First, illustrate the problem as shown in the attached picture. Next, let's find the distance traveled by planes A and B after 2.9 h.
Distance of A: 650 m/h * 2.9 h = 1,885 m
Distance of B: 560 m/h * 2.9 h = 1,624 m
Then, we use the cosine law to determine the distance x. The angle should be: 85 - 60.5 = 24.5°
x² = 1,885² + 1,624² - 2(1,885)(1,624)(cos 24.5°)
x = √619381.3183
<em>x = 787 m</em>
Answer:![a=2.42 m/s^2](https://tex.z-dn.net/?f=a%3D2.42%20m%2Fs%5E2)
Explanation:
Given
mass ![m_1=3.50\times 10^{3} kg](https://tex.z-dn.net/?f=m_1%3D3.50%5Ctimes%2010%5E%7B3%7D%20kg)
![m_2=1.00\times 10^{3} kg](https://tex.z-dn.net/?f=m_2%3D1.00%5Ctimes%2010%5E%7B3%7D%20kg)
(inclination)=![30^{\circ}](https://tex.z-dn.net/?f=30%5E%7B%5Ccirc%7D)
![\mu =0.21](https://tex.z-dn.net/?f=%5Cmu%20%3D0.21)
Let T be the tension in the rope
From Diagram
-----------------1
where ![f_r=friction\ force](https://tex.z-dn.net/?f=f_r%3Dfriction%5C%20force)
![f_r=\mu m_g\cos \theta](https://tex.z-dn.net/?f=f_r%3D%5Cmu%20m_g%5Ccos%20%5Ctheta%20)
For block ![m_2](https://tex.z-dn.net/?f=m_2)
-----------2
From 1 & 2
![m_1g\sin \theta -m_2a-\mu m_1g\cos \theta =m_1a](https://tex.z-dn.net/?f=m_1g%5Csin%20%5Ctheta%20-m_2a-%5Cmu%20m_1g%5Ccos%20%5Ctheta%20%3Dm_1a)
![m_1g(\sin \theta -\mu \cos \theta )=(m_1+m_2)a](https://tex.z-dn.net/?f=m_1g%28%5Csin%20%5Ctheta%20-%5Cmu%20%5Ccos%20%5Ctheta%20%29%3D%28m_1%2Bm_2%29a)
![\frac{9.8\times 3.5\times 10^3}{4.5\times 10^3}(0.5-\0.21\cos 30)=4.5a](https://tex.z-dn.net/?f=%5Cfrac%7B9.8%5Ctimes%203.5%5Ctimes%2010%5E3%7D%7B4.5%5Ctimes%2010%5E3%7D%280.5-%5C0.21%5Ccos%2030%29%3D4.5a)
![a=2.42 m/s^2](https://tex.z-dn.net/?f=a%3D2.42%20m%2Fs%5E2)
Answer : The change in momentum of an object is equal to the impulse that acts on it.
Explanation :
Change in momentum : The change in momentum of an object is the product of the mass and the change in velocity of an object.
The formula of change in momentum is,
![\Delta p=m\times \Delta v](https://tex.z-dn.net/?f=%5CDelta%20p%3Dm%5Ctimes%20%5CDelta%20v)
Impulse : An impulse of an object is the product of the force applied on an object and the change in time. Impulse is also equivalent to the change in momentum of an object.
![J=F\times \Delta t](https://tex.z-dn.net/?f=J%3DF%5Ctimes%20%5CDelta%20t)
Proof :
![J=F\times \Delta t\\\\J=(m\times a)\times \Delta t\\\\J=m\times (a\times \Delta t)\\\\J=m\times \Delta v=\Delta p](https://tex.z-dn.net/?f=J%3DF%5Ctimes%20%5CDelta%20t%5C%5C%5C%5CJ%3D%28m%5Ctimes%20a%29%5Ctimes%20%5CDelta%20t%5C%5C%5C%5CJ%3Dm%5Ctimes%20%28a%5Ctimes%20%5CDelta%20t%29%5C%5C%5C%5CJ%3Dm%5Ctimes%20%5CDelta%20v%3D%5CDelta%20p)
Hence, the change in momentum of an object is equal to the impulse that acts on it.
Get a sample of food. place onto a test tube. add sudan 111 reagand to it. If there are lipids red flat globules will appear.
Answer:
<em>Si hay rozamiento y el valor de la fuerza de roce es 10 N</em>
Explanation:
<u>Fuerza Neta</u>
La fuerza neta sobre un cuerpo es la suma vectorial de todas las fuerzas actuantes sobre el mismo.
Si conocemos el módulo de la fuerza neta F y la masa m del cuerpo, aplicamos la segunda ley de Newton para relacionarlas con la aceleración a:
![F=m.a](https://tex.z-dn.net/?f=F%3Dm.a)
Tenemos los datos cinemáticos de la situación, según la cual el cuerpo adquiere una velocidad (desde el reposo) de 4 m/s en 5 s.
Utilizamos la fórmula:
![v_f=v_o+a.t](https://tex.z-dn.net/?f=v_f%3Dv_o%2Ba.t)
Y despejamos la aceleración:
![\displaystyle a=\frac{v_f-v_o}{t}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%3D%5Cfrac%7Bv_f-v_o%7D%7Bt%7D)
![\displaystyle a=\frac{4-0}{5}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%3D%5Cfrac%7B4-0%7D%7B5%7D)
![a=0.8 \ m/s^2](https://tex.z-dn.net/?f=a%3D0.8%20%5C%20m%2Fs%5E2)
Podemos calcular la aceleración real que el cuerpo adquiere, producto de una fuerza efectiva igual a:
![F_e=25\ Kg\cdot 0.8 \ m/s^2](https://tex.z-dn.net/?f=F_e%3D25%5C%20Kg%5Ccdot%200.8%20%5C%20m%2Fs%5E2)
![F_e=20\ N](https://tex.z-dn.net/?f=F_e%3D20%5C%20N)
Si se está aplicando una fuerza de
y solo 20 N producen movimiento, entonces se está perdiendo en rozamiento una fuerza:
![F_r=F_a-F_e=30 - 20=10](https://tex.z-dn.net/?f=F_r%3DF_a-F_e%3D30%20-%2020%3D10)
![F_r=10\ N](https://tex.z-dn.net/?f=F_r%3D10%5C%20N)
Si hay rozamiento y el valor de la fuerza de roce es 10 N