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AlladinOne [14]
3 years ago
9

Work is being done in which of these situations? all motions are at a constant velocity

Physics
1 answer:
fredd [130]3 years ago
7 0
Where the force is not perpendicular to the path of motion 

are you missing the the situations ?

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If an astronaut had a mass of 30 kg on the moon what would his mass be on earth?
Novay_Z [31]

Explanation:

A one-kilogram mass is still a one-kilogram(as mass is an intrinsic property of the object) but the downward force due to gravity, and therefore it's weight, is only one-sixth of what the object would have on the Earth. So man of mass 180 pounds weights only about 30 pounds-force when visiting the moon

hope it help..... pls add me as brainlist.

Have a nice day

7 0
3 years ago
Sam makes six star-shaped cookies out of dough. He puts all of the cookies on a balance and measures the total weight as 36 gram
son4ous [18]

Answer:

The answer is "36 grams".

Explanation:

In this question, the weight of the ball is not mentioned but is the weight of the cookies is declared, which is equal to 36 grams, and all the cookies are squeezes into the ball and after that, it calculates the overall weight so, let assume that ball weight is =0 and then the overall weight is:

=\text{weight of ball + cookies weight}\\\\=0+36 \ grams \\\\=36 \ grams

5 0
3 years ago
Write answers with significant figures:<br>a) 17.35 g +8.498 g​
Genrish500 [490]

Answer:

25.9 g

Explanation:

= 17.35

8.498

________+

= 25.848 g = 25.85 g = 25.9 g

*so sorry if wrong

6 0
2 years ago
__________energy might also be released during a chemical reaction
pentagon [3]
Kinetic energy i think
7 0
3 years ago
A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the cen
NeTakaya

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

F_1 + F_2 = W

F_1 + F_2 = (1530\times 9.81)

F_1 + F_2 = 15009.3 N

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

F_1(1.15) = F_2(2.70 - 1.15)

F_1 = 1.35 F_2

now from above equation

F_2 + 1.35F_2 = 15009.3

now we have

F_2 = 6392.85 N

now the other force is given as

F_1 = 8616.45 N

4 0
3 years ago
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