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Ket [755]
3 years ago
11

What is impossible for a machine to do

Physics
1 answer:
maksim [4K]3 years ago
6 0

Answer:

Its impossible for a machine to work without an energy source

Explanation:

pls give brainliest

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Please help me ejnfmwkefwefm i dont wanna fail
Rasek [7]

Answer:B

Explanation:

6 0
3 years ago
YALL THIS QUESTION IS DUE IN 2 HOURS AND I HAVE 3 MORE PAGED TO DO PLEASE HELP ME WITH THIS PHYSICS HOMEWORK PROBLEM!!!! I ATTAC
goldfiish [28.3K]

Answer:

I dont know :)

Explanation:

7 0
3 years ago
When an object's final velocity is less than its initial velocity, however, it has ________________ acceleration.
algol [13]

Answer:

The body has negative acceleration PR a deceleration.

Explanation:

HOPE THAT THIS IS HELPFUL.

HAVE A GREAT DAY.

4 0
3 years ago
If the value of static friction for a couch is 400N, what could be a possible value of its kinetic friction? *
Dafna11 [192]

Answer:

kinetic friction may be greater than 400 N or smaller than 400 N

Explanation:

As we know that maximum value of static friction on the rough surface is known as limiting friction and the formula of this limiting friction is known as

F_s = \mu_s N

now when object is sliding on the rough surface then the friction force on that surface is known as kinetic friction and the formula of kinetic friction is known as

F_k = \mu_k N

now we know that

\mu_k < \mu_s

so here value of limiting static friction force is always more than kinetic friction

also we know that

initially when body is at rest then static friction value will lie from 0 N to maximum limiting friction

and hence kinetic friction may be greater than static friction or if the static friction is maximum limiting friction then kinetic friction is smaller than static friction

so kinetic friction may be greater than 400 N or smaller than 400 N

5 0
3 years ago
What is the specific heat of an object if it takes 1200 J to raise the temperature of a 20 kg object by 6 degrees C?
ch4aika [34]

Answer:

c=10\ J/kg^{\circ} C

Explanation:

Given that,

Heat required, Q = 1200 J

Mass of the object, m = 20 kg

The increase in temperature, \Delta T=6^{\circ} C

We need to find the specific heat of the object. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{1200}{20\times 6}\\\\c=10\ J/kg^{\circ} C

So, the specific heat of the object is 10\ J/kg^{\circ} C.

5 0
3 years ago
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