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IrinaVladis [17]

3 years ago

9

In an experiment, you use 0.457 L from one container and 0.432 L from another container of acetic acid in a reaction. How many M

ILLILITERS (mL) of acetic acid are you using?

1 answer:

uysha [10]3 years ago

4 0

Answer: 0.889 Milliliters

Explanation:

Added

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Why the Earth is dependent on the Sun?

makvit [3.9K]

The Earth revolves (orbits) around the Sun in one year. The Earth's rotation axis is tilted relative to the plane of its orbit around the Sun. This tilt of the Earth is responsible for the seasons as the Earth orbits the Sun. The Sun provides energy that sustains all life on Earth.

4 0

3 years ago

A mass of 0.34 kg is fixed to the end of a 1.4 m long string that is fixed at the other end. Initially at rest, he mass is made

frozen [14]

At time $t$ seconds, the mass has angular speed

$\omega = \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) t$

and hence linear speed

$v = (1.4\,\mathrm m) \omega = (1.4\,\mathrm m) \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) t$

After 8 s, its linear speed is

$v = (1.4\,\mathrm m) \left(3.31\dfrac{\rm rad}{\mathrm s^2}\right) (8\,\mathrm s) = 37.072 \dfrac{\rm m}{\rm s} \approx 37 \dfrac{\rm m}{\rm s}$

and has centripetal acceleration with magnitude

$a = \dfrac{v^2}{1.4\,\rm m} \approx 981.667\dfrac{\rm m}{\mathrm s^2} \approx 980 \dfrac{\rm m}{\mathrm s^2}$

To maintain this linear speed, by Newton's second law the required centripetal force should have magnitude

$F = (0.34\,\mathrm{kg}) a \approx 333.767\,\mathrm N \approx \boxed{330 \,\mathrm N}$

5 0

2 years ago

A stun gun or TASER is designed to put out a few seconds worth of electric pulses that impress a

iren [92.7K]

Answer:

Resistance = 400000 Ohms

Explanation:

Given the following data;

Voltage = 1200 V

Current = 3 mA = 3/1000 = 0.003 A

To find the resistance;

Voltage = current * resistance

Resistance = 1200/0.003

Resistance = 400000 Ohms

5 0

3 years ago

What force is required to accelerate a 1840 kg car from 4.77 m/s to 23.5 m/s,

neonofarm [45]

A :-) for this question , we should apply
a = v - u by t
Given - u = 4.77 m/s
v = 23.5 m/s
t = 5.18 m/s
Solution -
a = v - u by t
a = 23.5 - 4.77
a = 28.27 m/s^2

.:. The acceleration is 28.27 m/s^2

7 0

3 years ago

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:

a)

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,

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

b)

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

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

c)

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:

$E=E_e = 14.3 J$

while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:

$E=U_f = mg h_f$

where $h_f$ is the maximum height reached. Solving for this quantity, we find

$h_f = \frac{E}{mg}=\frac{14.3}{(2.7)(9.8)}=0.54 m$

which is the initial height: this is correct, because the total mechanical energy is conserved, so the block must return to its initial position.

Learn more about kinetic and potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

brainly.com/question/6536722

#LearnwithBrainly

5 0

3 years ago