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never [62]
3 years ago
9

PLZ HELP ME

Physics
1 answer:
bonufazy [111]3 years ago
4 0

Answer:

1. The stone will strike the ground 49.46 m from the base of the cliff

2. A) Approximately 0.542 seconds

B)  Approximately 3.69 m/s

3. A) The time the ball spends in the air is approximately 4.0775 s

B) The horizontal range is approximately  141.25 m.

Explanation:

1. The time it takes the stone to land is given by the equation, t = √(h/(1/2 × g)

∴ t = √(30/(1/2 × 9.81)) ≈ 2.473 seconds

The horizontal distance covered by the stone in that time = 20 × 2.473 ≈ 49.46 m

The stone will strike the ground 49.46 m from the base of the cliff

2. A) The time the ball spends in the air = t = √(h/(1/2 × g)

∴ t = √(1.44/(1/2 × 9.81)) ≈ 0.542 seconds

B) The initial horizontal velocity, u = Horizontal distance/(Time) = 2/0.542 ≈ 3.69 m/s

The initial horizontal velocity ≈ 3.69 m/s

3. A) The time the ball spends in the air is given by the following equation;

t = 2 × u × sin(θ)/g = 2 × 40 × sin(30)/9.81 ≈ 4.0775 s

t ≈ 4.0775 s

B) The horizontal range, R, of the  ball is given by the equation for the range of a projectile as follows;

Range, R = \dfrac{u^2 \times sin (2 \cdot \theta) }{g}

Substituting the known values, gives;

Range, R = \dfrac{40^2 \times sin (2 \times 30^{\circ}) }{9.81} \approx 141.25 \ m

The horizontal range ≈ 141.25 m.

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A(n)_______is a reflected<br> sound wave.
strojnjashka [21]

Answer:

A reflected sound wave can be one of two things, an echo or a reverberation. Reverberation happens when sound bounces off surfaces and reaches back to the ear within 0.1 seconds. Echoes happen when sound waves bounce back to the ear after more than 0.1 seconds.

Explanation:

hehe

3 0
3 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
2 years ago
a 1150 kg car is on a 8.70 hill. using x-y axis tilted down the plane, what is the x-component of the normal force(unit=N)
rodikova [14]

The x-component of the normal force is equal to <u>1706.45 N.</u>

Why?

To solve the problem, and since there is no additional information, we can safely assume that the x-axis is parallalel to the hill surface and the y-axis is perpendicular to the x-axis. Knowing that, we can calculate the components of the normal force (or weight for this case), using the following formulas:

N_{x}=W*Sin(\alpha)=mg*Sin(\alpha)\\\\N_{y}=W*Cos(\alpha)=mg*Cos(\alpha)

Now, using the given information, we have:

mass=m=1150Kg\\\alpha=8.70\°\\g=9.81\frac{m}{s^{2}}

Calculating, we have:

N_{x}=mg*Sin(\alpha)

N_{x}=1150Kg*9.81\frac{m}{s^{2}}*Sin(8.70\°)\\\\N_{x}=11281.5\frac{Kg.m}{s^{2} }*Sin(8.70\°)=1706.45\frac{Kg.m}{s^{2} }=1706.45.23N

Hence, we have that the x-component of the normal force is equal to  <u>1706.45 N.</u>

Have a nice day!

3 0
4 years ago
Which of the following would add more resistance? Check all that apply.
Blababa [14]

Answer:C. E.

Explanation:

4 0
3 years ago
A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init
Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

h=\frac{gt^2}{2}

For second stone

h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}---2

Equating 1 &2 we get

\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}

\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

26.46t-35.721-52.92t+142.884=0

t=4.05 s

4 0
3 years ago
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