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marusya05 [52]
3 years ago
5

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

Physics
1 answer:
ElenaW [278]3 years ago
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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You have a sealed glass jar full of air. If you put it in the freezer, what happens to the gas pressure in the jar?
xz_007 [3.2K]
The pressure inside decreases
8 0
3 years ago
Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m. What is the force (in N) of
Lostsunrise [7]

Answer:

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Explanation:

F₂₁ = \frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}

Where;

F₂₁ is the vector force on q₁ due to q₂

K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²

r₂₁ is the unit vector

|r₂₁| is the magnitude of the unit vector

|q₁| is the absolute charge on point charge one

|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| = \sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}

(|r₂₁|)² = 148.25

F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }

      = 0.050938(0.19107i + 0.54933j) N

      = (0.00973i + 0.02798j) N

Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N

7 0
3 years ago
A team of dogs accelerates a 290kg dogsled from 0 to 6.0m/s in 3.0 s. Assume that the acceleration is constant.Part AWhat is the
Mademuasel [1]

Answer:

(a) a=2m/sec^2

(b) 5220 j

(c) 1740 watt

(d) 3446.66 watt

Explanation:

We have given mass m = 290 kg

Initial velocity u = 0 m/sec

Final velocity v = 6 m/sec

Time t = 3 sec

From first equation of motion

v = u+at

So a=\frac{v-u}{t}=\frac{6-0}{3}=2m/sec^2

(a) We know that force is given by

F = ma

So force will be F=290\times 2=580N

(b) From second equation of motion we know that

s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 2\times 3^2=9m

We know that work done is given by

W = F s = 580×9 =5220 j

(c) Time is given as t = 3 sec

We know that power is given as

P=\frac{W}{t}=\frac{5220}{3}=1740Watt

(d) Time t = 1.5 sec

So P=\frac{W}{t}=\frac{5220}{1.5}=3466.66Watt

5 0
4 years ago
Dimension equation of work
kkurt [141]

Answer:

Explanation:

Work

Other units Foot-pound, Erg

In SI base units 1 kg⋅m2⋅s−2

Derivations from other quantities W = F ⋅ s W = τ θ

Dimension M L2 T−2

Idk if this is what u are looking for but i hope this help.:)

3 0
3 years ago
Incident rays parallel to the axis of a concave mirror reflect parallel to the axis.
coldgirl [10]
No they don't.  Incident rays parallel to the axis of a concave mirror
reflect from the mirror's surface and converge at its focal point.
3 0
3 years ago
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