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Korvikt [17]
2 years ago
11

A student picks up four books from the floor, walks across the room through some distance with the book

Physics
1 answer:
Marta_Voda [28]2 years ago
8 0

Work is performed while initially lifting books off the floor. The work is due to gravity.

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A constant force is applied to an object, causing the object to accelerate at 10 m/s^2. What will the acceleration be if: a) The
Liula [17]
What you need to know is that the force is

F=ma

The force is the product of mass and acceleration

this means that the acceleration is

a=F/m

a) The force is halved?
this means that f will be \frac{F}{2} now:

a=\frac{F}{2m}

So the accelaration will also he halved (it's the original acceleratation divided by 2)


 b) The object's mass is halved?
a=\frac{F}{m/2}=a=\frac{F2}{m}

which is the original acceleration times two!! so it will double


c) The force and the object's mass are both halved?
now we have

a=\frac{F/2}{m/2}=a=\frac{2F}{2m}=a=\frac{F}{m}

so they will cancel each other out and the acceleration will stay the same!











5 0
3 years ago
A(n) 0.2 kg object is swung in a vertical circular path on a string 0.1 m long. The acceleration of gravity is 9.8 m/s2 . If a c
Leya [2.2K]

Answer:

T=83.37N

Explanation:

Since the object is under a circular motion, according to Newton's second law, when the object is at the top of the circle we have:

\sum F_y: T-mg=F_c

Where F_c is the centripetal force and is given by:

F_c=ma_c=m\frac{v^2}{r}

Replacing and solving for T:

T=m\frac{v^2}{r}+mg\\T=0.2kg\frac{(6.38\frac{m}{s})^2}{0.1m}+0.2kg(9.8\frac{m}{s^2})\\T=83.37N

8 0
3 years ago
How much force is needed to accelerate a vehicle with a mass of 1000 kg at a rate<br> of 5 m/s2?
Mashcka [7]
The force needed to accelerate a vehicle with a mass of 1000kg at a rate of 5m/s2 would be 5000
8 0
2 years ago
An elements atomic number is 85 how many protons would an atom of this element have?
garik1379 [7]

85 protons are in astatine have.

5 0
3 years ago
Read 2 more answers
A refrigerator removes 55.0 kcal of heat from the freezer and releases 73.5 kcal through the condenser on the back.How much work
sammy [17]

Here refrigerator removes 55 kcal heat from freezer

Refrigerator releases 73.5 kcal heat to surrounding

So here we can use energy conservation principle by II Law of thermodynamics

the law says that

Q_1 = Q_2 + W

here we know that

Q_1 = heat released to the surrounding

Q_2 = heat absorbed from freezer

W = work done by the compressor

now using above equation we can write

73.5 = 55 + W

W = 73.5 - 55

W = 18.5 kcal

So here compressor has to do 18.5 k cal work on it

5 0
3 years ago
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