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Bingel [31]
3 years ago
14

A kitten sits in a lightweight basket near the edge of a table. A person accidentally knocks the basket off the table. As the ki

tten and basket fall, the kitten rolls, turns, kicks, and catches the basket in its claws. The basket lands on the floor with the kitten safely inside. If air resistance is negligible, what is the acceleration of the kitten-basket system while the kitten and basket are in midair? A The acceleration is directed downward with magnitude less than g because the basket is light. B The acceleration is directed downward with magnitude equal to g because the system is a projectile. C The acceleration fluctuates because of the rolling, turning, and kicking motion of the kitten. D The acceleration cannot be determined without knowing how hard the basket is pushed.
Physics
1 answer:
Lesechka [4]3 years ago
4 0

Answer:

the answer is B

Explanation: this was actually an ap exam question a few years back. the reason for answer B is that the only force being applied to the kitten is the force of gravity after being pushed.

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A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters​ [cm]. The rover picks up the rock and lifts it
Makovka662 [10]

Answer: 5166.347

Explanation:

The specific gravity of a solid SG (also called relative density) is the ratio of the density of that solid \rho_{rock} to the density of water \rho_{water}=1 kg/m^{3} (normally at 4\°C):

SG=\frac{\rho_{rock}}{\rho_{water}} (1)

On the other hand, the density of the rock is calculated by:

\rho_{rock}=\frac{m_{rock}}{V_{rock}} (2)

Where:

m_{rock} is the mass of the rock

V_{rock}=\frac{4}{3} \pi r^{3} is the volume of the rock, since is spherical

Well, we already know the value of \rho_{water}, but we need to find \rho_{rock} in order to find the rock's specific gravity; and in order to do this, we firsly have to find m_{rock} and then calculate V_{rock}:

In the case of the mass of the rock, we can calclate it by the following equation:

W_{rock}=m_{rock}g_{mars} (3)

Where:

W_{rock} is the weight if the rock in mars

g_{mars}=3.7 m/s^{2} is the acceleration due gravity in Mars

Isolating m_{rock}:

m_{rock}=\frac{W_{rock}}{g_{mars}} (4)

m_{rock}=\frac{W_{rock}}{3.7 m/s^{2}} (5)

To find W_{rock} we can use the following equation of the potential gravitational energy U:

U=W_{rock}H (6)

Where:

U=2 J=2 Nm is the potential energy

H=20 cm \frac{1m}{100 cm}=0.2 m is the height at which the rock has the mentioned potential energy

Isolating W_{rock}:

W_{rock}=\frac{U}{H} (7)

W_{rock}=\frac{2 Nm}{0.2 m} (8)

W_{rock}=10 N (9)

Substituting (9) in (5):

m_{rock}=\frac{10 N}{3.7 m/s^{2}} (10)

m_{rock}=2.702 kg (11)

Substituting (11) in (2):

\rho_{rock}=\frac{2.702 kg}{V_{rock}} (12) At this point we only need to find the volume of the rock, knowing its diameter is d=10 cm, hence its radius is r=\frac{d}{2}=5 cm

V_{rock}=\frac{4}{3} \pi (5 cm)^{3} (13)

V_{rock}=523.59 cm^{3} \frac{1 m^{3}}{(100 cm)^{3}}=0.000523 m^{3} (14)

Substituting (14) in (12):

\rho_{rock}=\frac{2.702 kg}{0.000523 m^{3}} (15)

\rho_{rock}=5166.34 kg/m^{3} (16)

Substituting (16) in (1):

SG=\frac{5166.34 kg/m^{3}}{1 kg/m^{3}} (17)

Finally we obtain the specific gravity of the​ rock:

SG=5166.347

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