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2 years ago

The highest points you can score in a B.B game. true or false. (help a bestie out )

Physics

2 answers:

vichka [17]2 years ago

8 0

Answer:

Uh true :)

Explanation:

ivanzaharov [21]2 years ago

4 0

Answer:

your question is incomplete.

most points by a player? most you *can* score? NBA, college, high school?

really, the possibilities are endless.

Wilt Chamberlain scored 100 points in a game.

There is no upper limit.

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A baseball (m=140g) traveling 32m/s moves a fielders

posledela

Answer:

Average force, F = 286.72 N

Explanation:

Given that,

Mass of the baseball, m = 140 g = 0.14 kg

Speed of the ball, v = 32 m/s

Distance, h = 25 cm = 0.25 m

We need to find the average force exerted by the ball on the glove. It is solved using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh$

F = mg

$\dfrac{1}{2}mv^2=Fh$

$F=\dfrac{mv^2}{2h}$

$F=\dfrac{0.14\times (32)^2}{2\times 0.25}$

F = 286.72 N

So, the average force exerted by the ball on the glove is 286.72 N. Hence, this is the required solution.

7 0

3 years ago

To practice tactics box 13.1 hydrostatics. in problems about liquids in hydrostatic equilibrium, you often need to find the pres

OLEGan [10]

Answer:

A. The pressure denoted as Pa and Pb at the surfaces of A and B in the tube is

PA= Pgas

PB= Patmos

B. The second sketch

C. The gas pressure is

Pgas= Patmos+ rho.g(h2-h1)

= 1atm + rho.g (h2-h1)

Explanation:

5 0

3 years ago

A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for

nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ( $W$ ), measured in newtons, is:

$W = m\cdot g$ (1)

Where:

$m$ - Mass of the block of ice, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 14\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the magnitudes of the weight and normal force of the block of ice are, respectively:

$N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$N = W = 137.298\,N$

And the pull force is:

$F_{pull} = 98\,N$

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

$m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}$ (2)

Where:

$v_{o}$ , $v_{f}$ - Initial and final speeds of the block, measured in meters per second.

$\Sigma F$ - Horizontal net force, measured in newtons.

$\Delta t$ - Impact time, measured in seconds.

Now we clear the final speed in (2):

$v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $m = 14\,kg$ , $\Sigma F = 98\,N$ and $\Delta t = 1.40\,s$ , then final speed of the ice block is:

$v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}$

$v_{f} = 9.8\,\frac{m}{s}$

The final speed of the block of ice is 9.8 meters per second.

6 0

2 years ago

Convert 6.95×105 cal to joules. recall that 1 cal=duncan takes a break from studying and goes to the gym to swim laps. if swimmi

quester [9]

For this problem, let's use the approach of dimensional analysis. This technique is done by cancelling out like units that appear both on the numerator and the denominator side. As a result, this technique will let you know that your final answer conforms to what parameter is asked. In this case, the final answer should be in kJ. We use the conversion: 1000 cal = 4.184 kJ The solution is as follows:

<span>6.95×10</span>⁵<span> cal * 4.184 kJ/1000 cal = 2,907.88 kJ</span>

3 0

3 years ago

The tropic of cancer is to the tropic of capricorn as the arctic circle is to the

Y_Kistochka [10]

Answer:

Antarctic Circle

Explanation:

The Tropic of Cancer, which is also referred to as the Northern Tropic, is the most northerly circle of latitude on Earth at which the Sun can be directly overhead. This occurs on the June solstice, when the Northern Hemisphere is tilted toward the Sun to its maximum extent.

Tropic of Capricorn Is it Southern Hemisphere counterpart, marking the most southerly position at which the Sun can be directly overhead.

7 0

3 years ago