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Zinaida [17]
3 years ago
7

Most motorcycles could win a drag race against most cars. UseNewton's 2nd law to explain this.​

Physics
1 answer:
shutvik [7]3 years ago
5 0

Answer:

is has be deducted that acceleration of an object depends only on the the mass of the object. car has a greater mass that a motorcycle.

Explanation:

the newton second law states that when a net force act on an object,the object accelerates in the direction of the net force. the acceleration is directly proportional to the net force and inversely to the mass.

mathematically, F = ma    .......................... (eqn 1)

where, F = force, m=mass and  a = acceleration

F = m(ΔV/t)    .......................... (eqn 2)

acceleration =change in velocity divide time

t = time

v= velocity

Ft = mΔV

also a∝1/m

this same second can also mean the rate of change in momentum of an object is directly proportional to the resultant force.

Drag race is a function of acceleration, therefore, a car weighs more than a motorcycle.

the reason while motorcycle moves more faster than a car is due to the weight of the motorcycle as compare to the car. if the same acceleration is applied, and there weigth are difference, this will affect the force (causing an increased resultant force for the car), whereas time used will be affected.

for example:

a motorcycle has a weight 0f 181kg

a car has a weight 0f 1590kg

from  a∝1/m

they both will moves at a different acceleration

the force will be different

Ft = mΔV

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The acceleration due to gravity on earth will decrease as which of the following occurs. The mass of the object decreases. The d
pentagon [3]

Answer:

The distance of the object to the center of the earth increases.

Explanation:

The acceleration due to gravity on Earth is given by:

g=\frac{GM}{r^2}

where

G is the gravitational constant

M is the Earth's mass

r is the distance of the object from the Earth's centre

We notice that:

- g does not depend on the mass of the object

- g is inversely proportional to r

This means that if the distance of the object from the Earth's centre increases, g decreases. So, the correct option is

The distance of the object to the center of the earth increases.

4 0
3 years ago
man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and h
-Dominant- [34]

Answer:

w₂ = 22.6 rad/s

Explanation:

This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.

Let's write the moment two moments

initial instant. Before releasing bricks

       L₀ = I₁ w₁

final moment. After releasing the bricks

       L_{f} = I₂W₂

       L₀ = L_{f}

       I₁ w₁ = I₂ w₂

       w₂ = I₁ / I₂ w₁

let's reduce the data to the SI system

     w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s

 

 let's calculate

       w₂ = 6.0/2.0   7.54

       w₂ = 22.6 rad/s

3 0
3 years ago
Given an electron beam whose electrons have kinetic energy of 10.0 kev , what is the minimum wavelength λmin of light radiated b
kolbaska11 [484]
To answer the problem we would be using this formula which isE = hc/L where E is the energy, h is Planck's constant, c is the speed of light and L is the wavelength 
L = hc/E = 4.136×10−15 eV·s (2.998x10^8 m/s)/10^4 eV 

= 1.240x10^-10 m 

= 1.240x10^-1 nm
3 0
3 years ago
Read 2 more answers
Earth's gravity acts upon objects with a steady force of __________.
luda_lava [24]
None of the choices is a force. 'a' and 'b' are speeds. 'C' and 'd' are accelerations. ... The steady force of gravity is 9.8 newtons PER KILOGRAM of mass. ... The question is written by someonewho very much wants to discourage anyone interested in Physics.
6 0
3 years ago
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A hot-air balloon is rising upward with a constant speed of 2.51 m/s. When the balloon is 3.16 m above the ground, the balloonis
icang [17]

Answer:

  t = 1.099 s

Explanation:

given,

constant speed = 2.51 m/s

height of balloon above ground = 3.16 m

time elapsed before it hit the ground = ?

Applying equation of motion to the compass

y = u t + \dfrac{1}{2}at^2

-3.16 = 2.51 t + \dfrac{1}{2}\times (-9.8)t^2

4.9 t^2 - 2.51 t - 3.16 = 0

using quadratic formula to solve the equation

t = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

t = \dfrac{-(-2.51)\pm \sqrt{2.51^2-4(4.9)(-3.16)}}{2\times 4.9}

  t = 1.099 s, -0.586 s

hence, the time elapses before the compass hit the ground is equal to 1.099 s.

8 0
3 years ago
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