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mariarad [96]
3 years ago
10

What type of change accurs when a substance stays the same

Physics
1 answer:
Thepotemich [5.8K]3 years ago
3 0
This would be a physical change because it can change back to its original form. This is like ripping paper. You can piece it back together and it still is paper.

The opposite of this is chemical change. Chemical change means the product has been changed completely like burning paper. The paper has now been turned to ash and it's impossible to change this back to its original form.
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Identify two factors you must know to describe the motion of an object along a straight line
Sveta_85 [38]

In order to describe motion along a straight line, you must state the speed and direction of the motion. Those two quantities, together, comprise what's known as "velocity".

6 0
3 years ago
A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of
Nastasia [14]

Answer:

Amplitude = 0.058m

Frequency = 6.25Hz

Explanation:

Given

Amplitude (A) = 8.26 x 10-2 m

Frequency (f) = 4.42Hz

Conversation of energy before split

½mv² = ½KA²

Make A the subject of formula

A = v\sqrt{\frac{m}{k} }

Conversation of energy after split

½(m/2)V'² = ½(m/2)V² = ½KA'²

½(m/2)V² = ½KA'²

Make A the subject of formula

First divide both sides by ½

(m/2)V² = KA'²

Divide both sides by K

\frac{m}{2K}V² = A'²

v\sqrt{\frac{m}{2k} } = A'

Substitute v\sqrt{\frac{m}{k} } for A in the above equation

A' = A/√2

A' = 8.26 x 10^-2/√2

A' = 0.05840702012600882

Amplitude after split = 0.058 (Approximated)

Frequency (f') = f√2

f' = 4.42√2

f' = 6.25082394568908011

Frequency after split = 6.25Hz (approximated)

8 0
3 years ago
What is the energy in joules of a photon with a frequency of 3.16e 12 s-1?
erica [24]
We have: Energy(E) = Planck's constant(h) × Frequency(∨)
Here, Planck's constant(h) = 6.626 × 10⁻³⁴ J/s
Frequency (∨) = 3.16 × 10¹² /s

Substitute the values into the expression:
E = (6.626 × 10⁻³⁴)(3.16 × 10¹²) J
E = 2.093 × 10⁻²¹ Joules

In short, Your Final answer would be 2.093 × 10⁻²¹ J

Hope this helps!
5 0
3 years ago
Read 2 more answers
A person fires a 38 gram bullet straight up into the air. It rises, then falls straight back down, striking the ground with a sp
Y_Kistochka [10]

Answer:

Force exerted = 25.41 kN

Explanation:

We have equation of motion

      v² = u²+2as

u = 345 m/s, s = 8.9 cm = 0.089 m, v = 0 m/s

     0² = 345²+2 x a x 0.089

       a = -668679.78 m/s²

Force exerted = Mass x Acceleration

Mass of bullet = 38 g = 0.038 kg

Acceleration = 668679.78 m/s²

Force exerted = 25409.83 N = 25.41 kN

4 0
3 years ago
A 55.6 kg ice skater is moving at 1.73 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole.
GarryVolchara [31]

Answer:

The force exerted by the rope on her arms is 273.7 N = 0.274 kN

Explanation:

Step 1: Data given

Mass of the ice skater = 55.6 kg

Velocity = 1.73 m/s

She then moves in a circle of radius 0.608 m around the pole.

Step 2:

Force exterted by the horizontal rope is the centripetal force acting on theice skater:

Fc = M*ac

⇒ with ac = v²/r

Fc = M * v²/r

Fc = 55.6 * 1.73²/0.608

Fc =273.69 N

The force exerted by the rope on her arms is 273.7 N = 0.274 kN

7 0
3 years ago
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