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scoray [572]
3 years ago
9

A basketball is tossed upwards with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m

, end text, divided by, start text, s, end text, end fraction. We can ignore air resistance. What is the maximum height reached by the basketball from its release point?
Physics
1 answer:
Hatshy [7]3 years ago
4 0

Answer:

The last one

Explanation:

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Which statement best describes how the temperature of the oceans surface water varies
lorasvet [3.4K]
If the temperature is high there is less water because it evaporates if it is cloudy it is more because it doesn't evaporate
8 0
3 years ago
determine the maximum angle theta for which the light rays incident on the end of the optical fiber of radius 1 mm are subhect t
Vesna [10]

Answer:

Explanation:

Let the critical angle be C .

sinC = 1 / μ where μ is index of refraction .

sinC = 1 /1.2

= .833

C = 56°

Then angle of refraction r = 90 - 56 = 34 ( see the image in attached file )

sin i / sinr = 1.2 , i is angle of incidence

sini = 1.2 x sinr = 1.2 x sin 34 = .67

i = 42°.  

7 0
3 years ago
4. Calculate the total resistance for two 180 ohm resistors connected in<br> parallel
solmaris [256]

Answer:

90 ohms

Explanation:

1/r = 1/180 + 1/180

1/r= 2/180

take the reciprocal of 2/180 which is 180/2 and its 90 ohms

3 0
3 years ago
What is the mass of an object if a force of 34 N produces an acceleration of 4.0 m/s squared​
mrs_skeptik [129]

Answer:

8.5 kg

Explanation:

F = ma

34 N = m(4.0m/s^2)

m = 8.5 kg

6 0
3 years ago
Ball A is dropped from the top of a building of height h at the same instant that ball B is thrown vertically upward from the gr
DiKsa [7]

Answer:

(2/3) times height collision occur

Explanation:

for ball A

from the kinematic equation

the distance of ball A is

x_A = v_0 t + \frac{1}{2} at^2

v_0* t= velocity ( time ) = distance

since ball is at height, the above equation changes as

x_A = H - \frac{1}{2} gt^2

for ball B

xB = v0 t - \frac{1}{2} gt^2

the condition of collision is

xA = xB

vA = - 2vB (given)

from the kinematic equation

the speed of the ball A is

v_A = u- gt

since initial speed of the ball A is zero

, so

v_A = -gt

the speed of the ball B is

v_B = v_0 - gt

sincev_A = - 2v_B

   -gt = -2 ( v_0 - gt)

-gt = -2 v_0 +2gt

3gt =2 v_0

t = \frac{2v_0}{3g}

since x_A = x_B

H -  \frac{1}{2} gt^2= v_0 t - \frac{1}{2} gt^2

H = v_0 t

= v_0 (2v_0/3g)

= \frac{2 v_0^2}{ 3g}

x_A = 2 (\frac{v_0^2}{ 3g})- \frac{1}{2} gt^2

  = 4 \frac{v_0^2}{9g} = (2/3) H

so, (2/3) times height collision occur

3 0
3 years ago
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