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Neporo4naja [7]
3 years ago
5

A student conducts an experiment in which a cart is pulled by a variable applied force during a 2sec time interval. In trial 1,

the student exerts the force on a cart of mass M. In trial 2, the student exerts the force on a cart of mass 3M. In trial 3, the student exerts the force on a cart of 5M. In which trial will the cart experience the greatest change in momentum from 0 sec to 2 sec?
Physics
1 answer:
Harlamova29_29 [7]3 years ago
8 0

Answer:

change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

Explanation:

Let's look for the speed of the car

     F = m a

    a = F / m

We use kinematics to find lips

    v = v₀ + a t

    v = v₀ + (F / m) t

The moment is defined by

    p = m v

The moment change

    Δp = m v - m v₀

Let's replace the speeds in this equation

    Δp = m (v₀ + F / m t) - m v₀

    Δp = m v₀  + F t - m v₀  

    Δp = F t

We see that the change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

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A body with mass of 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time tequals0 the bod
11Alexandr11 [23.1K]

Answer:

X(t) = 13/13 cos(12t+α)

C =13/13

π/6 s

Explanation:

(A) A body with mass 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time t = 0 the body is pulled 1 m in to the right, stretching the spring, 3 set in motion with an initial velocity of 5 m/s to the left.  

(a) Find X(t) in the form c • cos(w_o*t— α)  

(b) Find the amplitude 3 Period of motion of the body 1  

mass: m = 200g =  0.200 kg  

displacement: ΔX = 20 cm =  0.20 m

Spring Constant: K =  9/0.20 = 45 N/m

IV:   X(0) = 1m V(0) = -5 m/s

Simple Harmonic Motion: c•cos(cosw_t— α) = X(t)  

Circular Frequency: w_o = √k/m= √36/(0.20) = 13 rad/s

X(0) = 1m =c_1

X'(0) = V(0) = c_2*w_o/w_o

        = -5/12 =   c_2

"radians Technically Unitless"  

Amplitude: c = √ci^2 + c^2 ==> √1^2 + (-5/12)^2 = 1 m =13/13 = c

X(t) = 13/13 cos(12t+α)

since, C>0 : damped forced vibration c_1>0, c_2>0

phase angle 2π+tan^-1(c_2/c_1)

                        =2π+tan^-1(-5/12/1)= 5.884

period: T =2π/w_o

                =π/6 s

6 0
4 years ago
You are loading a toy dart gun, which has two settings, the more powerful with the spring compressed twice as far as the lower s
Amiraneli [1.4K]

Answer:

option A

Explanation:

given,

compression for lower setting = x

work done to compress in lower setting = 5 J

compression in higher setting, x' = 2 x

work done in higher setting = ?

Work done in compression of spring at lower setting

W = \dfrac{1}{2}kx^2

5 = \dfrac{1}{2}kx^2............(1)

Work done in compression of spring at higher setting

W' = \dfrac{1}{2}kx'^2

W'= \dfrac{1}{2}k(2x)^2

W'= 4\times \dfrac{1}{2}kx^2

W'= 4\times W

from equation (1)

W'= 4\times 5

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Work take for the higher setting is equal to 20 J

Hence, the correct answer is option A

4 0
3 years ago
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