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3 years ago

Why was bowling one of the first racially integrated sports

Physics

1 answer:

expeople1 [14]3 years ago

3 0

Answer:

bowling was one of the first racially integrated sports because bowling alleys were primarily located in urban areas. Racial integration was inevitable since it was promoted by US Armed Forces during 1940s and its image as sport for the common man made it a choice of activity for Americans!!

Explanation:

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What is the density of an object with mass of 60 g and a volume of 2cm3

Artemon [7]

Answer:

D=mass /volume

=60/0.002

=30000

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2 years ago

Which of the outer planets would you most likely visit? Why? What would you see on your visit?

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3 years ago

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How are astronomers able to observe and study black holes?

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3 years ago

Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting li

erica [24]

Answer:

(a) 10 s

(b) 236.5 m

(c) Kathy's speed = 47.3 m/s

Stan's speed = 42.9 m/s

Explanation:

<u>Given:</u>

$u_k$ = initial speed of Kathy = 0 m/s

$u_s$ = initial speed of Stan = 0 m/s
$a_k$ = acceleration of Kathy = $4.73\ m/s^2$

$a_s$ = acceleration of Stan = $3.9\ m/s^2$

<u>Assumptions:</u>

$v_k$ = final speed of Kathy when see catches Stan
$v_s$ = final speed of Stan when Kathy catches him
$s_k$ = distance traveled by Kathy to catch Stan
$s_s$ = distance traveled by Stan when Kathy catches him
$t_k$ = time taken by Kathy to catch Stan = $t$
$t_s$ = time interval in which Kathy catches Stan = $t+1$

Part (a):

Kathy will catch Stan only if the distances traveled by each of them are equal at the same instant.

$\therefore s_s=s_k\\\Rightarrow u_st_s+\dfrac{1}{2}a_st_s^2=u_kt_k+\dfrac{1}{2}a_kt_k^2\\ \Rightarrow (0)(t+1)+\dfrac{1}{2}(3.9)(t+1)^2=(0)(t)+\dfrac{1}{2}(4.73)t^2\\ \Rightarrow \dfrac{1}{2}(3.9)(t+1)^2=\dfrac{1}{2}(4.73)t^2\\\Rightarrow (3.9)(t+1)^2=(4.73)t^2\\\Rightarrow \dfrac{(t+1)^2}{t^2}=\dfrac{4.73}{3.9}\\\textrm{Taking square root in both sides}\\\dfrac{t+1}{t}= 1.1\\\Rightarrow t+1=1.1t\\\Rightarrow 0.1t = 1\\\Rightarrow t = 10\\$

Hence, Kathy catches Stan after 11 s from the Stan's starting times.

Part (b):

Distance traveled by Kathy to catch Stan will be distance the distance traveled by her in 10 s.

$s_s = u_kt_k+\dfrac{1}{2}a_kt_k^2\\\Rightarrow s_s= (0)(t)+\dfrac{1}{2}(4.73)t^2\\\Rightarrow s_s= \dfrac{1}{2}(4.73)(10)^2\\\Rightarrow s_s= 236.5$

Hence, Kathy traveled a distance of 236.5 m to overtake Stan.

Part (c):

$v_k = u_k+a_kt_k\\\Rightarrow v_k = 0+(4.73)(t)\\\Rightarrow v_k = (4.73)(10)\\\Rightarrow v_k =47.3$

The speed of Kathy at the instant she catches Stan is 47.3 m/s.

$v_s = u_s+a_st_s\\\Rightarrow v_s = 0+(3.9)(t+1)\\\Rightarrow v_s = (3.9)(10+1)\\\Rightarrow v_s =42.9$

The speed of Stan at the instant Kathy catches him is 42.9 m/s.

7 0

3 years ago

A force is acting on a 3-kg object during 5 seconds. If the object initially has a speed of 1.5m/s and after applying the force

uranmaximum [27]

Answer:

1.5 N

Explanation:

You've left us to guess what the question is. I will Assume it is what's the force?

Givens

m = 3 kg

vi = 1.5 m/s

vf = 4 m/s

t = 5 seconds

Formula

F = m * (vf - vi)/t

Solution

F = 3 * (4 - 1.5) / 5

F = 1.5 N

3 0

2 years ago