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Y_Kistochka [10]
3 years ago
8

A force of 10 N acts to the left and a force of 5 N acts to the right on an object of mass 1 kg. If its initial speed is zero ,

what is its velocity after 1 second ?
Physics
1 answer:
dimaraw [331]3 years ago
5 0

Answer: -5 m/s

Explanation:

First find the net force by summing up the forces:

-10N + 5N = -5N

Use Newton's Second Law of motion to solve for the acceleration:

F = ma

-5N = 1kg(a)

a = -5 m/s^2

Use the first kinematic equation to solve for the final velocity:

Vf = Vi + at

Vf = 0 + (-5)(1)

Vf = -5 m/s

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A rocket's acceleration is 6.0 m/s 2 . Assuming it starts at 0 m/s, how long will it take for the rocket to reach a velocity of
kupik [55]
At=v-u, where u=0

t= v/a=42/6 = 7
Answer: 7 seconds
8 0
3 years ago
A 37.5 kg box initially at rest is pushed 4.05 m along a rough, horizontal floor with a constant applied horizontal force of 150
vodomira [7]

Answer:

a) 607.5 J

b) 160.531875 J

c)  0 J

d)  0 J

e) 2.925 m\s

Explanation:

The given data :-

  • Mass of the box ( m ) = 37.5 kg.
  • Displacement made by box ( x ) = 4.05 m.
  • Horizontal force ( F ) = 150 N.
  • The co-efficient of friction between box and floor ( μ ) = 0.3
  • Gravitational force ( N ) = m × g = 37.5 × 9.81 = 367.875

Solution:-

a) The work done by applied force ( W )

W = force applied × displacement = 150 × 4.05 = 607.5 J

b)  The increase in internal energy in the box-floor system due to friction.

Frictional force ( f ) = μ × N = 0.3 × 367.875 = 110.3625 N

Change in internal energy = change in kinetic energy.

ΔU = ( K.E )₂ - ( K.E )₁

Since the initial velocity is zero so the  ( K.E )₁ = 0  

ΔU = ( K.E )₂ = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

c) The work done by the normal force .

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

d)  The work done by the gravitational force.

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

e) The change in kinetic energy of the box

( K.E )₂ - ( K.E )₁ = ( K.E )₂ - 0 = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

f) The final speed of the box

( K.E )₂ = 160.531875 J = 0.5 × 37.5 × v²

v² = 8.56

v = 2.925 m\s.

5 0
4 years ago
Please answer B, C, E, and D
Natasha_Volkova [10]
B- the biosphere needs the hydrosphere (water) to survive. They drink the water.
C- the biosphere use the géosphère as land/ home.
D-The géosphère depends on the hydrosphere for water. Without the water, the géosphère would eventually turn dry and nothing will be able to grow
E- The animals in the hydrosphere need air to survive. The gills on sea animals are used to filter out the water from the oxygen and use the oxygen to breathe. As the fish opens its mouth, water runs over the gills, and blood in the capillaries picks up oxygen that's dissolved in the water.
3 0
2 years ago
Read 2 more answers
What variable is represented on the y-axis?
gladu [14]

Distance/ Time which means Distance is on horizontal and time is on vertical

8 0
3 years ago
4-12. The morning inspection of the tank farm finds a leak in the turpentine tank. The leak is repaired. An investigation finds
bonufazy [111]

Answer:

a) V=759.8727\ ft^3

b) \dot V=1.403\times 10^{-3}\ ft^3.s^{-1}

c) t\approx29.541\ s

Explanation:

Given:

  • diameter of hole in the tank, d'=0.1\ in=\frac{1}{120}\ ft
  • position of the hole form the tank bottom, h' =7\ ft
  • initial level of turpentine in the tank before the leakage, h_i=17.3\ ft
  • level of turpentine in the tank after the repair of leakage, h_f=13\ ft
  • diameter of the tank, d=15\ ft
  • density of turpentine oil, \rho=55\ lbm.ft^3

a)

Now, volume of turpentine spilled:

V=A.(h_i-h_f)

where:

A= area of the cross section of the tank's volume

V=\pi .\frac{d^2}{4} \times(h_i-h_f)

V=\pi\times\frac{15^2}{4} \times(17.3-13)

V=759.8727\ ft^3

b)

When the tank was full the liquid level was highest:

so velocity form the height of the hole will be given as:

v=\sqrt{2g.(h_i-h')}

v=\sqrt{2\times 32.12\times (17.3-7) }

v=25.722\ ft.s^{-1}

<u>Now we have the flow rate of the spillage given by:</u>

\dot V=(\pi.\frac{d'^2}{4}) \times v

\dot V=\pi\times \frac{(\frac{1}{120})^2 }{4} \times 25.722

\dot V=1.403\times 10^{-3}\ ft^3.s^{-1}

c)

Total time the leak was active can be calculated as:

t=\frac{V}{\dot V}

t=\frac{759.8727}{25.722}

t\approx29.541\ s

7 0
4 years ago
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