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3 years ago

A ford F-250 pulls a 1000-kg car with a net force of 2000 N. What is the acceleration of the car A=F/m

Physics

1 answer:

vfiekz [6]3 years ago

5 0

Answer:

2ms^2

Explanation:

A = 2000 divided by 1000

A= 2

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A cd has a mass of 17 g and a radius of 6.0 cm. When inserted into a player, the cd starts from rest and accelerates to an angul

GuDViN [60]

Answer:

the net toque is τ=8.03* 10⁻⁴ N*m

Explanation:

Assuming the disk has constant density ρ, the moment of inertia I of is

I = ∫r² dm

since m = ρ*V = ρπR² h , then dm= 2ρπh r dr

thus

I = ∫r²dm = ∫r²2ρπh r dr =2ρπh ∫r³ dr = 2ρπh (R⁴/4- 0⁴/4)= ρπhR⁴ /2= mR²/2

replacing values

I = mR²/2= 0.017 kg * (0.06 m)²/2 = 3.06 *10⁻⁵ kg*m²

from Newton's second law applied to rotational motion

τ= Iα , where τ=net torque and α= angular acceleration

since the angular velocity ω is related with the angular acceleration through

ω= ωo + α*t → α =(ω-ωo)/t = (21 rad/s-0)/0.8 s = 26.25 rad/s²

therefore

τ= Iα= 3.06 *10⁻⁵ kg*m²*26.25 rad/s² = 8.03* 10⁻⁴ N*m

3 0

3 years ago

Помогите плз!

Softa [21]

Все написано в скобках правильно

3 0

3 years ago

Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of

kondaur [170]

Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

g = acceleration due to gravity, g = 32 ft/s²

$m=\dfrac{F}{g}$

$m=\dfrac{160\ lb}{32\ ft/s^2}$

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

$m=\dfrac{F}{g}$

$m=\dfrac{1.9\ lb}{32\ ft/s^2}$

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

$W=2300\ kg\times 32\ ft/s^2=73600\ lb$

(b) A 22 g song sparrow, m = 22 g = 0.022 kg

$W=0.022\ kg\times 32\ ft/s^2=0.704\ lb$

Hence, this is the required solution.

5 0

3 years ago

A mass of 3.6 kg oscillate on a horizontal spring with a spring constant of 160 N/m.

Darya [45]

Answer:

48.7 J

Explanation:

For a mass-spring system, there is a continuous conversion of energy between elastic potential energy and kinetic energy.

In particular:

- The elastic potential energy is maximum when the system is at its maximum displacement

- The kinetic energy is maximum when the system passes through the equilibrium position

Therefore, the maximum kinetic energy of the system is given by:

$KE=\frac{1}{2}mv^2$

where

m is the mass

v is the speed at equilibrium position

In this problem:

m = 3.6 kg

v = 5.2 m/s

Therefore, the maximum kinetic energy is:

$KE=\frac{1}{2}(3.6)(5.2)^2=48.7 J$

6 0

3 years ago

URGANT!!!!!!! NEED ANSWERS TODAY!!!!

natima [27]

2H2 + O2 = 2H2O + Energy

5 0

2 years ago