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2 years ago

Physical science help

Physics

2 answers:

mojhsa [17]2 years ago

4 0

The answer is pulley

VMariaS [17]2 years ago

3 0

The flagpole used a PULLEY at the top.

The cutting edge of the axe is a WEDGE.

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Light is emitted by electrons when they drop from one energy level to a lower level. Which transition results in the emission of

natulia [17]

Answer:

Level 4 to level 2

Explanation:

Electrons in an atom are contained in specific energy levels (1, 2, 3, and so on) having different distances from the nucleus. When light is emitted by electrons from one energy level to a lower level, level 4 to level 2 has the greatest energy.

Hence, the correct option is "Level 4 to level 2".

3 0

3 years ago

Una partícula se mueve en el plano XY efectúa un desplazamiento mientras actúa sobre ella una fuerza constante. X= (4i + 3j) m,

dsp73

Answer:

a) La magnitud del desplazamiento es de 5 m

La magnitud de la fuerza es 20 N

b) El trabajo realizado por la fuerza es de 100 J

c) El ángulo entre la fuerza y el plano es 0 °

Explanation:

a) La magnitud del desplazamiento se encuentra por la relación;

$\left | X \right | = \sqrt{X_{x}^{2}+X_{y}^{2}}$

Lo que da;

$\left | X \right | = \sqrt{4^{2}+3^{2}} = 5 \ m$

De manera similar, la magnitud de la fuerza, F, se encuentra como sigue;

$\left | F \right | = \sqrt{F_{x}^{2}+F_{y}^{2}}$

Lo que da;

$\left | F \right | = \sqrt{16^{2}+12^{2}} = 20 \ N$

b) El trabajo, W, realizado por la fuerza = Fuerza, F × Distancia, X

∴ Ancho = 20 N × 5 m = 100 N · m = 100 J

c) La dirección de la fuerza viene dada por la siguiente fórmula;

$tan^{-1} \left (\dfrac{F_y}{F_x} \right ) = tan^{-1} \left (\dfrac{12}{16} \right ) = 38.9^{\circ}$

La dirección del plano viene dada por la siguiente fórmula;

$tan^{-1} \left (\dfrac{X_y}{X_x} \right ) = tan^{-1} \left (\dfrac{3}{4} \right ) = 38.9^{\circ}$

Por tanto, el ángulo entre la fuerza y el plano = 0 °

La fuerza actúa a lo largo del plano.

6 0

3 years ago

Help in my home work!<br>convert 80°c to fahrenheit?

Vsevolod [243]

Answer:

176 degrees fahrenheit

Explanation:

80°C × 9/5) + 32 = 176°F

Mark me brainliest!!

7 0

2 years ago

A ball is thrown with a velocity of 10 m/s at a 25º from the horizontal. How far does the ball travel?

Stels [109]

Answer:

use a=v-u/t formula it will be correct

5 0

3 years ago

The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge

lesya [120]

Answer:

Loss, $\Delta E=-10.63\ J$

Explanation:

Given that,

Mass of particle 1, $m_1=m =0.66\ kg$

Mass of particle 2, $m_2=7.4m =4.884\ kg$

Speed of particle 1, $v_1=2v_o=2\times 6=12\ m/s$

Speed of particle 2, $v_2=v_o=6\ m/s$

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

$m_1v_1+m_2v_2=(m_1+m_2)V$

$V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}$

$V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}$

V = 6.71 m/s

Initial kinetic energy before the collision,

$K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)$

$K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)$

$K_i=135.43\ J$

Final kinetic energy after the collision,

$K_f=\dfrac{1}{2}(m_1+m_2)V^2$

$K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2$

$K_f=124.80\ J$

Lost in kinetic energy,

$\Delta K=K_f-K_i$

$\Delta K=124.80-135.43$

$\Delta E=-10.63\ J$

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0

3 years ago