All categories

2 years ago

55. A projectile of mass 2.0 kg is fired in the air at an angle of 40.0° to the horizon at a speed of 50.0 m/s. At

1 answer:

Crank2 years ago

3 0

The total moment in an islated system remains constant through time. a) V₂ = 109.43 m/s and V₃ = 33.33 m/s. b) h = 56.67 m. c) d = 480 m.

<h3>What is the law of conservation of momentum?</h3>

First let us remember that momentum or lineal momentum is a motion quantity. In physics it is the fundamental quantity that characterizes the motion of any object.

The momentum is a vectorial quantity that can be calculated as the product of the object mass by its lineal velocity ⇒ mv

The total lineal moment of a system constituted by a group of objects is the sum of the vectorial moments of each of the objects.

In an isolated system -the one that does not interact with the exterior environment-, the moment remains constant through time. This is the law of conservation of momentum. The initial moment is equal to the final moment ⇒ Pinitial = Pfinal

Objects involved in crushes, explosions, collitions, and others, are considered to be isolated systems.

.......................

So, in the exposed example we know that,

Projectile mass ⇒2kg.
Velocity = 50 m/s.
Angle⇒ 40°.
Smaller masses ⇒ 1 kg with 10m/s speed, 0.7 kg, and 0.3 kg.

a) We need to calculate the speed of the 0.7 and 0.3 kg objects.

So, 0.7 kg object moves in the original forward direction, meaning that we can use the following formula to calculate its velocity,

m₀v₀ = m₂v₂

Where

m₀ = projectile mass = 2kg
v₀ = horizontal component of projectile velocity
m₂ = object mass = 0.7 kg
v₂ = object velocity = ??

So first, we need to get the horizontal component of projectile velocity, V₀.

V₀ = V cosθ

V₀ = 50 cos40º

V₀ = 38.3 m/s

Now we can calculate the velocity of the 0.7 kg object.

m₀v₀ = m₂v₂

2 kg x 38.3 m/s = 0.7 kg x v₂

v₂ = (2 x 38.3) / 0.7

v₂ = 109.43 m/s

Now we need to calculate the velocity of the 0.3 kg object. Since this part goes exactly in the opposite direction of the 10 kg object, their addition will equal zero. So we can use the following equation to calculate the velocity of the 0.3 kg object.

m₁v₁ + m₃v₃ = 0

Where

m₁ = 1 kg object mass
v₁ = 1 kg object velocity = 10m/s
m₃ = 0.3 kg object mass
v₃ = object velocity = ??

m₁v₁ + m₃v₃ = 0

(1 kg x 10m/s) + (0.3 kg x v₃ ) = 0

(0.3 kg x v₃ ) = - (1 kg x 10m/s)

v₃ = - (1 kg x 10m/s) / 0.3

v₃ = - 33.33 m/s

The - sign is indicating the opposite direction concerning the 1 kg object.

ANSWER:

The speed of the 0.7 kg object is 109.43 m/s
The speed of the 0.3 kg objects is 33.33 m/s

.........................

b) We need to get the high from the break-up point where the 0.3-kg piece go before rest

At rest,

1/2 m₃v₃² = m₃gh

Where,

m₃ = 0.3 kg object mass
v₃ = object velocity = 33.33m/s
g = gravity = 9.8 m/s²
h = high = ??

1/2 m₃v₃² = m₃gh

h = 1/2 (v₃²) /g

h = 1/2 (33.33²) / 9.8

h = 56.67 m

ANSWER: The high from the break-up point where the 0.3-kg piece go before coming to rest is 56.67 m.

........................

c) Finally, we need to get where does the 0.7-kg piece land

ANSWER: The distance at which the 0.7 piece lands, concerning the point from which it was fired is 480 m.

You can learn more about law of conservation of momentum at

brainly.com/question/17140635

brainly.com/question/1113396

#SPJ1

You might be interested in

PLEASE ANSWER 50 POINTS SPENT I'LL GIVE BRAINLIEST TO THE PERSON WHO ANSWERS ALL! (or most)

Afina-wow [57]

Answer:

1) The matter absorbs or reflects the light

2) Lens

3) Concave- curves inwards. Diverges light

b.Convex- curves outward. Converges light

4) The image is real if the distance of the object from the lens is greater than the focal length and virtual if it is less than the focal length

5) Lens and, for convex lenses, on the distance between the lens and the object.

6) Index of refraction?

Explanation:

I hope this helped you, sorry if anything is wrong

6 0

3 years ago

A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr

rjkz [21]

Answer:

Part A

I = 1.4 mW/m²

Part B

β = 91.46 dB

Explanation:

Part A

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^{2}}$

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

$I = \frac{34}{4\pi (44)^{2}}$

I = 0.0013975 W/m²

≈ I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

Part B

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value. It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;

β $= 10log_{10}\frac{I}{I_{0}}$ dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

∴ β $= 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}$

β $= 10log_{10} (1.4 * 10^{9})$

β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.

4 0

3 years ago

Please help I need this fast

Ghella [55]

The solution has reacted.

5 0

3 years ago

On your first trip to Planet X you happen to take along a 280 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You'r

arsen [322]

Answer:

5.31143691523 m/s²

Explanation:

m = Mass = 280 g

x = Displacement of spring = 21.7 cm

Time period

$T=\dfrac{14}{11}\\\Rightarrow T=1.27\ s$

Angular velocity is given by

$\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{1.27}\\\Rightarrow \omega=4.94739\ rad/s$

$\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow k=\omega^2m\\\Rightarrow k=4.94739^2\times 0.28\\\Rightarrow k=6.85346698739\ N/m$

From Hooke's law

$mg=kx\\\Rightarrow g=\dfrac{kx}{m}\\\Rightarrow g=\dfrac{6.85346698739\times 0.217}{0.28}\\\Rightarrow g=5.31143691523\ m/s^2$

The acceleration due to gravity on the planet is 5.31143691523 m/s²

Yes, I have been able to satisfy my curiosity.

7 0

3 years ago

Two cars cover the same distance in a straight line. Car A covers the distance at a constant velocity. Car B starts from rest an

SpyIntel [72]

Answer:

a) 2.43 m/s

b) 4.83 m/s

c) 0.023 m/s²

Explanation:

a) Both cars cover a distance of 510 m in 210 s. Since car A has no acceleration

Speed = Distance / Time

$\text{Speed}=\frac{510}{210}=2.43\ m/s$

Velocity of car A is 2.43 m/s

t = Time taken = 210 seconds

u = Initial velocity

v = Final velocity

s = Displacement = 510 m

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 510=0\times 210+\frac{1}{2}\times a\times 210^2\\\Rightarrow a=\frac{510\times 2}{210^2}\\\Rightarrow a=0.023\ m/s^2$

Acceleration of car B is 0.023 m/s²

$v=u+at\\\Rightarrow v=0+0.023\times 210\\\Rightarrow v=4.83\ m/s$

Final velocity of car B is 4.83 m/s

6 0

3 years ago