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3 years ago

a snail can move approximately 0.30 meters per minute.How many meters can the snail cover in 15 minutes ?

1 answer:

8 0

Sometimes if you have the word per, you need to multiply. In this problem, they are asking you to find the amount of meters a snail can travel per minute.They are giving you the amount of minutes, so in this case you multiply 15 times 0.30. The answer is 4.5 meters.

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Determine the mass of a sphere with a radius of 8 cm having a density of 7500 kg / m3

Aliun [14]

Answer:

Explanation:

V=2140CM3

M=VOLUME*DENSITY

7500kg/m3=7.5g/cm3

2140cm3*7.5g/cm3=16050

16050g

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2 years ago

How should the student change the circuit to give negative values for current and

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Flip the cell.

Explanation:

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2 years ago

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3 years ago

If the weight of the ruler is one Newton ,Gc cannot have a value more than 25cm

nevsk [136]

If the length of the ruler is 50 cm, the center of gravity cannot be greater than 25 cm.

The given parameters:

Weight of the ruler = 1 N

<h3>What is center of gravity (CG)?</h3>

Center of gravity is the point at which the weight of an object is concentrated.

Let the length of the ruler = L

The center of the gravity of the ruler is calculated as follows;

$X_{CG} = \frac{W(L_0) + W(L -X_{CG})}{W} \\\\X_{CG} = \frac{1(0) + 1(L -X_{CG})}{1}\\\\X_{CG} = L - X_{CG}\\\\X_{CG } + X_{CG} = L\\\\2X_{CG} = L\\\\X_{CG} = \frac{L}{2} \\\\when , \ L = 50 \ cm\\\\X_{CG} = \frac{50}{2} \\\\X_{CG} = 25 \ cm$

Thus, if the length of the ruler is 50 cm, the center of gravity cannot be greater than 25 cm. This may change if the length of the ruler changes because the center of gravity of uniform ruler depends on the length of the ruler.

Learn more about center of gravity here: brainly.com/question/6765179

4 0

2 years ago

On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs accou

Feliz [49]

Answer:

Angular velocity, $N_f = 242.36 rpm$

Explanation:

The mass of the skater, M = 74.0 kg

Mass of each arm, $m_{a} = 0.13 * \frac{M}{2}$ ( since it is 13% of the whole body and each arm is considered)

$m_{a} = 0.13 * 37\\m_a = 4.81 kg$

Mass of the trunk, $m_{t} = M - 2m_{a}$

$m_t = 74 - 2(4.81)\\m_{t} = 64.38 kg$

Total moment of Inertia = (Moment of inertia of the arms) + (Moment of inertia of the trunks)

$(I_{T} )_i = 2(\frac{m_{a}L^2 }{12} + m_a(0.5L + R)^2) + 0.5 m_t R^2$

$(I_{T} )_i = 2(\frac{4.81 * 0.7^2 }{12} + 4.81(0.5*0.7 + 0.175)^2) + 0.5 *64.38* 0.175^2\\(I_{T} )_i = 3.052 + 0.986\\(I_{T} )_i = 4.038 kgm^2$

The final moment of inertia of the person:

$(I_{T} )_f = \frac{1}{2} MR^{2} \\(I_{T} )_f = \frac{1}{2} * 74*0.175^{2}\\(I_{T} )_f = 1.133 kg.m^2$

According to the principle of conservation of angular momentum:

$(I_{T} )_i w_{i} = (I_{T} )_f w_{f}\\w_{i} = 68 rpm = (2\pi * 68)/60 = 7.12 rad/s\\4.038 * 7.12 =1.133* w_{f}\\w_{f} = 25.38 rad/s\\w_{f} = \frac{2\pi N_f}{60} \\25.38 = \frac{2\pi N_f}{60}\\N_f = (25.38 * 60)/2\pi \\N_f = 242.36 rpm$

3 0

3 years ago