Answer:
0.687 m/s
Explanation:
Initial energy = final energy
1/2 mu² = mgh + 1/2 mv²
1/2 u² = gh + 1/2 v²
Given u = 2.00 m/s, g = 9.8 m/s², and h = 0.180 m:
1/2 (2.00 m/s)² = (9.8 m/s²) (0.180 m) + 1/2 v²
v = 0.687 m/s
Answer:
<em>1,378.9ms²</em>
Explanation:
Given the following
Distance S = 70.6m
Time t = 0.32secs
Initial velocity = 0m/s
Required
Acceleration
Using the equation of motion
S = ut+1/2at²
Substitute
70.6 = 0+1/2a(0.32)²
70.6 = 0.0512a
a = 70.6/0.0512
a = 1,378.9
<em>Hence the acceleration is 1,378.9ms²</em>
Answer:
P = 1 x 10⁸ Pa
Explanation:
given,
radius = 2.0 ×10⁻¹⁰ m
Temperature
T = 300 K
Volume of gas molecule =


V = 33.51 x 10⁻³⁰ m³
we know,
P V = 1 . k T
k = 1.38 x 10⁻²³ J/K
P(33.51 x 10⁻³⁰) = 1 . (1.38 x 10⁻²³) x 300
P = 1.235 x 10⁸ Pa
for 1 significant figure
P = 1 x 10⁸ Pa
Answer:
The angle between the magnetic field and the wire’s velocity is 19.08 degrees.
Explanation:
Given that,
Potential difference, V = 53 mV
Length of the wire, l = 12 cm = 0.12 m
Magnetic field, B = 0.27 T
Speed of the wire, v = 5 m/s
Due to its motion, an emf is induced in the wire. It is given by :

Here,
is the angle between magnetic field and the wire’s velocity

So, the angle between the magnetic field and the wire’s velocity is 19.08 degrees.