<span>The correct option is C. Gravity, and the complete sentence is: "The force of gravity is the force at which the Earth attracts another object towards itself". In fact, the force of gravity between two objects is given by
</span>
<span>
where G is the gravitational constant, m1 and m2 the masses of the two objects, r their separation. If we take the Earth as one of the two objects, then m1 represents the Earth's mass, m2 the mass of the object and r the distance between the center of Earth and the object, and F is the gravitational force at which the Earth attracts the object.</span>
The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.
<h3>How does test charge affect electric field?</h3>
As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.
Adjusting the amount of charge on the test charge will not change the electric field force.
<h3>What is a test charge used for?</h3>
The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.
To learn more about test charge, refer
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It is given that for the convex lens,
Case 1.
u=−40cm
f=+15cm
Using lens formula
v
1
−
u
1
=
f
1
v
1
−
40
1
=
15
1
v
1
=
15
1
−
40
1
v=+24.3cm
The image in formed in this case at a distance of 24.3cm in left of lens.
Case 2.
A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that
u=∞
f=15cm
Now, using mirror’s formula
v
1
+
u
1
=
f
1
v
1
+
∞
1
=
15
1
v=+15cm
The image is formed at a distance of 15cm in left of mirror
Answer:
a =3.33 m/s²
Explanation:
given,
initial speed of Plane, u = 0 m/s
final speed of plane, v = 60 m/s
time of the acceleration, t = 18 s
average acceleration of the plane, a = ?
average acceleration is equal to change in velocity per unit time.
a =3.33 m/s²
Hence, average acceleration of the plane is equal to a =3.33 m/s²
Answer: 0.067 s
Explanation:s = Ut + 1/2at^2
0.6 = 9t + 0.5 *10 *t^2
Where a = g =10m/s/s
Solving the quadratic equation
5t^2 + 9t - 0.6=0,
t= 0.067 s and - 1.7 s
Of which 0.067 s is a valid time
