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ser-zykov [4K]
3 years ago
8

A projectile is launched from the ground at an angle of 60o above the horizontal. At what point in its trajectory does it have t

he maximum value of acceleration? Neglect air resistance. It has the maximum value of acceleration just after it leaves the ground It has the maximum value of acceleration just before it lands on the ground Its acceleration is constant everywhere in its trajectory. It has the maximum value of acceleration halfway to the top of its trajectory. It has the maximum value of acceleration at the top of its trajectory.
Physics
1 answer:
Elodia [21]3 years ago
5 0

Answer:

It's constant everywhere in its trajectory.

Explanation:

the projectile was launched with an initial velocity, the only acceleration that is affecting the projectile's velocity is gravity.

The acceleration of gravity is practically equal everywhere on earth, so during its trajectory, we have to take into consideration only the acceleration because of gravity.

This is only correct because the projectile was launched with an initial velocity and it's not accelerating from rest and then falls.

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Calculate the resistance in a circuit with a 16 volt battery and 4 amps of current.
GrogVix [38]

Answer:

4 ohms

Explanation:

16v=Ω(4A)

4 0
3 years ago
Two particles are attracted to each other by the gravitational force between them. Particle 1 has a mass of 12kg, Particle 2 has
svlad2 [7]

Answer: 1.4 x 10^-8N

Explanation:

Given that,

Mass of Particle 1 (m1) = 12kg

Mass of Particle 2 (m2) = 25kg

distance between particles (r) = 1.2m. Gravitational force (F) =

Apply the formula for gravitational force:

F = Gm1m2/r²

where G is the gravitational constant with a value of 6.7 x 10^-11 Nm2/kg2

Then, F = (6.7 x 10^-11 Nm²/kg² x 12kg x 25kg) / (1.2m)²

F = (2.01 x 10^-8Nm²) / (1.44m²)

F = 1.396 x 10^-8N (Rounded to the nearest tenth as 1.4 x 10^-8N)

Thus, the magnitude of the gravitational force acting on the particles is 1.4 x 10^-8 Newton

8 0
4 years ago
In billiards, the 0.165 kg cue ball is hit toward the 0.155 kg eight ball, which is stationary. The cue ball travels at 5.8 m/s
Damm [24]

Answer:

another ball velocity = 3.92 m/s and with 30° clockwise from initial direction

Explanation:

given data

mass m1 = 0.165 kg

mass m2 = 0.155 kg

before collision velocity v1 = 5.8 m/s

before collision velocity v2 = 0

angle =  35.0° from initial direction

after collision 1st ball velocity v3 = 3.2 m/s

to find out

after collision another ball velocity v4

solution

we consider here ball move in x axis and after collision 1st ball move upside of x axis with angle 35 degree and other ball move downside with x axis with angle θ

so from conservation of momentum we say

m1v1 = m1v3cos35 + m2v4cosθ   with x axis    .............1

m1v3sin35 = m2v4sinθ                   with y axis  .............2

so from 1 equation

0.165 × 5.8 = 0.165(3.2)cos35 + 0.155(v4)cosθ

v4 cosθ  = 3.38                                            .................3

form 2 equation

0.165(3.2)sin35 = 0.155(v4)sinθ  

v4 sinθ = 1.95                                              ......................4

so magnitude of another ball velocity is square and adding equation 3 and 4

another ball velocity = √(3.39²+1.96²)

another ball velocity = 3.92 m/s

and direction is tanθ = 1.96/3.39

θ = 30° clockwise from initial direction

3 0
3 years ago
Mae prefers to complete graphs or charts to show the information she has learned. What kind of learner is Mae likely to be?
yan [13]

Answer:

Prob. D

Explanation:

I am also a Visual Learner.

3 0
3 years ago
Read 2 more answers
An 0.80-m aluminum bar is held with its length parallel to the east-west direction and dropped from a bridge. Just before the ba
Dimas [21]

Answer:

a) B=2.9891\times 10^{-5}\,T

b)  west end of the bar is positive.

Explanation:

Given:

  • emf induced in the bar, \epsilon=5.5\times 10^{-4}\,V
  • length of the bar, l=0.8\,m
  • velocity of the bar at the given instant, v=23\,m.s^{-1}

(a)

The magnitude of the horizontal component of the Earth's magnetic field(B):

We know:

\epsilon=B.l.v

5.5\times 10^{-4}=B\times 0.8\times 23

B=2.9891\times 10^{-5}\,T

(b)

Using Fleming's left hand rule we determine that the current is flowing towards east end of the bar i.e. west end of the bar is positive.

4 0
3 years ago
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