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3 years ago

When an airplane is sitting on the ramp, which of Newton's Laws does it exhibit?

1 answer:

4 0

It shows the first law of Newton.

The first law states that :-

'If an object is at rest it will be at rest or if it is at motion, it will be in the state of motion until an external force is applied'

Since the airplane is at rest, it will be at rest showing the first law

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A car traveling in a straight line has a velocity of 5.0 m/s. After an acceleration of 0.75 m/s/s, the cars velocity is 8.0. In

Bogdan [553]

Vs - velocity on beginning
ve - velocity on ending. You've got:
$v_s = 5 \frac{m}{s} \\ v_e=8 \frac{m}{s} \\ \hbox{Then:} \\ \Delta v=v_e - v_s = 8 \frac{m}{s} - 5\frac{m}{s}=3 \frac{m}{s} \\ a=0,75 \frac{m}{s^2} \\ \hbox{And from formula:} \\ a=\frac{\Delta v}{\Delta t} \qquad \Rightarrow \qquad \Delta t= \frac{\Delta v}{a} \\ \hbox{Substitute:} \\ \Delta t=\frac{3\frac{m}{s}}{0,75 \frac{m}{s^2}}= \frac{3}{\frac{3}{4}} s= 3 \cdot \frac{4}{3} s= 4 s$
So he needed 4 second.

3 0

3 years ago

A heater marked 2000W/3000W is switched on for 4 hours. For the first hour it is on the highest setting and for the last 3 hours

Leni [432]

Answer:

9 kW-hr

Explanation:

3000 W = 3 kW for 1 hr = <u>3 kW - hr </u>

2000 W = 2 kW for 3 hr = <u> 6 kW -hr</u>

<u />

total power = <u> 3 + 6 = 9 kW-hr</u>

3 0

2 years ago

Based on your knowledge of the gravitational force near the surface of the earth and on Newton's second law, explain the sign an

yaroslaw [1]

Your answer would be A. 9.8m/s2

8 0

3 years ago

What is the maximum frictional force in the knee joint of a person who supports 76. 0 kg of her mass on that knee?

Bad White [126]

The maximum frictional force in the knee joint of a person who supports 76. 0 kg of her mass on that knee 119.28N.

Given,

m=76kg, μ=0.16, g=9.81 m/ $s^{2}$

frictional force = mμg= 76*0.16*9.81= 119.28 N

<h3>Frictional force</h3>

An opposing force to the relative motion of two bodies in contact is known as frictional force. Always acting in the opposite direction from the direction of motion, frictional force is applied to a moving body. Because it resists motion, it aids in lowering the moving object's speed. The force is one of touch. Four broad categories can be made for the force of friction depending on the sort of motion that occurs between the two objects. Static friction is the force of friction between an object and the surface it is put on.

What is the maximum frictional force in the knee joint of a person who supports 76. 0 kg of her mass on that knee?

Learn more about frictional force here:

brainly.com/question/13707283

#SPJ4

4 0

2 years ago

A bead slides without friction around a loop the-loop. The bead is released from a height of 17.6 m from the bottom of the loop-

solong [7]

Answer:

N₁ = 393.96 N and N = 197.96 N

Explanation:

In This exercise we must use Newton's second law to find the normal force. Let's use two points the lowest and the highest of the loop

Lowest point, we write Newton's second law n for the y-axis

N -W = m a

where the acceleration is ccentripeta

a = v² / r

N = W + m v² / r

N = mg + mv² / r

we can use energy to find the speed at the bottom of the circle

starting point. Highest point where the ball is released

Em₀ = U = m g h

lowest point. Stop curl down

$Em_{f}$ = K = ½ m v²

Emo = Em_{f}

m g h = ½ m v²

v² = 2 gh

we substitute

N = m (g + 2gh / r)

N = mg (1 + 2h / r)

let's calculate

N₁ = 5 9.8 (1 + 2 17.6 / 5)

N₁ = 393.96 N

headed up

we repeat the calculation in the longest part of the loop

-N -W = - m v₂² / r

N = m v₂² / r - W

N = m (v₂²/r - g)

we seek speed with the conservation of energy

Em₀ = U = m g h

final point. Top of circle with height 2r

$Em_{f}$ = K + U = ½ m v₂² + mg (2r)

Em₀ = Em_{f}

mgh = ½ m v₂² + 2mgr

v₂² = 2 g (h-2r)

we substitute

N = m (2g (h-2r) / r - g)

N = mg (2 (h-r) / r 1) = mg (2h/r -2 -1)

N = mg (2h/r - 3)

N = 5 9.8 (2 17.6 / 5 -3)

N = 197.96 N

Directed down

3 0

3 years ago