Answer:
stop, drop and roll.
Explanation:
This is because rolling on the ground can help put out the fire by depriving it of oxygen.
Explanation:
Let magnitude of the two forces be x and y.
Resultant at right angle R1= √15N) and at
60 degrees be R2= √18N.
Now, R1 = √(x² + y²) = √15,
R2= √(x² + y² +2xycos50) = √18.
So x² + y² = 15,
and x² + y² + 1.29xy = 18,
therefore 1.29xy = 3,
y = 3/1.29x.
y = 2.33/x
Now, x2 + (2.33/x)2 = 15,
x² + 5.45/x² = 15
multiply through by x²
x⁴ + 5.45 = 15x²
x⁴ - 15x2 + 5.45 = 0
Now find the roots of the equation, and later y. The two values of x will correspond to the
magnitudes of the two vectors.
Good luck
Answer: KE = 62.5J
Explanation:
Given that
Mass of object = 5kg
kinetic energy KE = ?
velocity of object = 5m/s
Since kinetic energy is the energy possessed by a moving object, and it depends on the mass (m) of the object and the velocity (v) by which it moves. Therefore, the object has kinetic energy.
i.e K.E = 1/2mv^2
KE = 1/2 x 5kg x (5m/s)^2
KE = 0.5 x 5 x 25
KE = 62.5J
Thus, the object has 62.5 joules of kinetic energy.
You've given the answer, right there in your question.
The "magnitude of gravity" is described in terms of the acceleration
due to it, and you just told us what that is.
We can also notice that the figure you gave is about 0.66 of the
acceleration due to gravity on the Earth's surface. That tells us that
the distance from the Earth's center at that height is about
(1 / √0.66) = 1.23 times
the Earth's radius, so the height is about 910 miles above the surface.
Answer:
The electric field will be zero at x = ± ∞.
Explanation:
Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.
We know that,
The electric field is
The electric field vector due to charge one
The electric field vector due to charge second
We need to calculate the electric field
Using formula of net electric field
Put the value into the formula
Put the value into the formula
If x = ∞, then the equation is be satisfied.
Hence, The electric field will be zero at x = ± ∞.