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3 years ago

A is a non-contact breaking of the basketball rules

1 answer:

7 0

A foul occurs when a player or coach breaks a rule covering either contact with an opposing player of unsportsmanlike action. Basketball is a so-called non-contact sport but basketball rules recognize that is not really possible to have 10 players running around a small area without making some contact.

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A mass m at the end of a spring oscillates with a frequency of 0.89 hz. when an additional 603 g mass is added to m, the frequen

MrMuchimi

The solution for this problem:

Given:

f1 = 0.89 Hz

f2 = 0.63 Hz

Δm = m2 - m1 = 0.603 kg

The frequency of mass-spring oscillation is:
f = (1/2π)√(k/m)
k = m(2πf)²

Then we know that k is constant for both trials, we have:
k = k

m1(2πf1)² = m2(2πf2)²

m1 = m2(f2/f1)²

m1 = (m1+Δm)(f2/f1)²

m1 = Δm/((f1/f2)²-1)

m 1 = 0.603/ (0.89/0.63)^2 – 1

= 0.609 kg or 0.61kg or 610 g

5 0

3 years ago

1.Convert 340 cm into m *(answer=0.34m)

Nataly [62]

Answer:

1.for the first one 100centimeters make 1 meter therefore

100cm-1m

340cm-x

340/100

=3.4

the answer is supposed to be 3.4, maybe there's a mistake with the question or the answer

2.the weight of a body is given by the formula

mass×gravity,in this case the mass is 75kg and the gravity is 9.8

weight=75×9.8

=735N

3.for this one the mass of a body is given by the formula

mass=weight/gravity

=420/9.8

=42.8kg

I hope this helps

4 0

3 years ago

Scientists often work on projects for a long time and fail to see sources of error in their research. Which process allows an ou

Marina CMI [18]

Answer:

Peer Review

Explanation: I took the test

6 0

3 years ago

Two cars are traveling along perpendicular roads, car A at 40 mi/hr, car B at 60 mi/hr. At noon, when car A reaches the intersec

Serggg [28]

Answer:

$\frac{dD}{dt} = -4 miles/hour$

negative sign indicates that the distance is decreasing with time

Explanation:

Let at any time t after noon that is 12 p.m.

distance traveled by car A = 40t

distance traveled by car B = 90-60t

then distance between the two cars at time t

$D^2= (40t)^2+(90-60t)^2$ ............1

also, at time 1 p.m.

distance $D^2= (40\times1)^2+(90-60\times1)^2$

D=50 Km

differentiating equation 1 w.r.t. t we get

$2D\frac{dD}{dt}= 2\times40t\times40+2(90-60t)(-60)$

put t= 1 and D= 50 we get

$2\times50\frac{dD}{dt}= 3200\times1-3600\times1$

$\frac{dD}{dt} = -4 miles/hour$

3 0

3 years ago

What must occur for work to be done on an object?

stiv31 [10]

The answer is D since work done must have displacement..if theres force but no displacement there will be no work done based on W=Fs

4 0

3 years ago

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A is a non-contact breaking of the basketball rules ​

A is a non-contact breaking of the basketball rules