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kotegsom [21]
3 years ago
9

A toy car is given a quick push so that it rolls up an inclined ramp. After it is released, it rolls up, reaches its highest poi

nt and rolls back down again. Friction is so small it can be ignored. What net force acts on the car?(A) net force of zero(B) Net constant force up the ramp(C) net increasing force up the ramp(D) net increasing force down the ramp(E) net constant force down the ramp(F) net decreasing force down the ramp(G) net decreasing force up the ramp
Physics
1 answer:
kicyunya [14]3 years ago
5 0

Answer: Option (E)

Explanation:

There are only two forces acting on the car: The force of gravity, which point down, and the normal force generated by the interaction of the car and the ramp, which points perpendicular to the ramp.

N=mgcos(\alpha ),

m:mass of the car ; \alpha:angle of the ramp.

g: gravity's acceleration.

N is constante since all parameters are constants.

So, since the force of gravity is also constant, the net force acting on the car is constant. Since velocity decreases over time and the car stars moving down, the force is pointing down the ramp.

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A force of 2 kN is applied to an object to make it move 3.6 m in the direction of the force. Select the correct value of work do
vaieri [72.5K]

Answer:

W= F × d

W= 2kn × 3.6

W= 7.2 J

Work is measured in Joules!

4 0
3 years ago
A motorcycle that is slowing down uniformly. The motorcycle covers 1 ????m=1000 m in 80 sec⁡. The motorcycle then covers the nex
Lynna [10]

Answer:

Part a)

acceleration = -0.042 m/s/s

Part b)

initial speed = 14.17 m/s

final speed = 5.77 m/s

Explanation:

Part a)

Let the initial velocity of the motorcycle is

v_i = v_o

now at the end of 80 s let the speed is

v_f = v_1

after another 120 s let the speed will be

v_f' = v_2

now we know that

d = \frac{v_i + v_f}{2} (t)

d = \frac{v_o + v_1}{2}(80)

1000 = 40(v_o + v_1)

also we know that

v_1 - v_o = a(80)

also we have

1000 = \frac{v_1 + v_2}{2}(120)

1000 = 60(v_1 + v_2)

now we can say

(v_2 + v_1) - (v_o + v_1) = \frac{50}{3} - \frac{50}{2}

also we know

v_2 - v_o = a(120 + 80)

-8.33 = 200 a

a = -0.042 m/s^2

Part b)

now we have

v_1 + v_o = 25

v_1 - v_o = (-0.042)(80)

v_1 = 10.83 m/s

so the starting velocity of the trip is

v_o = 25 - 10.83 = 14.17 m/s

now speed after t = 200 s is given as

v_2 = v_o + at

v_2 = 14.17 - (0.042)(200)

v_2 = 5.77 m/s

5 0
3 years ago
An electron moves with velocity v⃗ =(5.8i−6.7j)×104m/s in a magnetic field B⃗ =(−0.81i+0.60j)T.
Minchanka [31]

Answer:

Fₓ = 0,  F_{y} = 0  and  F_{z}<em> = - 3.115 10⁻¹⁵   N</em>

Explanation:

The magnetic force given by the expression

       F = q v xB

the bold are vectors,  the easiest analytical way to determine this force in solving the determinant

   F = q \left[\begin{array}{ccc}i&j&k\\5.8&-6.7&0\\-0.81&0.6&0\end{array}\right]  10^{4}

   F = 1.6 10⁻¹⁵ [ i( 0-0) + j (0-0) + k^( 5.8 0.60 - 0.81 67) ]

   F =i^0 + j^0   - k^  3.115 10⁻¹⁵   N

   

Fₓ = 0

F_{y} = 0

F_{z}<em> = - 3.115 10⁻¹⁵   N</em>

6 0
3 years ago
A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th
svp [43]

a) x(t)=2.0 sin (10 t) [m]

The equation which gives the position of a simple harmonic oscillator is:

x(t)= A sin (\omega t)

where

A is the amplitude

\omega=\sqrt{\frac{k}{m}} is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s

The amplitude, A, can be found from the maximum velocity of the spring:

v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m

So, the equation of motion is

x(t)= 2.0 sin (10 t) [m]

b)  t=0.10 s, t=0.52 s

The potential energy is given by:

U(x)=\frac{1}{2}kx^2

While the kinetic energy is given by:

K=\frac{1}{2}mv^2

The velocity as a function of time t is:

v(t)=v_{max} cos(\omega t)

The problem asks as the time t at which U=3K, so we have:

\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}

However, \frac{m}{k}=\frac{1}{\omega^2}, so we have

(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\

with two solutions:

\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s

\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s

c) 3 seconds.

When x=0, the equation of motion is:

0=A sin (\omega t)

so, t=0.

When x=1.00 m, the equation of motion is:

1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s

The period of a pendulum is:

T=2 \pi \sqrt{\frac{L}{g}}

where L is the length of the pendulum. By using T=0.628 s, we find

L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m






5 0
3 years ago
What properties of sound determine the volume of sound? Is this affected by the motion of the sound source?
Alik [6]
The amplitude determines the volume of the sound. This isn't affected by the motion of the sound source.
Hope this helps
6 0
3 years ago
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