Ask question

Ask question

All categories

2 years ago

5

Martin is conducting an experiment. His first test gives him a yield of 5.2 grams. His second test gives him a yield of 1.3 gram

s. His third test gives him a yield of 8.5 grams. On average, his yield is 5.0 grams, which is close to the known yield of 5.1 grams of substance. Which of the following are true?

1 answer:

Bogdan [553]2 years ago

4 0

Complete Question

Martin is conducting an experiment. His first test gives him a yield of 5.2 grams. His second test gives him a yield of 1.3 grams. His third test gives him a yield of 8.5 grams. On average, his yield is 5.0 grams, which is close to the known yield of 5.1 grams of substance. Which of the following are true?

A His results are accurate but not precise.

B His results are neither accurate nor precise.

C His results are both accurate and precise

D His results are precise but not accurate.

Answer:

Correct option is A

Explanation:

From the question we are told that

The yield of the first test $k = 5.2 \ g$

The yield of the second is $u = 1.3 \ g$

The third yield is $p = 8.5 \ g$

The average yield $A = 5.0 \ g$

The know yield is $A_S = 5.1 \ g$

From the data given we see that

$A_S \ne A$

Since his average yield is closer to the known yield then the answer is accurate

But since the yield for each test are not repeated the answer is not precise

So the answer is accurate but not precise

You might be interested in

A hose has been clamped so that the area at the clamp is only one quarter the area of the rest of the hose. When we ignore the v

Bad White [126]

Answer:

1

Explanation:

A hose has been clamped so that the area at the clamp is only one quarter the area of the rest of the hose. When we ignore the viscosity of water, the ratio of the volume of water delivered per unit time when the clamp is on to the volume of water delivered per unit time without the clamp is 1 as continuity says the same amount of water must flow out

4 0

2 years ago

A typical raindrop is much more massive than a mosquito and falling much faster than a mosquito flies. How does a mosquito survi

seropon [69]

Answer:

Part a)

$v = 7.94 m/s$

Part b)

$a = 992.6 m/s^2$

Explanation:

Part a)

As we know that we can use momentum conservation for this

so we will have

$m_1v_1 = (m_1 + m_2)v$

$(50m)8.1 = (50m + m)v$

$v = 7.94 m/s$

Part b)

As we know that acceleration is rate of change in velocity

so we have

$a = \frac{v_f - v_i}{t}$

so we have

$a = \frac{7.94 - 0}{8 \times 10^{-3}}$

$a = 992.6 m/s^2$

5 0

1 year ago

Read 2 more answers

suppose that 273 g of one of the substances listed above displaces 26 mL of water. What is the substance?

guajiro [1.7K]

<span>The unknown substance is silver. I don't see a list of available substances, but let's see if there's something reasonable available that will match. First, let's calculate the density of the unknown substance. Density is mass per volume, so 273 g / 26 mL = 10.5 g/mL Looking up a list of elements sorted by density, I see the following: 10.07 Actinium 10.22 Molybdenum 10.5 Silver 11.35 Lead And silver at 10.5 g/ml is a very nice match for the unknown substances' density of 10.5 g/ml.</span>

6 0

2 years ago

Read 2 more answers

On a warm summer day (31 ∘c), it takes 4.60 s for an echo to return from a cliff across a lake. on a winter day, it takes 5.00 s

xenn [34]

The question is missing, but I guess the problem is asking for the distance between the cliff and the source of the sound.

First of all, we need to calculate the speed of sound at temperature of $T=31^{\circ}C$ :
$v=(331+0.60 T) m/s = (331+0.6 \cdot 31) m/s = 349.6 m/s$

The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):
$v= \frac{2L}{t}$
where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff:
$L= \frac{vt}{2}= \frac{(349.6 m/s)/4.60 s)}{2}= 804.1 m$

6 0

2 years ago

The Millersburg Ferry (m = 13000.0 kg loaded) is travelling at 11 m/s when the engines are put in reverse. The engineproduces a

Pie

Explanation:

It is given that,

Mass of Millersburg Ferry, m = 13000 kg

Velocity, v = 11 m/s

Applied force, F = 10⁶ N

Time period, t = 20 seconds

(a) Impulse is given by the product of force and time taken i.e.

$J=F.\Delta t$

$J=10^6\ N\times 20\ s$

$J=2\times 10^7\ N-s$

(b) Impulse is also given by the change in momentum i.e.

$J=\Delta p=p_f-p_i$

$J=p_f-p_i$

$p_f=J+p_i$

$p_f=2\times 10^7\ N-s+13000\ kg\times 11\ m/s$

$p_f=20143000\ kg-m/s$

(c) For new velocity,

$v_f=\dfrac{p_f}{m}$

$v_f=\dfrac{20143000\ kg-m/s}{13000\ kg}$

$v_f=1549.46\ m/s$

Hence, this is the required solution.

3 0

2 years ago

Read 2 more answers