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2 years ago

Which of the following is equivalent to 2,100 grams?

2 answers:

6 0

<span>D is the correct answer. 10 decigrams are equivalent to 1 gram, so 2,100 grams are equivalent to 2100 x 10. This makes 21,000. </span>

Reptile [31]2 years ago

6 0

21,000 decigrams is was hard for me too. ha ha ha

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An object has a kinetic energy of 175 J and a momentum of magnitude 25.0 kg m/s. Find the

DedPeter [7]

Answer:14 m/s

Explanation:

Kinetic energy(ke)=175J

Momentum(M)=25kgm/s

Speed=v

Mass=m

Ke=(m x v x v)/2

175=(mv^2)/2

Cross multiply

175 x 2=mv^2

350=mv^2

Momentum=mass x velocity

25=mv

m=25/v

Substitute m=25/v in 350=mv^2

350=25/v x v^2

350=25v^2/v

v^2/v=v

350=25v

v=350/25

v=14 m/s

5 0

3 years ago

The mass of an object changes as the distance from the center of gravity changes.

Alchen [17]

Answer:

true

Explanation:

this is the answer to this question

8 0

2 years ago

Average speed equals distance divided by time. on a journey to the

Tanzania [10]

The average speed will be 2.38×10⁶ m/sec.The average speed of an object indicates the pace at which it will traverse a distance. The metric unit of speed is the meter per second.

<h3>What is the average speed?</h3>

The total distance traveled by an object divided by the total time taken is the average speed.

The speed calculated at any particular instant of time is known as the instantaneous speed.

Given data;

Distance travelled = 4.12x10¹⁶ meter

Time period= 1.73x10¹⁰ sec

The average speed is found as

$\rm V_{avg}= \frac{d}{t} \\\\\ \rm V_{avg}= \frac{4.12 \times 10^{16}}{1.73 \times 10^{10} } \\\\\ V_{avg}=2.38 \times 10^6 \ m/sec$

Hence, the average speed will be 2.38×10⁶ m/sec.

To learn more about the average speed, refer to the link;

brainly.com/question/12322912

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6 0

1 year ago

Two balls of equal size are dropped from the same height from the roof of a building. the mass of ball a is twice that of ball b

Sindrei [870]

The mass of ball a is twice the mass of ball b:
$m_a = 2 m_b$
This means that the initial potential energy of ball a ( $U_a = m_a gh=2 m_b gh$ ) is twice the potential energy of ball b ( $U_b = m_b g h$ ):
$U_a = 2 U_b$
When the two balls reach the ground, the potential energy of each ball has converted into kinetic energy (since now their altitude is h=0), because the total mechanical energy of each ball must be conserved. Therefore:
$K_a = U_a$
$K_b = U_b$
and so the kinetic energy of ball a must be twice the kinetic energy of ball b:
$K_a = 2 K_b$

8 0

2 years ago

Read 2 more answers

A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat

Sever21 [200]

Answer:

Approximately $325$ (rounded down,) assuming that $g = 9.81\; {\rm N \cdot kg^{-1}}$ .

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, ${\rm J}$ ):

$\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}$ .

Height of the weight (should be in meters, ${\rm m}$ ):

$\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}$ .

Energy required to lift the weight by $\Delta h = 0.2\; {\rm m}$ without acceleration:

$\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}$ .

At an efficiency of $0.25$ , the actual amount of energy required to raise this weight to that height would be:

$\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}$ .

Divide $2.551 \times 10^{5}\; {\rm J}$ by $784\; {\rm J}$ to find the number of times this weight could be lifted up within that energy budget:

$\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}$ .

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

4 0

1 year ago