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blondinia [14]
3 years ago
6

I need this right ASAP!!! plz plz plz

Physics
2 answers:
UkoKoshka [18]3 years ago
8 0

Answer:

The answer is D

Nucleus is neutral

KiRa [710]3 years ago
5 0
The answer is D
electrically neutral region at the center of the atom.
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Can someone help with my physics homework? please
Murrr4er [49]

Answer:

a) 19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) 24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

Explanation:

a) If force applied in the cart is uniform, that is, constant in magnitude and direction and is parallel to distance travelled by the car, the work done on the cart is defined by the following equation:

W = F\cdot \Delta s (1)

Where:

F - Force applied by the cart, measured in newtons.

\Delta s - Distance travelled by the car, measured in meters.

W - Work done on the cart, measured in joules.

If we know that F = 888\,N and \Delta s = 22\,m, then the work done on the cart is:

W =(888\,N)\cdot (22\,m)

W = 19536\,J

19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) If we know that F = 1111\,N and \Delta s = 22\,m , then the work on the cart is:

W = (1111\,N)\cdot (22\,m)

W = 24442\,N

24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The gas consumption is directly proportional to the square of velocity and mass of the cart and, hence, to the work done on the cart. In consequence, we conclude that the bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

3 0
3 years ago
A student wants to design an experiment to study the transformation of mechanical energy. Which object can be used to investigat
Vsevolod [243]
I guess it’s the 3rd one
7 0
3 years ago
Read 2 more answers
A runner of mass 56.0kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its cent
SCORPION-xisa [38]

Answer: -0.84 rad/sec (clockwise)

Explanation:

Assuming no external torques act on the system (man + turntable), total angular momentum must be conserved:

L1 = L2

L1 = It ω + mm. v . r = 81.0 kg . m2 .21 rad/s – 56.0 kg. 3.1m/s . 3.1 m  

L1 = -521.15 kg.m2/sec (1)

(Considering to the man as a particle that is moving opposite to the rotation of  the turntable, so the sign is negative).

Once at rest, the runner is only a point mass with a given rotational inertia respect from the axis of rotation, that can be expressed as follows:

Im = m. r2 = 56.0 kg. (3.1m)2 = 538.16 kg.m2

The total angular momentum, once the runner has come to an stop, can be written as follows:

L2= (It + Im) ωf = -521.15 kg.m2/sec  

L2= (81.0 kg.m2 + 538.16 kg.m2) ωf = -521.15 kg.m2/sec  

Solving for ωf, we get:

ωf = -0.84 rad/sec  (clockwise)

5 0
3 years ago
two groups of students carry out experiments to investigate the relationships between force and extenstion
IrinaK [193]

Answer:

They investigated Hooke's law.

Explanation:

{ \tt{force \:  \alpha  \: extension }} \\ { \boxed{ \bf{F = kx}}} \\ x \: is \: extension \\ F \: is \: force \\ k \: is \: constant : k = 1

5 0
3 years ago
What is the gravitational force between two 1 kg objects that are 1 m apart?
lapo4ka [179]

Answer:

F=6.67\times 10^{-11}\ N

Explanation:

Given that,

The masses of two objects, m₁ = m₂ = 1 kg

The distance between masses, d = 1 m

We need to find the gravitational force between two masses. The force is given by the relation as follows :

F=G\dfrac{m_1m_2}{d^2}\\\\F=6.67\times 10^{-11}\times \dfrac{1\times 1}{(1)^2}\\F=6.67\times 10^{-11}\ N

So, the force between two masses of 1 kg is 6.67\times 10^{-11}\ N.

3 0
3 years ago
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