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mihalych1998 [28]
3 years ago
14

A scientist is performing an experiment on a powder. Which of the following is a sign that a chemical change has definitely occu

rred during his experiment? The powder catches on fire.
The powder releases bubbles.

The powder heats up.

The powder breaks apart into smaller particles.
Physics
1 answer:
Kitty [74]3 years ago
8 0
B. is the correct answer :)
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Rhythms that occur faster and slower than the beat are. Select one:. a. incorrect. b. not synchronized with the time signature.
zlopas [31]
Rhythms that occur faster and slower than the beat are b.<span>not synchronized with the time signature. The synchronization follows the same beat or rhythm. If the time signature say is lower than the original, then the rhythm should be faster. Otherwise, the rhythm is slower than the original one.</span>
4 0
3 years ago
Which of these is exhibiting kinetic energy?1) a rock on a mountain ledge2) a space station orbiting Earth 3) a person sitting o
jarptica [38.1K]

The answer would be:

A space station orbiting Earth.

7 0
3 years ago
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In which situation are both the speed and velocity of the car changing?
lubasha [3.4K]
Answer is A because the speed and velocity would change. Think of it as GTA, your going 30+ miles per hour and you take a left turn, the speed and velocity would change in an instant.. 
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6 0
3 years ago
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The net external force on a golf cart is 411 N north of tha cart has a total mass of 281 kg what is tha cart acceleration
Alisiya [41]

1.46m/s²

Explanation:

Given parameters:

Mass of cart = 281kg

Net external force = 411N

Unknown:

Acceleration of the cart = ?

Solution:

According to newton's second law " the net force on a car is the product of its mass and acceleration";

            Force = mass x acceleration

Since acceleration of the cart is unknown;

     Acceleration = \frac{force}{mass} = \frac{411}{281}

  Acceleration = 1.46m/s²

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

8 0
3 years ago
A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What
AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, \alpha=3.3 rad/s^2

Diameter of the wheel, d=21 cm

Radius of wheel, r=\frac{d}{2}=\frac{21}{2} cm

Radius of wheel, r=\frac{21\times 10^{-2}}{2} m

1m=100 cm

Magnitude of total linear acceleration, a=1.7 m/s^2

We have to find the linear speed  of a  at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration,a_t=\alpha r

a_t=3.3\times \frac{21\times 10^{-2}}{2}

a_t=34.65\times 10^{-2}m/s^2

Radial acceleration,a_r=\frac{v^2}{r}

We know that

a=\sqrt{a^2_t+a^2_r}

Using the formula

1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}

Squaring on both sides

we get

2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}

\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}

v^4=r^2\times 2.7699

v^4=(10.5\times 10^{-2})^2\times 2.7699

v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}

v=0.418 m/s

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0
3 years ago
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