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lorasvet [3.4K]
4 years ago
11

Which best describes a force?

Physics
2 answers:
Alika [10]4 years ago
6 0

Answer:

D is the answer

Explanation:

lora16 [44]4 years ago
6 0
I’m going to say it is D
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The resistance would increase.
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3 years ago
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class 7 Physics answera plot measuring 2 km in length and 1 by 2 km in breadth has area of how many hectare ​
motikmotik

Answer:

Area of plot in hectare = 100 hectare

Explanation:

Given;

Length of plot = 2 km

Width of plot = 1/2 km = 0.5 km

Find:

Area of plot in hectare

Computation:

Area of plot = Length x width

Area of plot = 2 x 0.5

Area of plot = 1 square kilometer

1 square kilometer = 100 hectare

Area of plot in hectare = 1 x 100

Area of plot in hectare = 100 hectare

7 0
3 years ago
49 POINTS!!!
Sladkaya [172]

Answer:

Explanation:

chemical kinetic

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Electromagnetic (light, x Ray's, microwaves, etc) kinetic

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3 years ago
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A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 44.1m/s^2 . The acce
ollegr [7]
The acceleration and distance is related to the following expression:
y=v0*t + a*t^2/2 ; v0=0 
y=44.1*100/2 = 2205m 
hence, the speed will be 
v=0 + a*t = 441m/s 
from that height it will just be subjected to the gravitational acceleration 
0=v_acc^2 -2g*y_free 
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7 0
3 years ago
Suppose we want to calculate the moment of inertia of a 56.5 kg skater, relative to a vertical axis through their center of mass
kirza4 [7]

Answer:

a. 0.342 kg-m² b. 2.0728 kg-m²

Explanation:

a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m

I = 1/2MR²

= 1/2 × 56.5 kg × (0.11 m)²

= 0.342 kgm²

So the moment of inertia of the skater is

b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)

I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m

I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²

= 0.1802 kg-m² + 0.6852 kg-m²

= 0.8654 kg-m²

The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².

So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²

5 0
3 years ago
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