Answer:
V = 90.51 m/s
Explanation:
From the given information:
Initial speed (u) = 0
Distance (S) = 391 m
Acceleration (a) = 18.9 m/s²
Using the relation for the equation of motion:
v² - u² = 2as
v² - 0² = 2as
v² = 2as
v = 121.57 m/s
After the parachute opens:
The initial velocity = 121.57 m/ss
Distance S' = 332 m
Acceleration = -9.92 m/s²
How fast is the racer can be determined by using the relation:
V = 90.51 m/s
Answer:
1 N
Explanation:
First the equation is momentum = Force / distance
20 cm = 0.2 m
5 N/m = F / 0.2 m
F = 1 N
Answer:
a)3312 x 10⁴ J
b)I = 57.5 A
c)9200 W
Explanation:
Given that
P =4600 W
Time t= 2 h = 2 x 3600 s= 7200 s
We know that
1 W = 1 J/s
a)
Energy stored in the battery = P .t
=4600 x 7200 J
=3312 x 10⁴ J
b)
We know that power P given as
P = V .I
V=Voltage ,I =Current
4600 = 80 x I
I = 57.5 A
c)
The energy supplied = 4600 x 2 = 9200 W
Answer:
Explanation:
Let the magnetic field be B = B₁i + B₂j + B₃k
Force = I ( L x B ) , I is current , L is length and B is magnetic field .
In the first case
force = - 2.3 j N
L = 2.5 i
puting the values in the equation above
- 2.3 j = 8 [ 2.5 i x ( B₁i + B₂j + B₃k )]
= - 20 B₃ j + 20 B₂ k
comparing LHS and RHS ,
20B₃ = 2.3
B₃ = .115
B₂ = 0
In the second case
L = 2.5 j
Force = I ( L x B )
2.3i−5.6k = 8 ( 2.5 j x (B₁i + B₂j + B₃k )
= - 20 B₁ k + 20B₃ i
2.3i−5.6k = - 20 B₁ k + 20B₃ i
B₃ = .115
B₁ = .28
So magnetic field B = .28 i + .115 B₃
Part A
x component of B = .28 T
Part B
y component of B = 0
Part C
z component of B = .115 T .
