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satela [25.4K]
3 years ago
12

Can someone help me plzzz..whoever answers the best will be marked as brainliest.....​

Physics
1 answer:
IrinaK [193]3 years ago
6 0

Answer:

1) 3 applications of pressure in daily life are :-

● The area of sharp edge of knife, scissor or handsaws are much less then blunt edge. So, for same total force pressure is more for sharp edges than the blunt one. Hence sharp knife, scissors etc, cuts easily than a blunt one.

●Broad handles in bags and suitcases are provided for the comfort. Broad handles have large area. So, the pressure exerted on hands and shoulders would be small while carrying the bags and the suitcases.

●Trucks carrying heavy loads have more than four tyres. More tyres in case of trucks increase the area of contact with the road. This results in reduced pressure on the tyres.

2) Area of the surface which is on ground = 1.5×1

= 1.5m^2

Mass of the block = 300kg

Force applied by the block = Mass × g = 300×10

= 3000N (where g = acceleration due to gravity )

Pressure = Force applied / Area of the surface

= 3000N / 1.5m^2

= 2000 Pa

3)

a) The above experiment signifies that more the area of the surface of an object , less the pressure an object applies.

b) B exerts the minimum pressure because the area of its surface to ground is greater than others & as it has more area of surface , it exerts less pressure. ( area is inversely proportional to pressure )

c) D exerts the maximum pressure because the area of its surface to ground is lesser than others & as it has less area of surface , it exerts more pressure. ( area is inversely proportional to pressure )

d) It depend upon the way an object is kept on ground. If an object is kept in such a way dat the area of the surface to the ground is more , then pressure will be least exerted .If an object is kept in such a way dat the area of the surface to the ground is less, then pressure will be exerted more .

e) Do it yourself . only i will suggest that make the tip of the cone ( which is to the ground ) more narrower.

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(4.65×10⁶ Pa) V / (21 + 273.15) K = (1.06×10⁶ Pa) (3V) / T

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b.

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A 30 kg student drops down from the monkey bars. The acceleration due to gravity is -9.8 m/s^2. Neglecting air drag, what is the
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The net force on the student is A) -294 N

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m = 30 kg is the mass of the student

g=-9.8 m/s^2 is the acceleration of gravity, where the negative sign means the direction is downward

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F=(30)(-9.8)=-294 N

And since this is the only force acting on the student, it is also the net force on him.

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3 years ago
Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m

The beetle can jump to a height of <u>2.1 m</u>



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