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2 years ago

5

A cat (5kg) has a potential energy of 8J. The cat is stuck on top of a bookshelf and then falls off the bookshelf. What is the v

elocity of the cat before it hits the ground?

1 answer:

kolbaska11 [484]2 years ago

8 0

Answer:

176J

Explanation:

the explanation is that before the cat hits the ground the amount of kinetic energy builds up making it so that there is more energy

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An ostrich can run at a speed of 43 mi/hr. How much ground can an ostrich cover if it runs at this speed for 15 minutes? (Hint:

Simora [160]

..... It would possibly she eenejjsjejeej 1.4

4 0

3 years ago

Your companion on a train ride through Illinois notices that telephone poles near the tracks appear to be passing by very quickl

Nikitich [7]

Answer: Relative motion

Explanation: If two objects are moving either towards or away from each other with both having their velocities in a reference frame and someone is outside this reference frame seeing the motion of the two objects.

The observer ( in his own frame of reference) will measure a different velocity as opposed to the velocities of the two object in their own reference frame. p

Both the velocity measured by the observer in his own reference frame and the velocity of both object in their reference is correct.

Velocities of this nature that have varying values based on motion referenced to another body is known as relative velocity.

Motion of this nature is known as relative motion.

<em>Note that the word reference frame is simply any where the motion is occurring and the specified laws of motion is valid</em>

<em />

For this example of ours, the reference frame of the companion is the train and the telephone poles has their reference frame as the earth.

The companion will measure the velocity of the telephone poles relative to him and the velocity of the telephone pole relative to an observer outside the train will be of a different value.

6 0

3 years ago

Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun

iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

$E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})$ (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

$E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV$

B. The energy of the emitted photon is given by the following formula:

$E=h\frac{c}{\lambda}$ (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

$2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J$

Next, you use the equation (2) and solve for λ:

$\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm$

C. The radius of the orbit is given by:

$r_n=n^2a_o$ (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

$r_5=5^2(2.380)=59.5$

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0

3 years ago

Calculate the induced electric field (in V/m) in a 40-turn coil with a diameter of 18 cm that is placed in a spatially uniform m

kozerog [31]

Answer:

ε = 6.617 V

Explanation:

We are given;

Number of turns; N = 40 turns

Diameter;D = 18cm = 0.18m

magnetic field; B = 0.65 T

Time;t = 0.1 s

The formula for the induced electric field(E.M.F) is given by;

ε = |-NAB/t|

A is area

ε is induced electric field

While N,B and t remain as earlier described.

Area = π(d²/4) = π(0.18²/4) = 0.02545

Thus;

ε = |-40 × 0.02545 × 0.65/0.1|

ε = 6.617 V

(we ignore the negative sign because we have to take the absolute value)

6 0

3 years ago

A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o

inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

$M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh$

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

$\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) = \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh$

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

$\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_i^2) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2 ) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 - \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 - \frac{3}{4}v_f^2\\\\$

$h = \frac{3}{4g}(v_1^2 -v_f^2)$

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

$h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m$

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0

3 years ago