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yaroslaw [1]
2 years ago
10

Х

Physics
1 answer:
Ganezh [65]2 years ago
3 0

Answer:

A. topsoil the answer

Explanation:

I think its a correct answer

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To measure specific heat, the student flows air with a velocity of 20 m/s and a temperature of 25C perpendicular to the length
san4es73 [151]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

3 0
3 years ago
30cm³ of brine of relative density 1.15 and 42cm³ of water are mixed. What is the density of the final solution​
Aleks04 [339]

Answer:

I think it's the most important part in this

7 0
3 years ago
A passenger on your boat falls overboard. what should you do first?
yarga [219]
The very first thing that you should do when a passenger on your boat falls overboard is to throw a PFD or also known as a Personal Flotation Device. This would include anything that can help the passenger to float. But this step would differ, only if the passenger is not wearing a lifevest. 
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3 years ago
A baseball pitcher throws a ball across home plate. The ball travels 13.40m in 0.357s Determine the
Tanya [424]

Answer: I'm not sure what it needs to be rounded to, but I got 37.53501401 m/s

Explanation: The formula for speed is speed = distance/time. You plug in the distance (13.40) and the time (0.357), then divide 13.40 by 0.357

I hope this helps! :)

4 0
2 years ago
A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?
Tcecarenko [31]
The weight of the meterstick is:
W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance 
d_1 = 0.50 m - 0.40 m=0.10 m
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
(mg) d_2 = 0.20 Nm
from which we find the value of d2:
d_2 =  \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
4 0
3 years ago
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