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saveliy_v [14]
3 years ago
15

A 10.0 kg mass is at the origin. A 20.0 kg mass is 0.500 m to the left of it, and a 30.0 kg mass is 1.25 m to the right of it, w

hat is the net gravitational force on the 10.o kg mass?
? x 10^-8 n
Physics
1 answer:
IRINA_888 [86]3 years ago
4 0

Explanation:

100N please mark me brainliest

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A veritical brass rod of circular section is loaded by placing a 10 kg wt on top of it .if it's length is 1 m. it's radius of cr
Inga [223]

Answer:

4.37 * 10^-4 J

Explanation:

Energy stored :

mgΔl / 2

m = mass = 10kg ; g = 9.8m/s² ; r = cross sectional Radius = 1cm = 1 * 10-2 m

Δl = mgl / πr²Y

Y = Youngs modulus = Y=3.5 ×10^10 ; l = Length = 1m

Δl = (10 * 9.8 * 1) / π * (1 * 10^-2)²* 3.5 ×10^10

Δl = 98 / 3.5 * π * 10^6

Δl = 0.00000891267

Energy stored :

mgΔl / 2

(10 * 9.8 * 0.00000891267) / 2

= 0.00043672083 J

4.37 * 10^-4 J

3 0
2 years ago
List the factors that affecting frictional force ?​
dexar [7]

Answer:

The frictional force between two bodies depends mainly on three factors: (I) the adhesion between body surfaces (ii) roughness of the surface (iii) deformation of bodies

7 0
2 years ago
1. What is the total distance traveled?<br> A 3.0m <br> B 4.0m <br> C 5.0m <br> D 6.0m
Ostrovityanka [42]

Answer:

c

Explanation:

7 0
2 years ago
An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17
Softa [21]

Answer:

distance cover is  = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

t_1 = 3.32 sec

t_2 = 5.08 sec

from equation of motion we know that

d_1 = vt_1 + \frac{1}{2} gt_1^2

where d_1 is distance covered in time t1

sod_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2=

d_1 = 110.78 m

d_2 = vt_2 + \frac{1}{2} gt_2^2

where d_2 is distance covered in time t2

d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2

d_2 = 213.31 m

distance cover is  = 213.31 - 110.78 = 102.53 m

3 0
3 years ago
How long does it take the projectile to reach its highest point in its trajectory?​
NeTakaya

Answer:

It takes about 88 seconds for the cannonball to reach its maximum height (ignoring air resistance).

Explanation:

3 0
3 years ago
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