Answer:
18N
Explanation:
Mechanical Advantage of crowbar (lever), MA = Length of effort arm / Length of resistance arm
Therefore, MA = 28 cm / 4.2 cm = 6.667
Force required to pull the nail, F = Friction force / MA
ie, F = 120 N / 6.667 = 18 N
Answer:
Vi = 8.28 m/s
Explanation:
This problem is related to the projectile motion.
As we know there are two components of motion associated with this, the horizontal component and vertical component.
The horizontal distance covered by the ball is
Vx*t = x
Vx*t = 5.3
Vx = 5.3/t eq. 1
Also we know that
Vx = Vicos(60)
Vx = Vi*0.5 eq. 2
equate eq. 1 and eq. 2
5.3/t = Vi*0.5
5.3/0.5 = Vi*t
Vi*t = 10.6 eq. 3
The vertical distance is
Vy = y1 + Vyi*t - 0.5gt²
also we know that
Vyi = Visin(60)
Vyi = Vi*0.866
It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance
3 = 1.9 + Vi*0.866*t - 0.5gt²
3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
1.1 = 0.866(Vi*t) - 4.9t²
0.866(Vi*t) = 4.9t² + 1.1
substitute Vi*t = 10.6 in above equation
0.866(10.6) = 4.9t² + 1.1
9.18 = 4.9t² + 1.1
4.9t² = 8.08
t² = 8.08/4.9
t² = 1.648
t = 1.28 sec
Finally, initial speed can be found by substituting the value of t into eq. 3
Vi*t = 10.6
Vi = 10.6/t
Vi = 10.6/1.28
Vi = 8.28 m/s
Answer:
Temperature when sound wave with wavelength 0.47m has frequency 741 hz is 188 degrees Centigrade.
Step-by-Step Explanation:
Given:
Speed of sound at 0 degrees centigrade = 235 ms
frequency = f = 741 hz
wavelength = w = 0.47m
speed of sound wave = wavelength * frequency = 741 * 0.47 = 348.27 m/s
Using the formula for speed of sound at a specific temperature:
Speed of sound at T degrees Centigrade = Speed of sound at 0 degrees + 0.6 * T
When speed of sound is 348 m/s, the temperature is given by:
348 = 235 + 0.6(T)
T = (348-235)/0.6
T = 188 degrees Centigrade
Answer:
6.8 eV
Explanation:
As you know, momentum (P) equals the mass of an object (M) times its velocity (V)
p=mv
Also, the momentum of the particle can be expressed in terms of the radius,
charge and field:
p=mv = qrB
K.Emax = p²/2m= q²r²B²/ 2m
K.Emax= (1.6x)² x (0.0104)² x (6.50 x)² / (2 x 9.1x)
K.E max= 1.16 x / 1.82 x
K.Emax= 6.37 J
work function 'Ф' = E- K.E max
since E= hc/ λ
Ф= hc/ λ - K.E max =>
Ф = 1.09 x J
Ф = 1.09 x / 1.6x
Ф =6.8 eV
Therefore, the work function of the metal is 6.8 eV