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Ganezh [65]
3 years ago
10

A world-class tennis player can serve a tennis ball at 130 mi/h mi/h (about 58 m/s m/s ). The length of a tennis court is 78 ft

ft . Approximately how long does an opponent have to react from the instant the ball is served until the ball reaches the opponent?.
Physics
1 answer:
riadik2000 [5.3K]3 years ago
8 0

Answer:

The time taken is   t = 0.4103 \ s

Explanation:

From the question we are told that

  The  speed is  v  =  130 \  mi/h =  58 \  m/s

   The  length of the tennis court l =  78 \ ft = \frac{78}{3.281}= 23.8 \ m

 Generally the time taken is mathematically represented as

      t =  \frac{ 23.8}{ 58}

=>        t = 0.4103 \ s

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Energy from solar radiation may be ________ or taken in by a surface or an object.
tester [92]

Answer:

Absorbed

Explanation:

I hope this helps you

7 0
3 years ago
An object of volume 0.0882 m3 is
vfiekz [6]

The density of the fluid is 776.3 m^{-3}

<u>Explanation:</u>

Buoyant force is the upward pushing force whenever an object is trying to get immersed in fluid. So this is the force given by the fluid on the object which is trying to get immersed. The buoyant force is found to be directly proportional to the product of density of the object, volume of the object. And here the acceleration due to gravity will be acting as proportionality constant.

      Buoyant force = Density \times Volume \times Acceleration

As, buoyant force is given as 671 N and volume is 0.0882 m^{3} and acceleration is known as 9.8 m/s^{2}. Then density is

   \text { Density }=\frac{\text { Buoyant force }}{\text {Volume } \times \text {Acceleration}}

Thus,

   \text { Density }=\frac{671}{0.0882 \times 9.8}=\frac{671}{0.86436}=776.296 \mathrm{kg} / \mathrm{m}^{3}

Density is 776.3 kg m^{-3}.

7 0
3 years ago
A bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes.
zaharov [31]

Given that,

Frequency emitted by the bat, f = 47.6 kHz

The speed off sound in air, v = 413 m/s

We need to find the wavelength detected by the bat. The speed of a wave is given by formula as follows :

v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{413}{47.6\times 10^3}\\\\\lambda=0.00867\ m

or

\lambda=8.67\ mm

So, the bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes i.e. 8.67 mm.

3 0
3 years ago
A person is trying to lift a crate that has a mass of 30 kg. The normal force of the floor is currently supplying 150N of force.
alexdok [17]

Here when an object is placed on the level floor then in that case there are two forces on the object

1). Weight of object downwards (mg)

2). Normal force due to floor which will counterbalance the weight (N)

so when no force is applied on the box at that time normal force is counter balanced by weight.

Now here it is given that A person tried to lift the box upwards

So now there are two forces on the box

1). Applied force of person

2). Normal force due to ground

So now these two forces will counter balance the weight of the crate

So we can write an equation for force balance like

F_g = F_n + F_a

given that

F_g = mg

here

m = 30 kg and

g = acceleration due to gravity = 10 m/s^2

F_n = 150 N

now from above equation

30*10 = 150 + F_a

F_a = 300 - 150 = 150 N

So force applied by the person must be 150 N

7 0
3 years ago
Froghopper insects have a typical mass of around 11.3 mg and can jump to a height of 58.8 cm. The takeoff velocity is achieved a
allochka39001 [22]

Answer:

2874.33 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s

Now H-h = 0.588 - 0.002 = 0.586 m

The final velocity will be the initial velocity

v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.586\\\Rightarrow a=-\frac{0.586\times -9.81}{0.002}\\\Rightarrow a=2874.33\ m/s^2

Acceleration of the frog is 2874.33 m/s²

6 0
3 years ago
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