Ask question

Ask question

All categories

3 years ago

15

The teacher prepares 2.50 liters (L) of a salt solution for a class experiment. How many quarts (qt) are in 2.50 L? (1 quart =0.

943 liters)

2 answers:

Sedbober [7]3 years ago

8 0

<u>Answer</u>

2.65 quarts

<u>Explanation</u>

1 quart = 0.943 liters

To find the how many quarts are there in 2.50 litres we divide the 2.50 litres with 0.943 litres.

Number of quarts = 2.50/0.943

= 2.651113468

To the nearest hundredths the answer = 2.65 quarts

elena55 [62]3 years ago

8 0

the answer would be 2.65 quarts

You might be interested in

When a piano tuner strikes both the A on the piano and a 440 Hz tuning fork, he hears a beat every 2 seconds. The frequency of t

AysviL [449]

Answer:

The frequency is $f = 439.5 \ Hz$

Explanation:

From the question we are told that

The frequency of the tuning fork is $f_t = 440 \ Hz$

The beat period is $T = 2 \ s$

Generally the beat frequency is mathematically represented as

$f_b = \frac{1}{T}$

$f_b = \frac{1}{2}$

$f_b = 0.5 \ Hz$

The beat frequency is also represented mathematically as

$f_b = f_t - f$

Where $f$ is the frequency of the piano

So

$f = 440 - 0.5$

$f = 439.5 \ Hz$

3 0

3 years ago

Pls help I will give brainliest

Firlakuza [10]

Answer: Pretty sure the answer is B but this kind of looks like a test question.

Explanation:

8 0

2 years ago

Read 2 more answers

According to the diagram shown below, the object is:

saveliy_v [14]

The answer is B-plus.

The object is accelerating to the right, and up.

4 0

3 years ago

Read 2 more answers

A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t

Irina-Kira [14]

Answer:

part (a) $\alpha\ =\ -2.5\ rad/s^2$

part (b) N = 79.61 rev

part (c) $\tau\ =\ 23.54\ Nm$

Explanation:

Given,

Initial speed of the wheel = $w_o\ =\ 50.0\ rad/s$
total time taken = t = 20.0 sec

part (a)

Let $\alpha$ be the angular acceleration of the wheel.

Wheel is finally at the rest. Hence the final angular speed of the wheel is 0.

$\therefore w_f\ =\ w_0\ +\ \alpha t\\\Rightarrow \alpha\ =\ -\dfrac{w_0}{t}\\\Rightarrow \alpha\ =\ -\dfrac{50}{20}\\\Rightarrow \alpha\ =\ -2.5\ rad/s^2$

part (b)

Let $\theta$ be the total angular displacement of the wheel from initial position till the rest.

$\therefore \theta\ =\ w_0t\ +\ \dfrac{1}{2}\alphat^2\\\Rightarrow \theta\ =\ 50\times 20\ -\ 0.5\times 2.5\times 20^2\\\Rightarrow \theta\ =\ 500\ rad$

We know, 1 revolution = $2\pi$ rad

Let N be the number of revolution covered by the wheel.

$\therefore N\ =\ \dfrac{\theta}{2\pi}\\\Rightarrow N\ =\ \dfrac{500}{2\times 3.14}\\\Rightarrow N\ =\ 79.61\ rev$

Hence the 79.61 revolution is covered by the wheel in the 20 sec.

part (c)

Given,

Mass of the pole = m = 4 kg
Length of the pole = L = 2.5 m
Angle of the pole with the horizontal axis = $\theta\ =\ 60^o$

Now the center of mass of the pole = $d\ =\ \dfra{L}{2}\ =\ \dfrac{2.5}{2}\ =\ 1.25\ m$

Weight component of the pole perpendicular to the center of mass = $F\ =\ mgcos\theta$

$\therefore \tau\ =\ F\times d\\\Rightarrow \tau\ =\ 4\times 9.81\times cos60^o\times 1.25\\\Rightarrow \tau\ =\ 23.54\ Nm$

3 0

3 years ago

A tennis ball is beaten downward with an initial speed of 5 m/s.

Anastasy [175]

Answer:

<h2>34.4m/s</h2>

Explanation:

Step one:

given data

initial speed u= 5m/s

time t= 3seconds

acceleration due to gravity g= 9.8m/s^2

Required:

the final velocity v

Step two:

applying the first equation of motion

v=u+gt-----------we used + because the ball is falling with gravity

v=5+9.8*3

v=5+29.4

v=34.4

The final velocity is 34.4m/s

4 0

3 years ago