When one "throws” a punch, the forearm applies force to the fist. Consider a(n) 0.75 kg fist that goes from rest to a

How can liquid pressure phenomena be used in daily life. Give one example with explanation of working?

N76 [4]

Answer:

An example in which liquid pressure phenomena can be used in daily life is in Water blasting

Explanation:

Water blasting refers application of pressurized water to remove materials from the surface of objects.

There are different varieties of water blasting, including;

Hydrocleaning; Cleaning enabled by the use of high pressure water

Hydrodemolition; Demolition or removal of concrete using pressurized water

Hydrojetting; The spraying of water under pressure on surfaces in order to remove surface contaminants.

8 0

3 years ago

Which statement best explains the path light takes as it travels? A. Light takes a curved path through matter and takes a straig

Zolol [24]

Answer:

it was "The light bends because it changes speed." for me i took the test

Explanation:

i show the thing. i hope this helps

4 0

2 years ago

Read 2 more answers

__________ is the most common type of stretching.

Verizon [17]

Answer: static stretching

Explanation:

e.g rubberband

6 0

3 years ago

Read 2 more answers

Pls help me with this problem!!

Olenka [21]

Answer:

v = 19.6 m/s.

Explanation:

Given that,

The radius of the circle, r = 5 m

The time period of the ball, T = 1.6s

We need to find the ball's tangential velocity.

The formula for the tangential velocity is given by :

$v=\dfrac{2\pi r}{T}$

Putting all the values in the above formula

$v=\dfrac{2\pi \times 5}{1.6}\\\\v=19.6\ m/s$

So, the tangential velocity of the ball is 19.6 m/s. Hence, the correct option is (c).

7 0

3 years ago

Suppose a car travels 106 km at a speed of 28 m/s and uses 1.9 gals of gasoline in the process. Only 30% of the gasoline goes in

USPshnik [31]

Answer:

a) The magnitude of the force is 968 N

b) For a constant speed of 30 m/s, the magnitude of the force is 1,037 N

Explanation:

<em>NOTE: The question b) will be changed in other to give a meaningful answer, because it is the same speed as the original (the gallons would be 1.9, as in the original).</em>

Information given:

d = 106 km = 106,000 m

v1 = 28 m/s

G = 1.9 gal

η = 0.3

Eff = 1.2 x 10^8 J/gal

a) We can express the energy used as the work done. This work has the following expression:

$W=F\cdot d$

Then, we can derive the magnitude of the force as:

$F=\frac{W}{d}=\frac{\eta\cdot (G\cdot Eff)}{d}=\frac{0.3*1.9*(1.8*10^8)}{106*10^3} =968\,N$

b) We will calculate the force for a speed of 30 m/s.

If the force is proportional to the speed, we have:

$F_2=F_1(\frac{v_2}{v_1} )=968(\frac{30}{28} )=968*1.0714=1,037\,N$

6 0

3 years ago