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Grace [21]
3 years ago
7

Rebecca was at the playground and found a rusted nail. What happened to the nail? A) It went through a physical change because t

he oxygen in water mixed with air. B) It went through a phase change because the oxygen in water mixed with air. C) It went through a chemical change because the oxygen in water mixed with air. D) It went through both chemical and physical changes.
Physics
1 answer:
BaLLatris [955]3 years ago
8 0
Buff dude skid did di d I’d
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What is an isotope ?
Step2247 [10]

<span>An isotope is a form of a chemical element whose atomic nucleus contains a specific number of neutrons in addition to the number of protons that distinctively defines the element. The nuclei of most atoms have neutrons as well as protons.</span>

8 0
3 years ago
A force does 30000 J of work along a distance of 9.5m. Find the applied force.
Elina [12.6K]

Answer: F=3158N

Explanation:

Work is the product of force applied and the distance the object moves along the force applied. Work is measured in joules and the equation is as follows

W = F x d

So that F = W/d

F = 30000 J / 9.5m

F = ~3158 N

8 0
3 years ago
Which statement is true? The speed of sound in air is inversely proportional to the temperature of the air. The speed of sound i
sasho [114]
<span> the speed of sound in air is directly proportional to the temperature of the air. The speed of sound depends on the temperature of the surrounding air, this can be represented by a speed of sound in air formula: v = 331m/s + 0.6m/s/C * T (where T is temperature)</span>
6 0
2 years ago
Read 2 more answers
A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed b
Vilka [71]

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=\frac{1}{2} kx^{2} +PE1

Energy at the point where the block will stop consists of only gravitational potential energy=PE2

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

⇒\frac{1}{2} kx^{2} +PE1=PE2

⇒PE2-PE1=\frac{1}{2} kx^{2}

Also PE2-PE2=mgh

where m is the mass of block

g is acceleration due to gravity=9.8 m/s

h is the difference in height between two positions

⇒mgh=\frac{1}{2} kx^{2}

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

∴2100*9.8*h=\frac{1}{2}*2200*0.11^{2}

⇒20580*h=13.31

⇒h=\frac{13.31}{20580}

⇒h=0.0006467m=6.5e-4

7 0
3 years ago
A race car travels 40 m/s around a banked (45° with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of
OLEGan [10]

Answer:

c)F_{net} = 0.640 kN

Explanation:

As we know that resultant force is the net force that is acting on the system

As per Newton's II law we know that net force is product of mass and acceleration

so we will have

F_{net} = ma

here we know

m = 80 kg

for circular motion acceleration is given as

a_c = \frac{v^2}{R}

a_c = \frac{40^2}{200} = 8 m/s^2

now we have

F_{net} = 80 \times 8

F_{net} = 640 N

F_{net} = 0.640 kN

7 0
3 years ago
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