Answer:
0.833
Explanation:
Power = energy / time
Power = force × distance / time
Power = force × velocity
P = (850 kg) (9.8 m/s²) (1.00 m/s)
P = 8330 W
P = 8.33 kW
The efficiency of the motor is therefore:
e = 8.33 kW / 10.0 kW
e = 0.833
Answer:
t = 2.13 10-10 s
, d = 6.39 cm
Explanation:
For this exercise we use the definition of refractive index
n = c / v
Where n is the refraction index, c the speed of light and v the speed in the material medium.
The refractive indices of ice and crown glass are 1.13 and 1.52, respectively, therefore the speed of the beam in the material medium is
v = c / n
As the beam strikes perpendicularly, the beam path is equal to the distance of the leaves, there is no refraction, so we can use the uniform motion relationships
v = d / t
t = d / v
t = d n / c
Let's look for the times on each sheet
Ice
t₁ = 1.4 10⁻² 1.31 / 3 10⁸
t₁ = 0.6113 10⁻¹⁰ s
Crown glass (BK7)
t₂ = 3.0 10⁻² 1.52 / 3.0 10⁸
t₂ = 1.52 10⁻¹⁰ s
Time is a scalar therefore it is additive
t = t₁ + t₂
t = (0.6113 + 1.52) 10⁻¹⁰
t = 2.13 10-10 s
The distance traveled by this time in a vacuum would be
d = c t
d = 3 10⁸ 2.13 10⁻¹⁰
d = 6.39 10⁻² m
d = 6.39 cm
The question is a little confusing. An electromagnetic wave is radiation. One doesn’t emit the other. Take another look at the question and write it again please.
Answer:
27.22 m/s
Explanation:
Let the speed of clay before impact is u.
the speed of clay and target is v after impact.
use conservation of momentum
momentum before impact momentum after impact
mass of clay x u = (mass of clay + mass of target) x v
100 x u = (100 + 500) x v
u = 6 v .....(1)
distance, s = 2.1 m
μ = 0.5
final velocity is zero. use third equation of motion
v'² = v² + 2as
0 = v² - 2 x μ x g x s
v² = 2 x 0.5 x 9.8 x 2.1 = 20.58
v = 4.54 m/s
so by equation (1)
u = 6 x 4.54 = 27.22 m/s
thus, the speed of clay before impact is 27.22 m/s.
Answer:
112 m/s², 79.1°
Explanation:
In the x direction, given:
x₀ = 0 m
x = 19,500 cos 32.0° m
v₀ = 1810 cos 20.0° m/s
t = 9.20 s
Find: a
x = x₀ + v₀ t + ½ at²
19,500 cos 32.0° = 0 + (1810 cos 20.0°) (9.20) + ½ a (9.20)²
a = 21.01 m/s²
In the y direction, given:
y₀ = 0 m
y = 19,500 sin 32.0° m
v₀ = 1810 sin 20.0° m/s
t = 9.20 s
Find: a
y = y₀ + v₀ t + ½ at²
19,500 sin 32.0° = 0 + (1810 sin 20.0°) (9.20) + ½ a (9.20)²
a = 109.6 m/s²
The magnitude of the acceleration is:
a² = ax² + ay²
a² = (21.01)² + (109.6)²
a = 112 m/s²
And the direction is:
θ = atan(ay / ax)
θ = atan(109.6 / 21.01)
θ = 79.1°
