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Strike441 [17]
3 years ago
8

As a mouse sits on a desk, it has a gravitational force of 1x10^-9N pulling it towards the keyboard. If a different mouse with t

riple the mass was used, instead what would be the new gravitational force?
Physics
1 answer:
Nikitich [7]3 years ago
8 0

Answer:

The new gravitational pull would be three times as strong as the first which would be 0.000000003 or 3 × 10^-9N.

Explanation:

Hope this helps. . . <3

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Suppose two square wave pulses that are in the same plane could be created on a string, one with a maximum displacement of 4 cm
butalik [34]

Answer:

2 cm and 6 cm.

Explanation:

The maximum displacement of first square wave pulse is 4 cm and the maximum displacement of second square wave pulse is 2 cm. We need to find a possible displacement of the string when the wave pules overlap.

It could be result in constructive interference or destructive interference.The possible displacement of the string are :

4 cm - 2 cm = 2 cm

4 cm + 2 cm = 6 cm

Hence, the correct options are (A) and (C).                                                                          

6 0
3 years ago
1) Radiation from the sun can be deflected or contained by
earnstyle [38]

Answer:

b

Explanation:

light colors deflect ligjt

7 0
3 years ago
Part
ArbitrLikvidat [17]

Answer:

Explanation:

the block will move to the right side with small velocity because the force from the left side greater than force from right side. Velocity will be less because of friction and gravitational attraction.

8 0
3 years ago
Thorium^+2
Bad White [126]

Answer:

chemical symbol: Th

atomic number:90

protrons :90

neutrons:142

group#:4

period#: 9

Explanation:

you take the atomic weight (232.038)and subtract the atomic number to get (90) which is your neutrons

6 0
3 years ago
You are crouched at the "start" line. Bang! The race begins and 7 seconds later you end up 60 yards down the track. What was you
Nutka1998 [239]

Answer:

a = 7.35 ft / s²

Explanation:

For this exercise we must use the kinematics relations

        x = v₀ t + ½ a t²

as the runner leaves the starting line his initial velocity is zero

        x = ½ a t²

        a = \frac{2x}{t^2}

let's reduce the distance to foot

        x = 60 yd (3ft / 1yd) = 180 ft

let's calculate

         a = 2 180 / 7²

         a = 7.35 ft / s²

6 0
3 years ago
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