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nlexa [21]
4 years ago
15

You throw a 20-n rock vertically into the air from ground level. you observe that when it is a height 16.0 m above the ground, i

t is traveling at a speed of 24.7 m/s upward.
Physics
1 answer:
timama [110]4 years ago
6 0

The problem is missing some parts. But here are the questions and answers.

a.       Use the work-energy theorem to find its speed just as it left the ground 

Initial Kinetic Energy = GPE + KE

1/2 * m * V^2 = m * g * h + 1/2 * m * v^2

V^2 = 2 * 9.8 * 15.6 + 25.6^2

=sqrt 968.96

V (initial speed) ≈ 31.130 m/s

 

b.      Use the work-energy theorem to look for its maximum height. 

H = 968.96 / 2 * 9.8 ≈ 49.44 <span>m</span>

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A compact car, mass 664 kg, is moving at 19 km/h toward the east. What is the car's momentum in kg x m/s?
Natasha2012 [34]

Answer:

3504.4 kgm/s

Explanation:

First, convert km/hr to m/s:

19 km/hr x 1000m/1km x 1/60min x 1/60s = 5.28 m/s

p = mv

p = (664 kg)(5.28 m/s) = 3504.4 kgm/s

6 0
3 years ago
What is force? Write state of gravitational force
lana66690 [7]
Force is a push or a pull upon an object, resulting the objects interaction with another object
5 0
4 years ago
A collision between two large spiral galaxies is likely to produce
Lera25 [3.4K]

A large elliptical galaxy. Hope this helped

5 0
4 years ago
Si una bala de 122 g que viaja a 350 m/s golpea a un policía con chaleco antibalas el cual soporta 120 J de energía, ¿la bala at
Zanzabum

Answer:

Puesto que la energia cinética traslacional es mucho mayor que la capacidad del chaleco antibalas, la bala atravesaría el chaleco antibalas.

Explanation:

Un chaleco antibalas soporta el disparo de una bala disipando la energía de esta última a través de su propio material. Si sabemos que el chaleco antibalas soporta 120 joules de energía, cabe saber si la energía cinética traslacional es igual o inferior a ese límite, significando que la bala no atravesaría el chaleco.

La energía cinética traslacional de la bala (K), in joules, queda expresada con la siguiente fórmula:

K = \frac{1}{2}\cdot m\cdot v^{2} (1)

Donde:

m - Masa de la bala, en kilogramos.

v - Rapidez de la bala, en metros por segundo.

Si sabemos que m = 0.122\,kg y v = 350\,\frac{m}{s}, entonces la energía cinética traslacional de la bala es:

K = 7472.5\,J

Puesto que la energia cinética traslacional es mucho mayor que la capacidad del chaleco antibalas, la bala atravesaría el chaleco antibalas.

4 0
3 years ago
A very long, straight wire carries a current of 19.0 A in the +k direction. An electron 1.9 cm from the center of the wire in th
Anestetic [448]

(a) 1.03\cdot 10^{-16} N, -k direction

First of all, let's find the magnetic field produced by the wire at the location of the electron:

B=\frac{\mu_0 I}{2 \pi r}

where

I = 19.0 A is the current in the wire

r = 1.9 cm = 0.019 m is the distance of the electron from the wire

Substituting,

B=\frac{(1.256\cdot 10^{-6})(19.0A)}{2 \pi (0.019 m)}=2\cdot 10^{-4} T

and the direction is +j direction (tangent to a circle around the wire)

Now we can find the force on the electron by using:

F=qvBsin \theta

where

q=1.6\cdot 10^{-19}C is the electron's charge

v=3.23\cdot 10^6 m/s is the electron speed

B=2\cdot 10^{-4} T is the magnetic field

\theta is the angle between the direction of v and B

In this case, the electron is travelling away from the wire, while the magnetic field lines (B) form circular paths around the wire: this means that v and B are perpendicular, so \theta=90^{\circ}, sin \theta=1. So, the force on the electron is

F=(1.6\cdot 10^{-19}C)(3.23\cdot 10^6 m/s)(2\cdot 10^{-4} T)(1)=1.03\cdot 10^{-16} N

The direction is given by the right hand rule:

- Index finger: direction of motion of the electron, +i direction (away from the wire)

- Middle finger: direction of magnetic field, +j direction (tangent to a circle around the wire)

- Thumb: direction of the force --> since the charge is negative, the sign must be reversed, so it means -k direction (anti-parallel to the current in the wire)

(b) 1.03\cdot 10^{-16} N, +i direction

The calculation of the magnetic field and of the force on the electron are exactly identical as before. The only thing that changes this time is the direction of the force. In fact we have:

- Index finger: direction of motion of the electron, +k direction (parallel to the current in the wire)

- Middle finger: direction of magnetic field, +j direction (tangent to a circle around the wire)

- Thumb: direction of the force --> since the charge is negative, the sign must be reversed, so it means +i direction (away from the wire)

(c) 0

In this case, the electron is moving tangent to a circle around the wire, in the +j direction. But this is exactly the same direction of the magnetic field: this means that v and B are parallel, so \theta=0, sin \theta=0, therefore the force on the electron is zero.

6 0
3 years ago
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