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3 years ago

Which of the following are true for an object in projectile motion?

1 answer:

4 0

The correct options are: B, C and D.

In fact:
- B is true, because the motion of the projectile can be decomposed in two independent motions on the x and y axis. On the x-axis, there is no acceleration; while on the y-axis, the projectile is accelerating towards the ground with $g=9.81 m/s^2$ (acceleration of gravity)
- C is true, because the two motions on the horizontal and vertical direction are independent. In particular, the horizontal motion is a uniform motion (constant velocity), while the vertical motion is a uniformly accelerated motion (because of the gravitational acceleration g acting on the projectile)
- D is true, in fact (if we neglect air resistance) gravity is the only force acting on the projectile.

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A stone is thrown vertically upwards with a speed of 30.0 m/s.

matrenka [14]

Y₀ = initial position of the stone at the time of launch = 0 m

Y = final position of stone = 20.0 meters

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = ?

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

v² = 30² + 2 (- 9.8) (20 - 0)

v = 22.5 m/s

Y₀ = initial position of the stone at the time of launch = 0 m

Y = maximum height gained

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = 0 m/s

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

0² = 30² + 2 (- 9.8) (Y - 0)

Y = 46 m

6 0

3 years ago

Determine the thrust produced if 1.5 x 10^3 kg of gas exits the combustion chamber each second, with a speed of 4.00 x 10^3 m/s.

ozzi

Answer:

The thrust is $6\times 10^6\ N$

Explanation:

Given that,

Mass of gas, $m=1.5\times 10^3\ kg$

The rate at which the gas is expelling, $\dfrac{dv}{dt}=4\times 10^{3}\ m/s$

We need to find the thrust produced by the gas.

We know that force is equal to the rate of change of momentum. So,

$F=\dfrac{p}{t}$

Also, p = mv

$F=\dfrac{mv}{t}$

So,

$F=1.5\times 10^3\times 4\times 10^3\\\\F=6\times 10^6\ N$

So, the thrust is $6\times 10^6\ N$

3 0

3 years ago

Answer fast !!!!!!! pleaseee

wariber [46]

Answer: Magnetic Fields

Explanation:

5 0

3 years ago

Read 2 more answers

You wind a small paper tube uniformly with 153 turns of thin wire to form a solenoid. The tube\'s diameter is 5.11 mm and its le

lina2011 [118]

The concept required to solve this problem is linked to inductance. This can be defined as the product between the permeability in free space by the number of turns squared by the area over the length. Recall that Inductance is defined as the opposition of a conductive element to changes in the current flowing through it. Mathematically it can be described as

$L = \frac{\mu N^2 A}{l}$