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Mama L [17]
3 years ago
7

Which solid dissolves faster in water? sugar cube or powdered sugar

Physics
1 answer:
Verdich [7]3 years ago
6 0

Answer: <em>Powdered sugar</em>

Powdered sugar dissolves faster compare to the sugar cube. Because sugar cube has less surface area (the granules are tightly packed) compared to powdered sugar  

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In a video game, a ball moving at 0.6 meter/second collides with a wall. After the collision, the velocity of the ball changed t
DENIUS [597]

Answer:

-5 m/s²

Explanation:

Acceleration = change in velocity / change in time

a = (v − v₀) / (t − t₀)

a = (-0.4 m/s − 0.6 m/s) / 0.2 s

a = -5 m/s²

The ball decelerates at a rate of -5 m/s².

6 0
3 years ago
An RV is traveling 60 km/h along a highway at night. A boy sitting near the driver of the RV turns a flashlight on and shines it
Varvara68 [4.7K]

Answer:

300,000,000 and the second question is the same.

Explanation:

Edge

5 0
3 years ago
A 7.5 kg block is placed on a table. if it's bottom surface area is 0.6m2, how much pressure does the block exert on the tableto
xeze [42]
The answer to this is B.
3 0
3 years ago
Two parallel plates that are initially uncharged are separated by 1.7 mm, have only air between them, and each have surface area
yaroslaw [1]

Answer:

5.63\cdot 10^{-6} C

Explanation:

The capacitor of a parallel-plate capacitor is given by:

C=\epsilon_0 \frac{A}{d}

where

A is the area of each plate

d is the separation between the plates

\epsilon_0 is the vacuum permittivity

The energy stored in a capacitor instead is given by

U=\frac{1}{2}\frac{Q^2}{C}

where

Q is the charge stored in each plate

Substituting the expression we found for C inside the last formula,

U=\frac{1}{2}\frac{Q^2 d}{\epsilon_0 A}

And re-arranging it

Q=\sqrt{\frac{2U\epsilon_0 A}{d}}

Now if we substitute

d=1.7 mm=0.0017 m\\A=16 cm^2 = 16\cdot 10^{-4} m^2\\U = 1.9 J

We find the charge stored on the capacitor:

Q=\sqrt{\frac{2(1.9)(8.85\cdot 10^{-12})(16\cdot 10^{-4})}{0.0017}}=5.63\cdot 10^{-6} C

7 0
3 years ago
A 0.50-kg object moves in a horizontal circular track with a radius of 2.5 m. An external forceof 3.0 N, always tangent to the t
kodGreya [7K]

The work done by the force is 47.1 J

Explanation:

The work done by a force in moving an object is given by

W=Fd cos \theta (1)

where

F is the magnitude of the force

d is the distance covered by the object

\theta is the angle between the direction of the force and the motion of the object

In this problem, the force applied to the object is

F = 3.0 N

This force is always tangential to the track: this means that at every instant, the force is parallel to the motion of the object, so

\theta=0

And the distance covered is equal to the circumference of the circle, which is:

d=2\pi r=2\pi (2.5 m)=15.7 m

where r = 2.5 m is the radius.

Now we can substitute into eq.(1) to find the work done:

W=(3.0)(15.7)(cos 0)=47.1 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
3 years ago
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