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dolphi86 [110]
4 years ago
9

Recall some things you already know about projectile motion. Does a force in the vertical direction affect the horizontal compon

ent of an object’s velocity? In this situation, can you use the horizontal velocity component to find the time required to travel some horizontal distance?
Physics
1 answer:
azamat4 years ago
7 0

Answer:

vertical force cannot change the velocity on the x-axis.   t =x/v₀ₓ

Explanation:

The force is a vector magnitude, so the forces on the x-axis affect the acceleration on this axis. Consequently a vertical force cannot change the velocity on the x-axis.

     F_{y} = m g

    Fₓ = 0

The horizontal velocity in projectile motion is constant, if we neglect the air resistance, so it can be used to find the time of a horizontal displacement

       x = v₀ₓ t

       t =x/v₀ₓ

The only magnitude that is the same for both movements is the time that is a scalar

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A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.79 rad/s . Its total moment of inertia is 1790
Nezavi [6.7K]

Answer:

w_2=0.467rad/s

Explanation:

Four people standing on the ground each of mass and usually this questions have to find the final angular velocity

m_t=4*70kg=280kg

The radius r=4.2/2=2.1m

Angular velocity  w_1=0.79rad/s

The moment of inertia total is I_t=1790 kg/m^2

Momento if inertia

I_1=m_t*r^2

I_1=280kg*(2.1m)^2=1234.8kg*m^2

Angular momentum

I_1*w_1=I_t*w_2

Solve to w2

w_2=\frac{I_1*w_1}{I_t}

w_2=\frac{1790kg*m^2*0.79rad/s}{3024.8kg*m^2}

w_2=0.467rad/s

8 0
3 years ago
Calcular la longitud del faldón de una Rampa de Acceso , que en planta tiene una longitud de 20 m y la pendiente es 27%.
seraphim [82]

La longitud del faldón de la rampa es de 5.4 m.

 

La pendiente expresada en porcentaje sigue la siguiente ecuación:

m=\frac{y}{x}*100 (1)

Donde:

  • y es la elevacion de la rampa (faldón)
  • x es la longitud de la ramapa (20 m)

Sabemos que la pendiente es de 27%. Por lo tanto, usando la ecuación 1, despejamos y.

27=\frac{y}{20}*100

y=\frac{27*20}{100}

y=5.4\: m        

La longitud del faldón es 5.4 m

Pudes ver más sobre el tema aquí:

brainly.com/question/8906330

5 0
3 years ago
What would Hubble's constant be if we found one galaxy moving away at 30,000 km/s at a distance of 600 Mpc?
lora16 [44]

Answer:

H₀ = 1.6 x 10⁻¹⁸ s⁻¹

Explanation:

The Hubble's Constant can be found by the following formula:

v = H_o D\\\\H_o = \frac{v}{D}

where,

H₀ = Hubble's Constant = ?

v = speed of galaxy = 30000 km/s = 3 x 10⁷ m/s

D = Distacance = 600 Mpc = (6 x 10⁸ pc)(3.086 x 10¹⁶ m/1 pc)

D = 18.52 x 10²⁴ m

Therefore,

H_o = \frac{3\ x\ 10^7\ m/s}{18.52\ x\ 10^{24}\ m}

<u>H₀ = 1.6 x 10⁻¹⁸ s⁻¹</u>

3 0
3 years ago
During a new moon, the moon is between the
mylen [45]
The moon is between the sun and earth.


The side where the light from the sun hits the moon is facing away from earth.
5 0
3 years ago
One isotope of a metallic element has the mass number 65 and 35 neutrons in the nucleus. the cation derived from the isotope has
Lina20 [59]
The mass number of an isotope is the sum of the numbers of protons and the numbers of the neutron. Given that the mass number is 65 and the number of neutrons is 35, the number of protons is 30. The atom is then Zinc (Zn). The charge is equal to +2, as it lacks 2 more electrons. 
7 0
3 years ago
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