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igomit [66]
2 years ago
8

Koala bears can eat only certain kind of Australian eucalyptus leaves.koalas are considered

Physics
1 answer:
DaniilM [7]2 years ago
3 0
<em>Hello there, and thank you for asking your question here on brainly.

<u>Answer: Koala bears are considered herbivores, or as in the scientific name, arboreal herbivorous marsupial, marsupial because it also carries it's babies around in a pouch. Koala bears are also native to Australia, which eucalyptus leaves are also native to.
</u>
Hope this helped you! ♥</em>
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While playing basketball in PE class, Logan lost his balance after making a lay-up and colliding with the padded wall behind the
olga2289 [7]

Answer:

a.) F = 3515 N

b.) F = 140600 N

Explanation: given that the

Mass M = 74kg

Initial velocity U = 7.6 m/s

Time t = 0.16 s

Force F = change in momentum ÷ time

F = (74×7.6)/0.16

F = 3515 N

b.) If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds

Change in momentum = 74×7.6 + 74×7.6

Change in momentum = 562.4 + 562.4 = 1124.8 kgm/s

F = 1124.8/0.0080 = 140600 N

6 0
2 years ago
A pendulum is used in a large clock. The pendulum has a mass of 2kg. If the pendulum is moving at a speed of 2.9 m/s when it rea
Vanyuwa [196]
You first us 1/2(mv^2) to solve for the potential energy and then put that in to PE=m*g*h and solve for hight

3 0
2 years ago
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A thermally isolated system is made up of a hot piece of aluminum and a cold piece of copper; the aluminum and the copper are in
blsea [12.9K]

Answer:

b) It is impossible to tell without knowing the masses.

Explanation:

The temperature change of a substance when it receives/gives off a certain amount of heat Q is given by

\Delta T= \frac{Q}{m C_s}

where

Q is the amount of heat

m is the mass of the substance

Cs is the specific heat capacity of the substance

In this case, we have a hot piece of aluminum in contact with a cold piece of copper: the amount of heat given off by the aluminum is equal to the amount of heat absorbed by the copper, so Q is the same for the two substances. However, we see that the temperature change of the two substances depends on two other factors: the mass, m, and the specific heat, Cs. So, since we know only the specific heat of the two substances, but not their mass, we can't tell which object will experience the greater temperature change.

4 0
2 years ago
16) A wheel of moment of inertia of 5.00 kg-m2 starts from rest and accelerates under a constant torque of 3.00 N-m for 8.00 s.
KiRa [710]

Answer:

57.6Joules

Explanation:

Rotational kinetic energy of a body can be determined using the expression

Rotational kinetic energy = 1/2Iω²where;

I is the moment of inertia around axis of rotation. = 5kgm/s²

ω is the angular velocity = ?

Note that torque (T) = I¶ where;

¶ is the angular acceleration.

I is the moment of inertia

¶ = T/I

¶ = 3.0/5.0

¶ = 0.6rad/s²

Angular acceleration (¶) = ∆ω/∆t

∆ω = ¶∆t

ω = 0.6×8

ω = 4.8rad/s

Therefore, rotational kinetic energy = 1/2×5×4.8²

= 5×4.8×2.4

= 57.6Joules

6 0
2 years ago
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4. A 505-turn circular-loop coil with a diameter of 15.5 cm is initially aligned so that
Basile [38]

The strength of the magnetic field is 4.8\cdot 10^{-5} T

Explanation:

According to Faraday's Law, the magnitude of the induced emf in the coil is equal to the rate of changeof the flux linkage through the coil:

\epsilon = \frac{N\Delta \Phi}{\Delta t} (1)

where

N = 505 is the number of turns in the coil

\Delta \Phi is the change in magnetic flux through the coil

\Delta t = 2.77 ms = 2.77\cdot 10^{-3} s is the time interval

\epsilon = 0.166 V

The coil is rotated from a position perpendicular to the Earth's magnetic field to a position parallel to it, so the final flux is zero, and the magnitude of the flux change is simply equal to the initial flux:

\Delta \Phi = B A cos \theta

where

B is the strength of the magnetic field

A is the area of the coil

\theta=0^{\circ} is the angle between the normal to the coil and the field

The area of the coil can be written as

A=\pi r^2

where

r=\frac{15.5 cm}{2}=7.75 cm = 7.75\cdot 10^{-2} m is its radius

Substituting everything into eq.(1) and solving for B, we find:

\epsilon= \frac{NB\pi r^2 cos \theta}{\Delta t}\\B=\frac{\epsilon \Delta t}{\pi r^2 cos \theta}=\frac{(0.166)(2.77\cdot 10^{-3})}{(505)\pi (7.75\cdot 10^{-2})^2(cos 0^{\circ})}=4.8\cdot 10^{-5} T

Learn more about magnetic fields:

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8 0
2 years ago
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