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attashe74 [19]
3 years ago
13

Can someone help?? 3/5

Physics
1 answer:
ycow [4]3 years ago
8 0

Answer:

Yes to answer is 3/5 of what

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What is the weight of an object with a mass of 6.0 kg on Earth?
gregori [183]
<h2>since weight is measured in newtons, convert the 6 kg to newtons</h2><h3>the formula to convert is kg x 9.807 = N</h3>
  • 6 x 9.807
  • = 58.842 N

hope that helps :))

3 0
3 years ago
Read 2 more answers
The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re
lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

\Delta x = \Delta y

Where:

\Delta x - Range of the bullet, measured in meters.

\Delta y - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

K_{1} = U_{2}

Where:

K_{1} - Kinetic energy at point 1, measured in joules.

U_{1} - Gravitational potential energy at point 2, measured in joules, and:

U_{2} = m\cdot g \cdot \Delta y

Where:

m - Mass of the bullet, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

K_{1} = m\cdot g \cdot \Delta y

\Delta y = \frac{K_{1}}{m\cdot g}

If K_{1} = 1066\,J, m = 0.25\,kg and g = 9.81\,\frac{m}{s^{2}}, the maximum height is now computed:

\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\Delta y = 434.791\,m

\Delta y = 0.435\,km

Lastly, the range of the bullet is 0.435 kilometers.

3 0
3 years ago
Please need in hurry
gulaghasi [49]

Explanation:

i) center of gravity (or mass)

ii) m = W/g = (160 N)/(9.8 m/s^2)

= 16.3 kg

3 0
3 years ago
One end of a horizontal spring with force constant 130.0 Ni'm is attached to a vertical wall. A 4.00-kg block sitting on the flo
tekilochka [14]

Answer:

Explanation:

When the spring is compressed by .80 m , restoring force by spring on block

= 130 x .80

= 104 N , acting away from wall

External force = 82 N , acting towards wall

Force of friction acting towards wall = μmg

= .4 x 4 x 9.8

= 15.68 N

Net force away from wall

= 104 -15.68 - 82

= 6.32 N

Acceleration

= 6.32 / 4

= 1.58 m / s²

It will be away from wall

Energy released by compressed spring = 1/2 k x²

= .5 x 130 x .8²

= 41.6 J

Energy lost in friction

= μmg x  .8

= .4 x 4 x 9.8 x .8

= 12.544 J

Energy available to block

= 41.6 - 12.544 J

= 29 J

Kinetic energy of block = 29

1/2 x 4 x v² = 29

v = 3.8 m / s

This will b speed of block as soon as spring relaxes. (x = 0 )

4 0
3 years ago
A force of 19 newtons is applied on a cart of 2 kilograms, and it experiences a frictional force of 1.7 newtons. What is the acc
Umnica [9.8K]
Frictional force always opposes applied force, so the net force on the cart would have to be 19N - 1.7N. The acceleration can then be solved by using the relation: F = ma. This is shown below:

Net force = 19 - 1.7 = 17.3 N

Acceleration = Force / mass
Acceleration = 17.3 / 2
Acceleration = 8.65 N/m
4 0
3 years ago
Read 2 more answers
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