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Sveta_85 [38]
3 years ago
11

Which would hold more water- a teaspoon (tsp) or a milliliter?

Physics
2 answers:
ki77a [65]3 years ago
4 0
A tsp

1 milliliter equals 0.202. US teaspoons. boi did that help
salantis [7]3 years ago
3 0

Answer:

A teaspoon

Explanation:

As we know that the capacity of hold something by a standard teaspoon is related with milliliters as,

1 teaspoon is equivalent to 4.92892 milliliters

From this it can be see that the 1 teaspoon is equivalent to approximately 5 milliliters.

Therefore, the water hold by 1 teaspoon is more than the water hold by a milliliter.

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A vehicle reaches a speed of 7.5 m/s over 15 seconds. What is its acceleration if it
denpristay [2]

acceleration = \frac{velocity}{time}  = \frac{7.5}{15}  = 0.5ms^-^2

So the answer is option b.

8 0
3 years ago
A car is travelling at 18m/s accelerates ti 30m/s in 3seconds. what's the acceleration of the car​
Luden [163]

Answer: a = 4 m/s²

Explanation:

a = Δv/t = (30 - 18) / 3 = 4 m/s²

3 0
3 years ago
A nautical mile is 6076 feet, and 1 knot is a unit of speed equal to 1 nautical mile/hour. How fast is a boat going 8 knots goin
mezya [45]

Answer:

The speed of the boat is equal to 13.50 ft/s.

Explanation:

given,

1  nautical mile = 6076 ft

1 knot = 1 nautical mile /hour

1 knot = 6076 ft/hr

speed of boat = 8 knots

 8 knots = 8 nautical mile /hour

               =8 \times \dfrac{6076\ ft}{1\nautical\ mile}\times \dfrac{1\ hour}{60\times 60\ s}

               = 13.50 ft/s

The speed of the boat is equal to 13.50 ft/s.

5 0
3 years ago
A train slows from 44 m/s to 19 m/s in 9 seconds. What is the train's acceleration? Show your work.
Korvikt [17]

Answer:

a = 2.77 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f} =v_{o} -a*t

where:

Vf = final velocity = 19 [m/s]

Vo = initial velocity = 44 [m/s]

a = acceleration [m/s²]

t = time = 9 [s]

Note: The negative sign in the above equation means that the train is decreasing its velocity.

19 = 44 - a*9

9*a = 44 -19

a = 2.77 [m/s²]

3 0
3 years ago
A 6.0-μF air-filled capacitor is connected across a 100-V potential source (a battery). After the battery fully charges the capa
Dovator [93]

Answer:

Change in Q = 2.1x 10^-3 C

Explanation:

We are given that

The Initialcapacitance C1 = 6.0μF

Initial charge oncapacitor

Q1 = C1 V

= 6.00 x 10^-6 x 100

= 6.00 x 10^-4 C

So the Final capacitance C2 will be

= K x C1 = 4.5 x 6.00 x 10^-6

= 2.7 x 10^ -5 F

So to get Finalcharge

We use Q2 = C2 x V

= 2.7 x 10^ - 5 x 100

= 27 x 10^ -4 C

So Charge flown in thecapacitor is change in Q

Which is = Q2 - Q1

= 27 x 10^-4 - 6.0 x 10^ -4

Change in Q = 2.1x 10^-3 C

5 0
3 years ago
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