Answer:
d = 1700 meters
Explanation:
During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound. The time between seeing the lighting and hearing the sound was 5 second, t = 5 seconds
Speed of sound, v = 340 m/s (say)
Let d is the distance of the colliding cloud from the observer. The distance covered by the object. It is given by :
![d=v\times t](https://tex.z-dn.net/?f=d%3Dv%5Ctimes%20t)
d = 1700 meters
So, the distance of the colliding cloud from the observer is 1700 meters. Hence, this is the required solution.
Answer:
15.2 s
Explanation:
Convert hp to W:
55.0 hp × 746 W/hp = 41,030 W
Power = energy / time
41030 W = 6.22×10⁵ J / t
t = 15.2 s
Answer:
PLEASE MARK AS BRAINLIEST!!
Explanation:
ANSWER IS IN THE IMG BELOW
Answer:
V=15.3 m/s
Explanation:
To solve this problem, we have to use the energy conservation theorem:
![U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}](https://tex.z-dn.net/?f=U_e%2BK_i%2BU_%7Bgi%7D%2BW_%7Bfriction%7D%3DK_f%2BU_%7Bgf%7D)
the elastic potencial energy is given by:
![U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J](https://tex.z-dn.net/?f=U_e%3D%5Cfrac%7B1%7D%7B2%7D%2Ak%2Ax%5E2%5C%5CU_e%3D%5Cfrac%7B1%7D%7B2%7D%2A1100N%2Fm%2A%284m%29%5E2%5C%5CU_e%3D8800J)
The work is defined as:
![W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J](https://tex.z-dn.net/?f=W_%7Bfriction%7D%3DF_f%2Ad%2Acos%28%5Ctheta%29%5C%5CW_%7Bfriction%7D%3D40N%2A2.5m%2Acos%28180%29%5C%5CW_%7Bfriction%7D%3D-100J)
this work is negative because is opposite to the movement.
The gravitational potencial energy at 2.5 m aboves is given by:
![U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J](https://tex.z-dn.net/?f=U_%7Bgf%7D%3Dm%2Ag%2Ah%5C%5CU_%7Bgf%7D%3D60kg%2A9.8%2A2.5%5C%5CU_%7Bgf%7D%3D1470J)
the gravitational potential energy at the ground and the kinetic energy at the begining are 0.
![8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s](https://tex.z-dn.net/?f=8800J%2B0%2B0%2B-100J%3D%5Cfrac%7B1%7D%7B2%7D%2A62kg%2Av%5E2%2B1470J%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B2%288800J-100J-1470J%29%7D%7B62kg%7D%7D%5C%5Cv%3D15.3m%2Fs)
Answer:
350 F to 100 F it take approx 87.33 min
Explanation:
given data
oven = 350◦F
cooling rack = 70◦F
time = 30 min
cake = 200◦F
solution
we apply here Newtons law of cooling
= -k(T-Ta)
=
(T(t) -Ta)
=
= -k(T-Ta)
-ky
= -ky
T(t) -Ta = (To -Ta)
T(t) = Ta+ (To -Ta)
put her value for time 30 min and T(t) = 200◦F and To =350◦F and Ta = 70◦F
so here
200 = 70 + ( 350 - 70 ) ![e^{-k30}](https://tex.z-dn.net/?f=e%5E%7B-k30%7D)
k = 0.025575
so here for T(t) = 100F
100 = 70 + ( 350 - 70 ) ![e^{-0.025575*t}](https://tex.z-dn.net/?f=e%5E%7B-0.025575%2At%7D)
time = 87.33 min
so here 350 F to 100 F it take approx 87.33 min