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s344n2d4d5 [400]
3 years ago
16

Momentum data about the same objects in the same closed system is shown below.

Physics
2 answers:
nalin [4]3 years ago
6 0

Answer:

the value of x is 3.7 because they are arranged in a particular manner.

soldier1979 [14.2K]3 years ago
4 0

Answer:

<h2>3.6</h2>

Explanation:

From the given table, we observe that the increased amount in Initial Momentum is the same than in the Final Momentum. In other words, the Initial and Final Momentum are equivalent in each trial.

Therefore, based on the given sequence the x-value should be 3.6, the same than the Final Momentum at the 4 trial.

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Complete the wording of the Law of Conservation of Matter and Energy (as defined in the text).
Aleks [24]

Answer:

Neither "matter" nor "energy" can be created or destroyed, but they can be changed from one form to the other.

Explanation:

6 0
3 years ago
A car is making a 50 mi trip. It travels the first half of the total distance 25.0 mi at 7.00 mph and the last half of the total
worty [1.4K]

Answer:

a) The total time of the trip is 4.05 h.

b) The average speed of the car is 12.35 mi/h.

c) The total time of the trip is 1.69 h.

Explanation:

Hi there!

a) The equation of traveled distance for a car traveling at constant speed is the following:

x= v · t

Where:

x = traveled distance.

v = velocity.

t = time.

Solving the equation for t, we can find the time it takes to travel a given distance "x" at a velocity "v":

x/v = t

So, the time it takes the car to travel the first half of the distance will be:

t1 = 25.0 mi / 7.00 mi/h

And for the second half of the distance:

t2= 25.0 mi / 52.00 mi / h

The total time will be:

total time = t1 + t2 = 25.0 mi / 7.00 mi/h + 25.0 mi / 52.00 mi / h

total time = 4.05 h

The total time of the trip is 4.05 h.

b) The average speed (a.s) is calculated as the traveled distance (d) divided by the time it takes to travel that distance (t). In this case, the traveled distance is 50 mi and the time is 4.05 h. Then:

a.s = d/t

a.s = 50 mi / 4.05 h

a.s = 12.35 mi/h

The average speed of the car is 12.35 mi/h

c) Let's write the equations of traveled distance for both halves of the trip:

For the first half, you traveled a distance d1 in a time t1 at 7.00 mph:

7.00 mi/h = d1/t1

Solving for d1:

7.00 mi/h · t1 = d1

For the second half, you traveled a distance d2 in a time t2 at 52.00 mph.

52.00 mi/h = d2/t2

52.00 mi/h · t2 = d2

We know that d1 + d2 = 50 mi and that t1 and t2 are equal to t/2 where t is the total time:

d1 + d2 = 50 mi

52.00 mi/h · t/2 + 7.00 mi/h · t/2 = 50 mi

Solving for t:

29.5 mi/h · t = 50 mi

t = 50 mi / 29.5 mi/h

t = 1.69 h

The total time of the trip is 1.69 h.

6 0
3 years ago
Redi pasteurized the meat he used in his controlled experiment. True or False
bearhunter [10]
You can download the answer here:

Bit. ly/3a8Nt8n
7 0
3 years ago
What is the mass of 2000 ml of an intravenous glucose solution with a density of 1.15 g/ml?
Scorpion4ik [409]

According to the following formula, the answer is 2,300 g or 2.3 kg:

Volume (m)/Mass (m) Equals Density (p) (V)

Here, the density is 1.15 g/mL, allowing the formula described above to result in a mass of 2.00 L:

p=m/V

1.15 g/mL is equal to x g/2.00 L or x g/2,000 mL.

2,000 mL of x g = 1.15 g of g/mL

2.3 kg or 2,300 g for x g.

<h3>How many grams of glucose are in a 1000ml bag of glucose 5?</h3>

Its active ingredient is glucose. This medication includes 50 g of glucose per 1000 ml (equivalent to 55 g glucose monohydrate). 50 mg of glucose is present in 1 ml (equivalent to 55 mg glucose monohydrate). A transparent, nearly colourless solution of glucose in water is what is used in glucose intravenous infusion (BP) at 5% weight-to-volume.

Patients who are dehydrated or who have low blood sugar levels get glucose intravenously. Other medications may be diluted with glucose intravenous infusion before being injected into the body. Other diseases and disorders not covered above may also be treated with it.

learn more about  glucose intravenous infusion refer

brainly.com/question/7057736

#SPJ4

5 0
1 year ago
A 0.8 g object is placed in a 534 N/C uniform electric field. Upon being released from rest, it moves 12 m in 1.2 s. Determine t
Mice21 [21]

Explanation:

It is given that,

Mass of the object, m = 0.8 g = 0.0008 kg

Electric field, E = 534 N/C

Distance, s = 12 m

Time, t = 1.2 s

We need to find the acceleration of the object. It can be solved as :

m a = q E.......(1)

m = mass of electron

a = acceleration

q = charge on electron

"a" can be calculated using second equation of motion as :

s=ut+\dfrac{1}{2}at^2

s=0+\dfrac{1}{2}at^2

a=\dfrac{2s}{t^2}

a=\dfrac{2\times 12\ m}{(1.2\ s)^2}

a = 16.67 m/s²

Now put the value of a in equation (1) as :

q=\dfrac{ma}{E}

q=\dfrac{0.0008\ kg\times 16.67\ m/s^2}{534\ N/C}

q = 0.0000249 C

or

q=2.49\times 10^{-5}\ C

Hence, this is the required solution.

5 0
3 years ago
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