

From equation 1

s = (v^2 - u^2)/2a ...2

Since the rock t moving up,

the acceleration = -g = -9.8m/s2

s = maximum height travelled

v = 0 (at maximum height velocity is zero)

Substituting into equation 2

s = (0 - 5^2)/(2×-9.8) = 1.28m

Therefore, the maximum height is 1.28 from his initial height Which is less than the 2.35m of the wall from his initial height. So the rock will not reach the top of the wall

b) Using equation 1:

u^2 = v^2 - 2as

v = 0

a = -9.8m/s

s = 2.35m. (distance between the initial height and wall top)

u^2 = 0 - 2(-9.8 × 2.35)

u^2 = 46.06

u = √46.06

u = 6.79m/s

Therefore, the rock must have a minimum initial speed of 6.79m/s

8 0
3 years ago
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