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frez [133]
3 years ago
12

A projectile is launched with an initial velocity of 60.0 m/s at an angle of 30.0° above the horizontal. How far does it travel?

Physics
2 answers:
Bezzdna [24]3 years ago
8 0

Answer:

Explanation: The initial velocity has two component v x  and v y. The initial velocity in direction of y is given.

igomit [66]3 years ago
3 0

Answer:

D

Explanation:

The range is how far the projectile can go;

R = U^2sin2A/g

Where U is the initial velocity

A is the angle of inclination

=60^2sin2(30)/10 = 3600 sin60 / 10

= 311.8m

The exact answer should 312m

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Ans of this question A test charge of 1 couloumb moved from 30cm against the field of intensity 50N/c find the energy store in i
UkoKoshka [18]

Answer:

A. Zero

Explanation:

Given data,

The charge of the test charge, q = 1 C

The distance the charge moved against the filed of intensity, x = 30 cm

                                                                                                        = 0.3 m

The electric field intensity, E = 50 N/C

The energy stored in the charge at 0.3 m is given by the formula,

                                V = k q/r

Where,                        

                                     = 9 x 10⁹ Nm²C⁻²

The charge is moved from the potential V₁ to V₂ at 30 cm

Substituting the given values in the above equation

                            V₁ = 9 x 10⁹ x 30 / 0.3

                                =  1.5 x 10¹² J

And,

                            V₂ = 1.5 x 10¹² J

The energy stored in it is,

                             W = V₂ - V₁

                                  = 0

Hence, the energy stored in the charge is, W = 0        

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