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suter [353]
3 years ago
6

a student stands on a metric scale, his weight is 490 N. then another student hands him a 5.00-kg mass to hold. the force shown

on the scale is now
Physics
1 answer:
Gwar [14]3 years ago
4 0

Answer:

539.04 N

Explanation:

5kg is approximately 49.04 newton's so add 49.04 to 490 and you get 539.04 N

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A paintball is shot horizontally in the positive x direction at time t after the ball is shot it is 4 cm to the right and 4 cm b
Artist 52 [7]

<u>Answer:</u>

At time 2t the paint ball is at 8 cm to the right and 16 cm to the bottom

<u>Explanation:</u>

 We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Considering the horizontal motion of paint ball

    Distance traveled during time t = 4 cm

    Initial velocity = u m/s

   Acceleration = 0 m/s^2

So 4 = u*t+\frac{1}{2} *0*t^2\\ \\ u = \frac{4}{t}

Now at time 2t,

  S= u*2t+\frac{1}{2} *0*(2t)^2\\ \\=\frac{4}{t} *2t\\ \\ =8cm

  So horizontal distance traveled in time 2t = 8 cm to the right

Now considering the vertical motion of paint ball

  Distance traveled during time t = 4 cm

    Initial velocity = 0 m/s

   Acceleration = -g m/s^2

4=0*t-\frac{1}{2} *g*t^2\\ \\ t^2=\frac{-8}{g}

At time 2t,

     S=0*2t-\frac{1}{2} *g*(2t)^2\\ \\ =-\frac{1}{2} *g*4*\frac{-8}{g}\\ \\ =16 cm

 So vertical distance traveled in time 2t = 16 cm to the bottom

 

6 0
3 years ago
PLEASE HELP I NEED THE CORRECT ANSWER TODAY
Natalija [7]

Answer:

The most common oxidation numbers for a given element

5 0
3 years ago
What factors do NOT affect friction between two objects? Explain how you know this.
Black_prince [1.1K]

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.

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4 0
2 years ago
A horizontal insulating rod of length 11.8-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge.
SCORPION-xisa [38]

Answer:

11.962337 × 10^-4 N

Explanation:

Given the following :

Length L = 11.8

Charge = 29nC = 29 × 10^-9 C

Linear charge density λ = 1.4 × 10^-7 C/m

Radius (r) = 2cm = 2/100 = 0.02 m

Using the relation:

E = 2kλ/r ; F =qE

F = 2kλq/L × ∫dr/r

F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))

2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4

In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214

Hence,

(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N

3 0
3 years ago
Which would be the most reliable source for information about the toxicity of an industrial chemical?
kvasek [131]
A scientific journal article that is peer reviewed. This is because it is more likely not have factual information and sources to that information.
3 0
3 years ago
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