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2 years ago

Define measurement with 10 points

Physics

1 answer:

Gemiola [76]2 years ago

6 0

Answer:

the process of comparing unknown quantities with known standards quantities are known as measurement

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When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not n

Harlamova29_29 [7]

Answer:

because the mass of the apple is very less compared to the mass of earth. Due to less mass the apple cannot produce noticable acceleration in the earth but the earth which has more mass produces noticable acceleration in the apple. thus we can see apple falling on towards the earth but we cannot see earth moving towards the apple.

6 0

2 years ago

stephen buys a new moped . he travels 4km south and then 6km east. how far does he need to go to get back where he started??

stiks02 [169]

Answer:

My answer is 7.2 km

Explanation:

When Stephen goes to the south and then to the east, he is drawing a right triangle, where the 4 km and 6 km sides are the cathetus of a right triangle.

Then we use the Pithagorean theorem to solve this problem. We need to find the hypotenuse.

c² = a² + b²

c² = 4² + 6²

c² = 16 + 36

c² = 52

c = 7.2 km

7 0

3 years ago

Vector A has a magnitude of 30 units. Vector B is perpendicular to vector Aand has a magnitude of 40 units. What would the magni

Fudgin [204]

Answer:

$|\vec A + \vec B| = 50 units$

Explanation:

As we know that magnitude of two vectors is given as

$|\vec A + \vec B| = \sqrt{A^2 + B^2 + 2AB cos\theta}$

here we know that

A = magnitude of vector A

B = magnitude of vector B

$\theta$ = angle between two vectors

so here we know that

A = 30 units

B = 40 units

angle = 90 degree

so we have

$|\vec A + \vec B| = \sqrt{30^2 + 40^2 + 2(30)(40)cos90}$

$|\vec A + \vec B| = \sqrt{30^2 + 40^2}$

$|\vec A + \vec B| = 50 units$

3 0

3 years ago

Can someone help me plzzz..<br>whoever answers the best will be marked as brainliest.....

IrinaK [193]

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

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.

6 0

3 years ago

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

5 0

3 years ago