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3 years ago

What is the standard used by scientist for mass

2 answers:

6 0

The standard measurement system used by the scientists is called: International System Of Units, or SI. The units of measurement of a few physical quantities are considered to be base units. (Also do you need the base units too cuz I can help you with that) :)

Serhud [2]3 years ago

6 0

The standard for measuring mass is
Kilograms or grams
Depending on the mass

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Pls help quickly

emmasim [6.3K]

Answer: frosted glass

Explanation:

8 0

2 years ago

Read 2 more answers

A person doing chin-up weighs 700.0 N, disregarding the weight of the arms. During the first 25.0 cm of the lift, each arm exert

dlinn [17]

First, we will get the resultant force:
The direction of the force due to the person's weight is vertically down.
weight of person = 700 newton

Assume that the force exerted by the arms has a vertically upwards direction.
Force exerted by arms = 2*355 = 710 newtons

Therefore, the resultant force = 710 - 700 = 10 newtons (in the vertically upwards direction)

Now, we will get the mass of the person.
weight = 700 newtons
weight = mass * acceleration due to gravity
700 = 9.8*mass
mass = 71.428 kg

Then we will calculate the acceleration of the resultant force:
Force = mass*acceleration
10 = 71.428*acceleration
acceleration = 0.14 m/sec^2

Finally, we will use the equation of motion to get the final speed of the person.
V^2 = U^2 + 2aS where:
V is the final velocity that we need to calculate
U is the initial velocity = 0 m/sec (person starts at rest)
a is the person's acceleration = 0.14 m/sec^2
S is the distance covered = 25 cm = 0.25 meters

Substitute with the givens in the above equation to get the final speed as follows:
V^2 = U^2 + 2aS
V^2 = (0)^2 + 2(0.14)(0.25)
V^2 = 0.07
V = 0.2645 m/sec

Based on the above calculations:
The person's speed at the given point is 0.2645 m/sec

4 0

3 years ago

Read 2 more answers

A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri

exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

$U_i +K_i = U_f + K_f$

where

$U_i$ is the initial potential energy, at the top

$K_i$ is the initial kinetic energy, at the top

$U_f$ is the final potential energy, at halfway

$K_f$ is the final kinetic energy, at halfway

The equation can be rewritten as

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m = 2.7 kg is the mass of the block

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 0.54$ is the initial height

u = 0 is the initial speed

$h_f = 0.25 m$ is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

$v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s$

The total mechanical energy of the block can be calculated from the initial conditions, and it is

$E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J$

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

$E=E_e = \frac{1}{2}kx^2$

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

$x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m$

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

$E=E_e = 14.3 J$