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SashulF [63]
3 years ago
14

Add a small ball to a graduated cylinder containing 10 milliliters of water.

Physics
1 answer:
malfutka [58]3 years ago
8 0
The volume of water will increase . If yu subtract the original volume from the new volume of water you will get the volume of the small ball.
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An object is thrown upwards with the speed of 40.0 km/hr in 5 second, calculate the distance in millimeters (mm).
sertanlavr [38]
V=(40km/hr)(hr/3600s)(1000000mm/km)
v=11111.1mm/s

v=d/t
d=vt
d=(11111.1mm/s)(5s)
d=55555mm
d=5.56x10^4mm
4 0
2 years ago
Give an example of a situation in which you would describe an
Viktor [21]
It would be A because it would make sense
5 0
3 years ago
A small block with a mass of 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. 6
Rina8888 [55]

Answer:

a) 0.147 N

b) 9.408 N

c) 9.261 N

Explanation:

The tension on the cord is the only force keeping the block in circular motion, thus representing the entirety of its centripetal force \frac{mv^{2} }{r}. Plugging in values for initial and final states and we get answers for a and b. The work done by the person causes the centripetal force to increase, and thus is the difference between the final tension and the initial tension.

7 0
2 years ago
If 36 grams of water is to be heated from 24.0°C to 48°C to make a cup of tea, how much heat must be added? The specific heat of
Vinvika [58]

We will have the following:

\begin{gathered} Q=mc\Delta T\Rightarrow Q=(36)(4.18)(48-24) \\  \\ \Rightarrow Q=3611.52 \end{gathered}

So, the heat to add is 3611.52 Joules.

3 0
1 year ago
A swimming pool is 25.0 ft. long, 18.5 ft. wide, and 9.0 ft. deep. When filled, the water level is 7.0 inches from the top. Disi
GarryVolchara [31]

Answer:

1.9841256 kg

Explanation:

Given;

Length of the swimming pool = 25.0 ft = 7.62 m   ( 1 ft = 0.3048 m )

Width of the swimming pool = 18.5 ft =  5.64 m

Depth of the pool = 9.0 ft =

Total depth of the water in the pool when filled = 9 ft - 7 inches = 2.56 m

now,

Volume of the water in the pool = Length × Width × Depth

or

Volume of the water in the pool = 7.62 × 5.64 × 2.56 = 110.2292 m³

also,

1 m³ = 1000 L

thus,

110.2292 m³ = 110229.2 L

also it is given that 18 mg of Cl is added to 1 liter of water

therefore,

In 110229.2 L of water Cl added will be = 110229.2 × 18 = 1984125.6 mg

or

= 1.9841256 kg

8 0
3 years ago
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