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Answer:
y maximum 3.54 m, value X 2.35 m
Explanation:
We have a projectile launch problem, let's calculate the maximum height of the projectile, where the vertical speed must be zero
Vyf² = Vyo² - 2g (Y-Yo)
Where Yo is the initial height of the ramp 1.5 m
0 = Vyo² -2g (Y-Yo)
Y-Yo = Voy² / 2g
Y = Yo + Voy² / 2g
Let's calculate the velocity components using trigonometry
Voy = vo without T
Vox = Vo cost
Voy = 7.3 sin 60
Vox = 7.3 cos 60
Voy = 6.32 m / s
Vox = 3.65 m / s
Let's calculate the maximum height
Y = 1.5 +6.32²/2 9.8
Y = 3.54 m
This is the maximum height from the ground
b) They ask us for the position of this point horizontally, we can calculate it looking for the time it took for the skateboarder to reach the highest point
Vfy = Voy - gt
0 = Voy - gt
t = Voy / g
t = 6.32 / 9.8
t = 0.645 s
Since there is no acceleration on the x-axis, we have a uniform movement, we can calculate the distance for this time
X = Vox t
X = 3.65 0.645
X= 2.35 m
According to Gauss' law, the electric field outside a spherical surface uniformly charged is equal to the electric field if the whole charge were concentrated at the center of the sphere.
Therefore, when you are outside two spheres, the electric field will be the overlapping of the two electric fields:
E(r > r₂ > r₁) = k · q₁/r² + k · q₂/r² = k · (q₁ + q₂) / r²
where:
k = 9×10⁹ N·m²/C²
We have to transform our data into the correct units of measurement:
q₁ = 8.0 pC = 8.0×10⁻¹² C
q₂ = 3.0 pC = 3.0×10<span>⁻¹² C
</span><span>r = 5.0 cm = 0.05 m
Now, we can apply the formula:
</span><span>E(r) = k · (q₁ + q₂) / r²
= </span>9×10⁹ · (8.0×10⁻¹² + 3.0×10⁻¹²) / (0.05)²
= 39.6 N/C
Hence, <span>the magnitude of the electric field 5.0 cm from the center of the two surfaces is E = 39.6 N/C</span>
Answer:
Answer. An atom that has fewer neutrons than protons and more electrons than protons is an anion. Ion are charged particles. ... Electrons are negatively charged particles, therefore, there excess will be define an ion as anion.
Explanation:
so yea
Answer:
a=8.06m/s^2
Explanation:
The box can be considered negligible body slidding down along a curved path defined by the parabola Y=Ax^2
Note:
When it's at A(x=2m, y=1.6m),
the speed Vb=8m/s and the increase in speed=4m/s^2
To find the acceleration,
Y=Ax^2
dy/dx=8x
d^2y/dx^2=8
p={[1+(dy/dx)^2]^3/2}/|d^2y/dx^2| .......1
substituting into 1, we have
p=8.39624m
an=v^2/p
an=8^2/8.39624=7.6224m/s^2
a=sqrt(at^2+an^2)
a=sqrt(4^2+7.62246^2)
a=8.06m/s^2