1) the weight of an object at Earth's surface is given by

, where m is the mass of the object and

is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is

2) On Mars, the value of the gravitational acceleration is different:

. The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth:

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as

<span>6) On Earth, the gravity acceleration is </span>

<span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is
</span>

<span>
</span>
Answer: 2, the nuclear strong force drops to practically nothing at large distances.
Explanation: The protons and neutrons in the nucleus share subatomic particles called pions. This exchange is what keeps the protons and neutrons stuck together in the nucleus. Despite the strong force being the strongest force, it has a very small range. This is because pions have very short lifespans. So, the strong force would have literally no effect at large distances.
Hope that helped! :)
Answer:
1.Theimage will be located at -0.13m or -13 cm
2.The height of the image will be 0.052m or 5.2cm
Explanation:
Given that;
Height of object, h=20 cm = 0.2m
Object distance in front of convex mirror, o,= 50 cm =0.5m
Radius of curvature, r, =34 cm =0.34m
Let;
Image distance, i,=?
Image height, h'=?
You know that focal length,f, is half the radius of curvature,hence
f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)
f= -0.17m
Apply the relationship that involves the focal length;


Re-arrange to get i

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror
Apply the magnification formula
magnification, m=height of image÷height of object

substitute the values to get the height of image h'

Answer:
f(t) = 28,7 [N]
Explanation: IMPORTANT NOTE: IN PROBLEM STATEMENT CHARGES ARE IN C (COULOMBS) AND IN THE DIAGRAM IN μC. WE ASSUME CHARGES ARE IN μC.
The net force on +q₂ is the sum of the force of +q₁ on +q₂ ( is a repulsion force since charges of equal sign repel each other ) and the force of -q₃ on +q₂ ( is an attraction force, opposite sign charges attract each other)
The two forces have the same direction to the right of charge q₂, we have to add them
Then
f(t) = f₁₂ + f₃₂
f₁₂ = K * ( q₁*q₂ ) / (0,1)²
q₁ = + 8 μC then q₁ = 8*10⁻⁶ C
q₂ = + 3,5 μC then q₂ = 3,5 *10⁻⁶ C
K = 9*10⁹ [ N*m² /C²]
f₁₂ = 9*10⁹ * 8*3,5*10⁻¹²/ 1*10⁻² [ N*m² /C²]* C*C/m²
f₁₂ = 252*10⁻¹ [N]
f₁₂ = 25,2 [N]
f₃₂ = 9*10⁹*3,5*10⁻⁶*2,5*10⁻⁶ /(0,15)²
f₃₂ = 78,75*10⁻³/ 2,25*10⁻²
f₃₂ = 35 *10⁻¹
f₃₂ = 3,5 [N]
f(t) = 28,7 [N]
Answer:
90 meters
Explanation:
Given:
x₀ = 0 m
v₀ = 0 m/s
v = 30 m/s
t = 6 s
Find:
x
x = x₀ + ½ (v + v₀)t
x = 0 + ½ (30 + 0)(6)
x = 90
The car travels 90 meters.