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ch4aika [34]
3 years ago
14

How much force is required to accelerate a 5 kg object at 5 m/s^2?

Physics
1 answer:
kramer3 years ago
3 0

Answer:25n

Explanation:

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Who was the first woman to compete and win a championship in the Olympics
natita [175]
Hélène de Pourtalès she was the first women
6 0
3 years ago
What is the wavelength of a wave if its frequency is 256 Hz and speed of the wave is 350 m/s?
DiKsa [7]

Answer:

Explanation:

The equation for this is

f=\frac{v}{\lambda} where f is the frequency, v is the velocity, and lambda is the wavelength. Filling in:

256=\frac{350}{\lambda} and

\lambda=\frac{350}{256} which means that

the wavelength is 1.37 m, rounded to the correct number of significant digits.

7 0
3 years ago
A body is placed on a rough inclined plane. Why does the frictional force decrease with the increase of angle of inclination
kondor19780726 [428]
The frictional force is directly proportional to the force that is perpendicular on the surface.

When the body is placed on a horizontal level with zero inclination, the only force acting on the body is the gravitational force which always pulls the body down. The gravitational force, in this case, is the perpendicular force to the surface. Accordingly, this entire force is used to generate friction

Now as the inclination of the surface increases, the gravitational force is no longer the perpendicular force of the body, its value decreases, which means only a part is used to generate frictional force. Consequently, frictional force decreases.

When the inclination reaches 90 degrees, the gravitational force does not act along the normal and accordingly, no friction force is generated.

7 0
4 years ago
A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t
Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

                       v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time  

                       V = at + U

using equation  v - u = at to get line equation for the graph of the motion of the train on the incline plane

                       V_{x} = mt + V_{o}      where m is the slope

Comparing equation (1) and (2)

V = V_{x}

a = m    

U = V_{o}

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

          a = -  1.40 m/s²

Sunstituting a = -  1.40 m/s² and  u = 22 m/s

                        V_{x} = -1.40t + 22

                            V_{x} = -1.40(7.30) + 22

                             V_{x} = -10.22 + 22

                             V_{x} = 11. 78 m/s

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

           Area of triangle +  Area of rectangle

          [\frac{1}{2} * (22 - 11.78) * (7.30)]  + [(11.78 - 0) * (7.30)]

                           = 37.303 + 85.994

                           = 123. 297 m

                           ≈ 123. 30 m

                 

4 0
3 years ago
Read 2 more answers
The energy levels of a particular quantum object are -11.7 eV, -4.2 eV, and -3.3 eV. If a collection of these objects is bombard
gogolik [260]

To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

By definition the energy balance is simply given by the change between the two states:

|\Delta E_{ij}| = |E_i-E_j|

Our states are given by

E_1 = -11.7eV

E_2 = -4.2eV

E_3 = -3.3eV

In this way the energy balance for the states would be given by,

|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\

|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV

|\Delta E_{23}| = |E_2-E_3|\\|\Delta E_{23}| = |-4.2-(-3.3)|\\|\Delta E_{23}| = 0.9eV

Therefore the states of energy would be

Lowest : 0.9eV

Middle :7.5eV

Highest: 8.4eV

8 0
4 years ago
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