Answer:
X=92.49 m
Explanation:
Given that
u= 21 m/s
h= 97 m
Time taken to cover vertical distance h
h= 1/2 g t²
By putting the values
97 = 1/2 x 10 x t² ( g = 10 m/s²)
t= 4.4 s
The horizontal distance
X= u .t
X= 21 x 4.4
X=92.49 m
Work done on the crate is 1411.2 J
Explanation:
Work done is defined as the product of force and the distance moved by the object. The unit of work done is in joules and denoted by the symbol J.
Work done = F * d
where F represents the force and d represents the distance moved by the object.
mass = 72 kg , distance moved by the object is given by 2.0 m
Force F = mass * gravity = 72 * 9.8
= 705.6 N =706 N.
Work done = 706 * 2.0 = 1412 J.
<u>Answer</u>
48 Volts
<u>Explanation</u>
The question can be solve using the turn rule of a transformer that states;
Np/Ns = Vp/Vs
Where Np ⇒ number of turns in the primary coil.
Ns ⇒number of turns in the seconndary coil
Vp ⇒ primary voltage
Vs ⇒secondary voltage
Np/Ns = Vp/Vs
10/4 = 120/Vp
Vp = (120 × 4)/10
= 480/10
= 48 Volts
2.0y i think this is guess but if it right then thats good
Answer:
E
= -4556.18 N/m
Explanation:
Given data
u = 3.6×10^6 m/sec
angle = 34°
distance x = 1.5 cm = 1.5×10^-2 m (This data has been assumed not given in
Question)
from the projectile motion the horizontal distance traveled by electron is
x = u×cosA×t
⇒t = x/(u×cos A)
We also know that force in an electric field is given as
F = qE
q= charge , E= strength of electric field
By newton 2nd law of motion
ma = qE
⇒a = qE/m
Also, y = u×sinA×t - 0.5×a×t^2
⇒y = u×sinA×t - 0.5×(qE/m)×t^2
if y = 0 then
⇒t = 2mu×sinA/(qE) = x/(u×cosA)
Also, E = 2mu^2×sinA×cosA/(x×q)
Now plugging the values we get
E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})
E
= -4556.18 N/m
