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Andrews [41]
3 years ago
12

A gun can fire a bullet at 540 m/s. If the gun is aimed at an angle of 55o above the horizontal and fired, what will be the hori

zontal and vertical components of the gun's velocity?
Physics
1 answer:
lisabon 2012 [21]3 years ago
5 0

Answer:

Explanation:

The <u>initial</u> vertical velocity is 540sin55° = 442.342103... 442 m/s

The <u>initial</u> horizontal velocity is 540cos55° = 309.731275... 310 m/s

In the real world, both initial velocities would be reduced by air resistance and vertical velocity will be altered by gravity.

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A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 14 s until its motor stops. Disreg
telo118 [61]

Answer:

The maximum height of the rocket will be 1.0 × 10⁴ m.

Explanation:

Hi there!

The height of the rocket at time "t" can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · a · t²  (when the rocket has an upward acceleration)

y = y0 + v0 · t + 1/2 · g · t²  (after the motor of the rocket stops)

Where:

y = height.

y0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration due to the motors.

g = acceleration due to gravity.

The velocity of the rocket can be calculated as follows:

v = v0 + a · t  (while the motor is running)

v = v0 + g · t  (after the motor stops)

Where "v" is the velocity of the rocket at time "t".

The rocket rises with upward acceleration for 14 s. After that, the rocket starts being accelerated in the downward direction due to gravity. But it will continue going up after the motor stops because the rocket has initially an upward velocity that will be reduced until it becomes zero and the rocket starts to fall.

Let´s find the height reached by the rocket while it was accelerated in the upward direction (the origin of the frame of reference is located at the launching point and upward is the positive direction):

y = y0 + v0 · t + 1/2 · a · t²

y = 0 m + 0 m/s · 14 s + 1/2 · 28 m/s² · (14 s)²

y = 2.7 × 10³ m

Now let´s find the velocity reached in that time:

v = v0 + a · t

v = 28 m/s² ·14 s

v = 3.9 × 10² m/s

Now, let´s find the maximum height reached by the rocket using the equations of height and velocity after the motor stops:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Notice that now v0 and y0 will be the velocity and height reached while the rocket was being accelerated in the upward direction, respectively.

Let´s find at which time the rocket reaches its maximum height. With that time, we can calculate the max-height.

At the maximum height, the velocity of the rocket is zero, then:

v = v0 + g · t

0 = 3.9 × 10² m/s - 9.8 m/s² · t

-3.9 × 10² m/s/ -9.8 m/s² = t

t = 40 s

After the motor stops, it takes the rocket 40 s s to reach the maximum height.

Using the equation of height:

y = y0 + v0 · t + 1/2 · g · t²

y = 2.7 × 10³ m +  3.9 × 10² m/s · 40 s - 1/2 · 9.8 m/s² · (40 s)²

y = 1.0 × 10⁴ m

The maximum height of the rocket will be 1.0 × 10⁴ m

4 0
3 years ago
The majority of water on Earth exists as salt water, and only a very small percentage exists as fresh water. More than half of t
serg [7]

Answer:

B Crystallization

Hope this helps♡♡

8 0
2 years ago
Which of the following statements about energy is false?
Artemon [7]

Answer:

c

Explanation:

5 0
3 years ago
A horse travels 1200 meters to the right in 60 seconds. What was its velocity?
just olya [345]

Answer:

a

Explanation:

If its not right choose D

6 0
3 years ago
How to change V=72km/hr to m/s
lapo4ka [179]
Multiply by (1000 meters / 1 km).
Then multiply by (1 hour / 3600 seconds).

Both of those fractions are equal to ' 1 ', because the top
and bottom numbers are equal, so the multiplications
won't change the VALUE of the 72 km/hr.  They'll only
change the units.

(72 km/hour) · (1000 meters / 1 km) · (1 hour / 3600 seconds)

= (72 · 1000 / 3600) (km·meter·hour / hour·km·second)

=  20 meter/second
7 0
4 years ago
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