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Katyanochek1 [597]
3 years ago
14

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

Physics
1 answer:
antiseptic1488 [7]3 years ago
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 particle moving along a straight line with constant acceleration has a velocity of 2.35 m/s at t=3.42 s, and a velocity of -8.
mrs_skeptik [129]

The particle's acceleration is 5.1 m/s²

<h3>What is Acceleration ?</h3>

Acceleration can be defined as the rate at which velocity is changing. It is a vector quantity and it is measured in m/s²

Given that a particle is moving along a straight line with constant acceleration has a velocity of 2.35 m/s at t=3.42 s, and a velocity of -8.72 m/s at t=5.59s

The given parameters are;

  • V1 = 2.35 m/s
  • V2 = - 8.72 m/s
  • T1 = 3.42s
  • T2 = 5.59s

Acceleration a = ΔV ÷ ΔT

a = (2.35 + 8.72) / (5.59 - 3.42)

a = 11.07 / 2.17

a = 5.1 m/s²

Therefore, the particle's acceleration is 5.1 m/s²

Learn more about Acceleration here: brainly.com/question/9069726

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4 0
2 years ago
a ball is thrown horizontally from a 20 m high building with a speed of 5.0 m/s. How far from the base of the building does the
kipiarov [429]
Given:
Dy= 20 m
Vi = 5.0 m/s horizontally
A=9.81 m/s^2

Find:
Horizontal displacement

Solution:
D=ViT+(1/2)AT^2
Dy=(1/2)AT^2
T^2=Dy/(1/2)A
T=sqrt(Dy/(1/2)A)
T=sqrt(20/4.905)
T=2.0s

Dx=ViT
Dx=(5.0)(2.0)
Dx=10. meters
7 0
3 years ago
How much Gravitational Potential Energy does a 1.48 kg book at rest on top of a 2.4 m tall desk have?
Naily [24]

Gravitational Potential Energy GPE = mg∆h. 12. A 5.0 kg mass is initially sitting on the floor when it is lifted onto a table 1.15 meters high at.

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2 years ago
flipper (the dolphin) is out in the open ocean hunting tuna avec. he emits his pulse at 22khz and .42 seconds later he hears it
marusya05 [52]
First we need to find the speed of the dolphin sound wave in the water. We can use the following relationship between frequency and wavelength of a wave:
v=\lambda f
where
v is the wave speed
\lambda its wavelength
f its frequency
Using \lambda = 2 cm = 0.02 m and f=22 kHz = 22000 Hz, we get
v=(0.02 m)(22000 Hz)=440 m/s

We know that the dolphin sound wave takes t=0.42 s to travel to the tuna and back to the dolphin. If we call L the distance between the tuna and the dolphin, the sound wave covers a distance of S=2 L in a time t=0.42 s, so we can write the basic relationship between space, time and velocity for a uniform motion as:
v= \frac{S}{t}= \frac{2L}{t}
and since we know both v and t, we can find the distance L between the dolphin and the tuna:
L= \frac{vt}{2}= \frac{(440 m/s)(0.42 s)}{2}=92.4 m
5 0
3 years ago
A train moves with a constant velocity for 15s for 155m. How fast is the train moving?
bonufazy [111]

Answer: 10.3m/s

Explanation:

In theory and for a constant velocity the physics expression states that:

Eq(1):  distance  = velocity times time <=> d = v*t    for v=constant.

If we solve Eq (1) for the velocity (v) we obtain:

Eq(2):  velocity  = distance divided by time <=> v = d/t

Substituting the known values for t=15s and d=155m we get:

v = 155 / 15 <=> v = 10.3

5 0
3 years ago
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