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Vinil7 [7]
3 years ago
10

A baseball is given an initial velocity with magnitude v at the angle beta above the surface of an incline which in turn incline

d at angle teta above horizontal calculate the distance measured along incline from the launch point to where the baseball strike the incline
​

Physics
1 answer:
lakkis [162]3 years ago
8 0

Explanation:

The maximum height of an object, given the initial launch angle and initial velocity is found with:h=v2isin2θi2g h = v i 2 sin 2 ⁡ θ i 2 g .

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A 1.2-kg mass suspended from a spring of spring constant 22 N.m-1 executes simple harmonic motion of amplitude 5 cm. What is the
iren2701 [21]

Answer:

a)  T = 1,467 s , b)    A = 0.495 m , c)  v = 4.97 10⁻² m / s

Explanation:

The simple harmonic movement is described by the expression

        x = A cos (wt + Ф)

Where the angular velocity is

       w = √ k / m

a) Ask the period

Angular velocity, frequency and period are related

      w = 2π f = 2π / T

      T = 2π / w

      T = 2pi √ m / k

      T = 2π √ (1.2 / 22)

      T = 1,467 s

      f = 1 / T

      f = 0.68 Hz

b) ask the amplitude

The mechanical energy of a harmonic oscillator

        E = ½ k A²

       A = √2 E / k

       A = √ (2 2.7 / 22)

       A = 0.495 m

c) the mass changes to 8.0 kg

As released from rest Ф = 0, the equation remains

         x = A cos wt

        w = √ (22/8)

        w = 1,658

         x = 3.0 cos (1,658 t)

Speed ​​is

         v = dx / dt

         v = -A w sin wt

The speed is maximum when without wt = ±1

         v = Aw

         v = 0.03    1,658

         v = 4.97 10⁻² m / s

6 0
2 years ago
An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the sp
laila [671]

Answer:

The centripetal acceleration changed by a factor of 0.5

Explanation:

Given;

first radius of the horizontal circle, r₁ = 500 m

speed of the airplane, v = 150 m/s

second radius of the airplane, r₂ = 1000 m

Centripetal acceleration is given as;

a = \frac{v^2}{r}

At constant speed, we will have;

v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1

a₂ = 0.5a₁

Therefore, the centripetal acceleration changed by a factor of 0.5

7 0
3 years ago
In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point
Misha Larkins [42]

Answer:

The charge on the oil drop is 3e

Explanation:

F = qE

Where;

F is the applied force in Newton

E is the electric field potential N/C

q is charge in C

Given;

Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ  X volume of the oil drop

volume of the oil drop (spherical) =  (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³

mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg

⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N

F = qE

q = F/E

q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C

In terms of e

1e = 1.6 X10⁻¹⁹ C

=  (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e

Therefore, the charge on the oil drop is 3e

7 0
2 years ago
How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium
Minchanka [31]

The height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

<h3>Pressure and temperature at equilibrium </h3>

The relationship between pressure and temperature can be used to determine the height risen by the water.

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where;

  • V₁ = AL
  • V₂ = A(L - y)
  • P₁ = Pa
  • P₂ = Pa + ρgh
  • T₁ = 20⁰C = 293 K
  • T₂ = 10⁰ C = 283 k

\frac{PaAL}{T_1} = \frac{(P_a + \rho gh)A(L-y)}{T_2} \\\\\frac{PaL}{T_1} = \frac{(P_a + \rho gh)(L-y)}{T_2} \\\\L-y = \frac{PaLT_2}{T_1(P_a + \rho gh)} \\\\y = L (1 - \frac{PaT_2}{T_1(P_a + \rho gh)})\\\\y = 4.2(1 - \frac{101325 \times 283}{293(101325\  +\  1000 \times  9.8 \times  100)} )\\\\y = 3.8 \ m

Thus, the height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

The complete question is below:

A diving bell is a 4.2 m -tall cylinder closed at the upper end but open at the lower end. The temperature of the air in the bell is 20 °C. The bell is lowered into the ocean until its lower end is 100 m deep. The temperature at that depth is 10°C. How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium?

Learn more about thermal equilibrium here: brainly.com/question/9459470

#SPJ4

3 0
1 year ago
Atomic math challenge will give brainly and thanks
nevsk [136]

Answer:

1. Hydrogen

Atomic # = 1

Atomic Mass = 1.00794  ( If you round it it's 1.008 )

# of protons = 1

# of neutrons = none

# of electrons = 1

8 0
3 years ago
Read 2 more answers
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