Answer:
Mass of the wooden Block is 20g.
Explanation:
The buoyant force equation will be used here
Buoyant Force= ρ*g*1/2V Here density used is of water
m*g= ρ*g*1/2V
Simplifying the above equation
2m= ρ*V Eq-1
Also we know from the question that
ρ*V = m + 0.020 Eq-2 ( Density = (Mass+20g)/Volume )
Equating Eq-1 & Eq-2 we get
2m = m+0.020
m = 0.020kg
m = 20g
Answer:
Constructive interference
Explanation:
Here, the medium is same, same wavelength, same frequency, same amplitude and same direction of propagation.
Let the intensity of waves be I which is same for both
The formula for the net intensity is
where, Ф be the phase difference
So, 
Here, IR is maximum so the interference is constructive in nature.
B4 the tackle:
<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>
<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>
<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>
<span>The vector triangle is right angled: </span>
<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>
<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>
<span>v(f) = 5.6 m/s (to 2 sig figs) </span>
<span>direction of v(f) is the same as the direction of the final momentum </span>
<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>
<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>
<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
Answer:
1,85 m / s²
Explanation:
De la pregunta anterior, se obtuvieron los siguientes datos:
Velocidad inicial (u) = 40 km / h
Hora inicial (t₁) = 0
Tiempo final (t₂) = 6 s
Velocidad final (v) = 0
Aceleración (a) =?
A continuación, convertiremos 40 km / ha m / s. Esto se puede obtener de la siguiente manera:
1 km / h = 0,2778 m / s
Por lo tanto,
40 km / h = 40 km / h × 0,2778 m / s / 1 km / h
40 km / h = 11,11 m / s
Por tanto, 40 km / h equivalen a 11,11 m / s.
Finalmente, determinaremos la aceleración del móvil durante el período en el que desaceleró. Esto se puede obtener de la siguiente manera:
Velocidad inicial (u) = 11,11 m / s
Hora inicial (t₁) = 0
Tiempo final (t₂) = 6 s
Velocidad final (v) = 0
Aceleración (a) =?
a = (v - u) / (t₂ - t₁)
a = (0 - 11,11) / (6 - 0)
a = - 11,11 / 6
a = –1,85 m / s²
Por tanto, la aceleración del móvil durante el período en el que se ralentizó es de –1,85 m / s²
C- 10ft. Hope this helped. Have a great day! :D
