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2 years ago

11

In a concave mirror, if an object is located behind the center of the curvature, What is the best description of the Image forme

d?

1 answer:

kipiarov [429]2 years ago

5 0

be either real or virtual.

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What type of friction present when you wrench on a car?

dlinn [17]

Answer:Rolling friction is friction that acts on objects when they are rolling over a surface. Rolling friction is much weaker than sliding friction or static friction. This explains why most forms of ground transportation use wheels, including bicycles, cars, 4-wheelers, roller skates, scooters, and skateboards.

Explanation:

4 0

2 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

2 years ago

"A student bikes to school by traveling first dN = 0.900" and "Similarly, let d⃗ W be the displacement vector corresponding to t

raketka [301]

Complete Question

A student bikes to school by traveling first dN = 0.900 miles north, then dW = 0.300 miles west, and finally dS = 0.100 miles south.

Similarly, let d⃗ W be the displacement vector corresponding to the second leg of the student's trip. Express d⃗ W in component form.

Express your answer as two numbers separated by a comma. Be careful with your signs.

Answer:

The value is $dT = ( -0.3, 0.8)$

Explanation:

From the question we are told that

The first displacement is $dN = 0.900 \ due \ North$ i.e positive y-axis

The second displacement is $dW = 0.300 \ miles \ due \ west$ i.e negative x-axis

The final displacement is $dS = 0.100 \ miles \ due \ south$ i.e negative y-axis

Generally dW in component for is

$dW = (-0.3 , 0)$

Generally the total displacement of the student is mathematically represented as

$dT = ( -0.3, (0.90 - 0.10))$

$dT = ( -0.3, 0.8)$

6 0

3 years ago

What two simple machines make up an axe?

serg [7]

It would be A since the axe head acts as a wedge and the handle is a basic lever

7 0

2 years ago

Do you wont to be friends

Angelina_Jolie [31]

Answer:

shore

Explanation:

8 0

3 years ago