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Firlakuza [10]
3 years ago
9

Two runners ran side by side, each holding one end of a horizontal pole. How would this affect the direction of the runners? Exp

lain.
Physics
1 answer:
zhuklara [117]3 years ago
6 0

Answer:

They will run parallel to each other as the none of a straight pole cannot be bent in such a way where one side can turn without the other turning.

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Saturn's moon Mimas has an orbital period of 82,800 s at a distance of 1.87x10^8m from Saturn. Using m central m= (4n^2d^3/GT^2)
FromTheMoon [43]
5.65×10^26kg here you go
5 0
4 years ago
Read 2 more answers
what is the average acceleration of a car if a car rest from 0m/s and reaches a final velocity is 50m/s in 10m/s
GuDViN [60]

Answer:

Assuming that you meant the final velocity of 50 m/s was reached in 10 s, the answer would be 5 m/s^2.

Explanation:

V_{f} = V_{i} + at

So we update that with the values that we have.

50 = 0 +a(10)

then simplify that using algebra to solve for a and we get 5 m/s^2

6 0
3 years ago
Suppose that, from measurements in a microscope, you determine that a certain bacterium covers an area of 1.50μm2. Convert this
s344n2d4d5 [400]
1.50*10^-6=0.0000015m^2
6 0
3 years ago
You throw a baseball with a mass of 0.5 kg. The ball leaves your hand with a speed of 35 m/s. Calculate the kinetic energy. (SHO
mixas84 [53]

Answer:

The kinetic energy of the baseball is 306.25 joules.

Explanation:

SInce the baseball can be considered a particle, that is, that effects from geometry can be neglected, the kinetic energy (K), in joules, is entirely translational, whose formula is:

K = \frac{1}{2}\cdot m\cdot v^{2} (1)

Where:

m - Mass, in kilograms.

v - Speed, in meters per second.

If we know that m = 0.5\,kg and v = 35\,\frac{m}{s}, then the kinetic energy of the baseball thrown by the player is:

K = \frac{1}{2}\cdot m \cdot v^{2}

K = 306.25\,J

The kinetic energy of the baseball is 306.25 joules.

6 0
3 years ago
I’m not sure how to solve this
spayn [35]

Answer:

Option 10. 169.118 J/KgºC

Explanation:

From the question given above, the following data were obtained:

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1.61 KJ

Mass of metal bar = 476 g

Specific heat capacity (C) of metal bar =?

Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:

1 kJ = 1000 J

Therefore,

1.61 KJ = 1.61 KJ × 1000 J / 1 kJ

1.61 KJ = 1610 J

Next, we shall convert 476 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

476 g = 476 g × 1 Kg / 1000 g

476 g = 0.476 Kg

Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1610 J

Mass of metal bar = 0.476 Kg

Specific heat capacity (C) of metal bar =?

Q = MCΔT

1610 = 0.476 × C × 20

1610 = 9.52 × C

Divide both side by 9.52

C = 1610 / 9.52

C = 169.118 J/KgºC

Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC

6 0
3 years ago
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