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jeka57 [31]
3 years ago
7

If there is a doubt on a boundary call in tennis, you should do what?

Physics
1 answer:
Ronch [10]3 years ago
3 0

Answer: Each player is responsible for all decisions in her half of the Court. She should be completely honest on all “calls” but, if in doubt, she should give her opponent the benefit of the doubt and play the ball as good. You should not play a let.

Explanation:

You might be interested in
A student weighing 700 N climbs at constant speed to the top of an 8 m vertical rope in 10 s. The average power expended by the
nasty-shy [4]

Answer:

The average power the student expended to overcome gravity is 560W, Watts is a units of work it is Joules /time or kg*m^{2}/ s^{3}

Explanation:

Weight = 700N

F= m*g*h \\m*g= 700N\\F= 700N*h\\F=700N*8m\\F= 5600 N*m \\F=5600 J

The power is the work in (Joules) or (N*m) in a determinate time (s) to get Watts (W) units for work

Work= \frac{5600 J}{10s} = \frac{5600 \frac{kg*m^{2} }{s^{2} } }{10 s}  \\Work= 560\frac{kg*m^{2} }{s^{3} } \\Work =560 W

3 0
3 years ago
Urgent if earths gravitational field strength is 10 N/kg how many newtons does a 30kg object weigh
egoroff_w [7]

weight of 30 Kg object= 300 N

Explanation:

Weight= m I

m= mass=30 kg

I= gravitational filed strength= 10 N/kg

weight=30 (10)

weight=300 N

5 0
3 years ago
4. A light string is attatched to a heavy rope, and the whole thing is pulled tight. A wave is sent along the light string. When
ale4655 [162]

Answer:

The correct answer to the question is (A)

When it hits the heavy rope, compared to the wave on the string, the wave that propagates along the rope has the same (A) frequency

Explanation:

The speed of a wave in a string is dependent on the square root of the tension ad inversely proportional  to the square root of the linear density of the string. Generally, the speed of a wave through a spring is dependent on the elastic and inertia properties of the string

v = \sqrt{ \frac{T}{\mu } } =  \sqrt{ \frac{T}{m/L } }

Therefore if the linear density of the heavy rope is four times that of light rope the velocity is halved and since

v = f×λ therefore  v/2 = f×λ/2

Therefore the wavelength is halved, however the frequency remains the same as continuity requires the frequency of the incident pulse vibration to be transmitted to the denser medium for the wave to continue as the wave is due to vibrating particles from a source for example

7 0
3 years ago
A 1.8 kg uniform rod with a length of 90 cm is attached at one end to a frictionless pivot. It is free to rotate about the pivot
Leokris [45]

Answer:

a. 32.67 rad/s²  b. 29.4 m/s²

Explanation:

a. The initial angular acceleration of the rod

Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.

So, Iα = WL

mL²α/3 = mgL

dividing through by mL, we have

Lα/3 = g

multiplying both sides by 3, we have

Lα = 3g

dividing both sides by L, we have

α = 3g/L

Substituting the values of the variables, we have

α = 3g/L

= 3 × 9.8 m/s²/0.9 m

= 29.4/0.9 rad/s²

= 32.67 rad/s²

b. The initial linear acceleration of the right end of the rod?

The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²

5 0
3 years ago
the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during
Ivanshal [37]

Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

Mass of the baseball after struck by the bat, M = 900 g

Velocity of the baseball after struck by the bat, v = 47 m/s

According to the conservation of momentum,

Mv+mu=Mv_1+mv_2

(900 x 47) + (200 x -30)  = (900 x v_1) + (200 x v_2)

36300 =  (900 x v_1) + (200 x v_2)

9v_1 + 2v_2 = 363 ..............(i)

9v_1 = 363 - 2v_2

v_1=\frac{363 - 2v_2}{9}

The mathematical expression for the conservation of kinetic energy is

\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2

\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2    ................(ii)

$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$  

21681 = 9v_1^2+2v_2^2

Substituting (i) in (ii)

21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2

(363-2v_2)^2+18v_2^2=195129

(363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0

22v_2^2-145v_2-63360=0

Solving the equation, we get

v_2=96 \ m/s, -30 \ m/s

The negative velocity is neglected.

Therefore, substituting 96 m/s for v_2 in (i), we get

v_1=\frac{363-(2 \times 96)}{9}

     = 19

Thus, only impulse of importance is used to find final velocity.

8 0
3 years ago
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