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2 years ago

4. You exert 375 J to move a 45-kg object 5 m. How much force is required? (Points : 3)

1 answer:

4 0

In physics, a force is said to do work if, when acting, there is a displacement of the point of application in the direction of the force. It is calculated by the formula W = f x d. Therefore, we calculate the problem above as follows:

W = f x d
375 = f x 5
f = 75 N <------SECOND OPTION

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Sand is pouring out of a pipe and is forming a conical pile on the ground. The radius of the pile is increasing at a rate of 2 f

Anton [14]

Answer:

$a)\ V=3053.628059 \ ft^3$

$b)\ V'=1017.876\ ft^3/day$

Explanation:

Rate of Change

The volume of a cone of radius r and height h is given by

$\displaystyle V=\frac{\pi r^2h}{3}$

The height is said to be 1/2 of the radius, thus

$\displaystyle V=\frac{\pi r^2\cdot r}{2\cdot 3}$

$\displaystyle V=\frac{\pi r^3}{6}$

a) Knowing r=18 feet, the volume is

$\displaystyle V=\frac{\pi 18^3}{6}$

$V=3053.628059 \ ft^3$

b) The rate of change of the volume is computed by taking the derivative of both sides respect to the time

$\displaystyle V'=3\frac{\pi r^2}{6}r'$

$\displaystyle V'=\frac{\pi r^2}{2}r'$

Where r' is the given rate of change of the radius: 2 feet/day.

Now we compute

$\displaystyle V'=\frac{\pi 18^2}{2}\cdot 2=1017.876$

$\boxed{V'=1017.876\ ft^3/day}$

6 0

3 years ago

A particle with mass 1.09 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.985

e-lub [12.9K]

Answer:

a) $f = 0.598\,hz$ , b) $v_{max} = 3.701\,\frac{m}{s}$ , c) $k = 15.385\,\frac{N}{m}$ , d) $U = 1.081\,J$ , e) $K = 6.382\,J$ , f) $v\approx 3.422\,\frac{m}{s}$

Explanation:

a) The frequency of oscillation is:

$f = \frac{76}{127\,hz}$

$f = 0.598\,hz$

b) The angular frequency is:

$\omega = 2\pi \cdot f$

$\omega = 2\pi \cdot (0.598\,hz)$

$\omega = 3.757\,\frac{rad}{s}$

Lastly, the speed at the equilibrium position is:

$v_{max} = \omega \cdot A$

$v_{max} = (3.757\,\frac{rad}{s} )\cdot (0.985\,m)$

$v_{max} = 3.701\,\frac{m}{s}$

c) The spring constant is:

$\omega = \sqrt{\frac{k}{m}}$

$k = \omega^{2}\cdot m$

$k = (3.757\,\frac{rad}{s} )^{2}\cdot (1.09\,kg)$

$k = 15.385\,\frac{N}{m}$

d) The potential energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$U = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.375\,m)^{2}$

$U = 1.081\,J$

e) The maximum potential energy is:

$U_{max} = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.985\,m)^{2}$

$U_{max} = 7.463\,J$

The kinetic energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$K = U_{max} - U$

$K = 7.463\,J - 1.081\,J$

$K = 6.382\,J$

f) The speed when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

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

$v = \sqrt{\frac{2\cdot K}{m} }$

$v = \sqrt{\frac{2\cdot (6.382\,J)}{1.09\,kg} }$

$v\approx 3.422\,\frac{m}{s}$

4 0

3 years ago

Huryyy

zmey [24]

The transfer is perpendicular

Surprised
I knew this

3 0

3 years ago

Read 2 more answers

Argon makes up 0.93% by volume of air. Calculate its solubility (in mol/L) in water at 20°C and 1.0 atm. The Henry's law constan

vladimir2022 [97]

Answer:

S= 1.40x10⁻⁵mol/L

Explanation:

The Henry's Law is given by the next expression:

$S = k_{H} \cdot p$ (1)

where S: is the solubility or concentration of Ar in water, $k_{H}$ : is Henry's law constant and p: is the pressure of the Ar

Since the argon is 0.93%, we need to multiply the equation (1) by this percent:

$S = 1.5 \cdot 10^{-3} \frac{mol}{L\cdot atm} \cdot 1.0atm \cdot \frac{0.93}{100} = 1.40 \cdot 10^{-5} \frac{mol}{L}$

Therefore, the argon solubility in water is 1.40x10⁻⁵mol/L.

Have a nice day!

8 0

2 years ago

A blue ball is thrown upward with an initial speed of 21.8 m/s, from a height of 0.9 meters above the ground. 2.7 seconds after

worty [1.4K]

I can think of two possible and logical questions for the problem given. First, you can calculate for the maximum height reached by the blue ball. Second, you can compute the length of time for the two balls to be at the same height. If so, the solution are as follows:

When the object is thrown upwards or when the object is dropped from a height, the only force acting upon it is the gravitational force. Because of this, it simplifies equations of motion.

1. For the maximum height, the equation is
H = v₀²/2g
where
v₀ is the initial speed
g is the acceleration due to gravity equal to 9.81 m/s²

For the blue ball, v₀ = 21.8 m/s. Substituting the values:
H = (21.8 m/s)²/2(9.81m/s²)
H = 24.22 m
The maximum height reached by the blue ball is 24.22 m + 0.9 = 25.12 m.

2. For this, you equate the y values of both balls:

y for red ball = y for blue ball
v₀t + 0.5gt² = v₀t + 0.5gt²
(10.4 m/s)t + 0.5(9.81 m/s²)(t²) + 26.6 m = (21.8 m/s)t + 0.5(9.81 m/s²)(t²) + 0.9 m
Solving for t,
t = 2.25 seconds

Thus, the two balls would be at the same height after 2.25 seconds.

3 0

3 years ago