Answer is
9.773m/s^2
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Given,
h=8848m
The value of sea level is 9.08m/s^2. So, Let g′ be the acceleration due to the gravity on Mount Everest.
g′=g(1 − 2h/h)
=9.8(1 - 6400000/17696)
=9.8(1 − 0.00276)
9.8×0.99724
=9.773m/s^2
Thus, the acceleration due to gravity on the top of Mount Everest is =9.773m/s^2
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hope this helps :)
Answer:
<em>The range is 35.35 m</em>
Explanation:
<u>Projectile Motion</u>
It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.
Being vo the initial speed of the object, θ the initial launch angle, and
the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:



The range is 35.35 m
Explanation:
→ Volume of cone = πr² × h/3
Here,
- Radius (r) = 13 cm
- Height (h) = 27 cm
→ Volume of cone = π(13)² × 27/3 cm³
→ Volume of cone = 169π × 9 cm³
→ Volume of cone = 1521π cm³
→ Volume of cone = 1521 × 22/7 cm³
→ Volume of cone = 33462/7 cm³
→ <u>Volume of cone = 4780.28 cm³</u>
Answer:
The new distance is d = 0.447 d₀
Explanation:
The electric out is given by Coulomb's Law
F = k q₁ q₂ / r²
This electric force is in balance with tension.
We reduce the charge of sphere B to 1/5 of its initial value (
=q₂ = q₂ / 5) than new distance (d = n d₀)
dat
q₁ = 
q₂ = 
r = d₀
In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points
F = k q₁ q₂ / d₀²
F = k q₁ (q₂ / 5) / (n d₀)²
.k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)
5 n² = 1
n = √ 1/5
n = 0.447
The new distance is
d = 0.447 d₀