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e-lub [12.9K]
3 years ago
14

Force is measured in what unit? What is this unit equal to in other units?

Physics
2 answers:
sergij07 [2.7K]3 years ago
8 0
Force is measured in Newton. The SI unit of force is the newton, symbol N
lozanna [386]3 years ago
8 0

Answer:

Explanation:

Nnewton=kgm/s²

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A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th
melomori [17]

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

5 0
3 years ago
A 76 N crate is hung from a spring
lutik1710 [3]

Answer: 0.169 (3 s.f.)

Explanation:

Force = 76 N

Spring constant = 450 N/m

Extension/displacement = x

Hooke's law states that: F = kx

Therefore, 76 = 450 X x

76/450 = x

0.169 (3 s.f.) = x

4 0
3 years ago
You walk 45 m to the north, then turn 90° to your right and walk another 45 m How far are you from where you originally started?
docker41 [41]
C^2=a^2+b^2
c^2=45^2+45^2
c^2=4050
c=63.64
c=64
The answer is D.
3 0
4 years ago
Which two variables have an effect on the electrical force between two electrically charged objects like two balloons?
V125BC [204]

Answer:

Correct answer: There are four variables rather than two - The amount of charge q₁, the amount of charge q₂, the distance between the spheres and the relative dielectric permeability εr of the medium.

Explanation:

I assume you meant two spherical objects.

The force between two charged objects according to Coulomb's law is calculated by the formula:

Fc =  1/ (4 π ε₀ εr) ·  (q₁ · q₂) / d²

where :

- ε₀ is the dielectric permeability of the vacuum

- εr is the dielectric permeability of the medium

- q₁ the amount of charge of the first sphere

- q₂ the amount of charge of the second sphere

- d is the distance between the spheres

God is with you!!!

5 0
4 years ago
A particle moves according to a law of motions=f(t),t≥0 Wheretismeasured in seconds andsin feet.a) Find the velocity at timet.b)
Ray Of Light [21]

Answer:

a) v(t) = \frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2}

b) 3s

c) t < 3s

d) 1.8ft

e) particle is speeding up when t > 5.385 s and vice versa, slowing down when t < 5.385 s

Explanation:

a)Suppose the equation for motion is:

f(t) = \frac{9t}{t^2+9}

Then the velocity is the derivative of the motion function

v(t) = \frac{df(t)}{dt} = (9t(t^2+9)^{-1})^'

From here we can apply product rule

v(t) = (9t)^{'}(t^2+9)^{-1} + (9t)((t^2+9)^{-1})^'

v(t) = \frac{9}{t^2+9} - \frac{9t(2t)}{(t^2+9)^2}

v(t) = \frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2}

b) The particle is at rest when v(t) = 0:

\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2} = 0

\frac{9}{t^2+9} = \frac{18t^2}{(t^2+9)^2}

Multiply 2 sides by (t^2+9)^2 we have:

t^2 + 9 = 2t^2

t^2 = 9

t = 3s

(c) The particle is moving in positive direction when v(t) > 0:

\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2} > 0

\frac{9}{t^2+9} > \frac{18t^2}{(t^2+9)^2}

Multiply 2 sides by (t^2+9)^2 we have:

t^2 + 9 > 2t^2

t^2 < 9

t < 3s

(d) As particle is moving in positive direction when t < 3s and negative direction when t > 3s, we can calculate the distance it's moving up to 3s and then after 3s

f(3) = \frac{9*3}{3^2+9} = \frac{27}{18} = 1.5 ft

At 3s, particle is changing direction to negative, so its position at 6s is

f(6) = \frac{9*6}{6^2+9} = \frac{54}{45} = 1.2 ft

Therefore from 3s to 6s it would have moved a distance of 1.5 - 1.2 = 0.3 ft

Then the total distance it has moved in the first 6 s is 1.5 + 0.3 = 1.8 ft

e) Acceleration is the derivative of velocity function:

a(t) = \frac{dv(t)}{dt} = (\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2})^'

a(t) = -\frac{9(2(t^2+9))(2t)}{(t^2+9)^2} - \frac{36t}{(t^2+9)^2} + \frac{18t^2(2(t^2+9))(2t)}{(t^2+9)^3}

a(t) = -\frac{36t}{t^2+9} - \frac{36t}{(t^2+9)^2} + \frac{72t^3}{(t^2+9)^2}

a(t) = -\frac{54t}{t^2+9} + \frac{72t^3 - 36t}{(t^2+9)^2}

Particle is speeding up when a(t) > 0:

-\frac{54t}{t^2+9} + \frac{72t^3 - 36t}{(t^2+9)^2} > 0

\frac{72t^3 - 36t}{(t^2+9)^2} > \frac{54t}{t^2+9}

as t \& (t^2 + 9)^2 \geq 0 we can multiply/divide both sides by it:

8t^2 - 4 > 6(t^2+9)

8t^2 > 6t^2 + 58

t^2 > 29

t > \sqrt(29) \approx 5.385 s

so particle is speeding up when t > 5.385 s and vice versa, slowing down when t < 5.385 s

8 0
4 years ago
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