Why do we consider eyes, ears, and fingers as physical sensor
Answer:
because
Explanation:
because we need eye to see if you don't have eyes how you regnise some one
if you have ears how can you know who's voice is who's
if don't have fingers how can you know what your are holding this is why they are called physical sensors
Answer:
because they stimulate our brain and the sensor of human body detect stimuli.
Explanation:
Which statement describes one way that nuclear fission differs from nuclear
fusion?
A. Nuclear fission occurs naturally only in the Sun and othling stars.
B. Nuclear fission occurs only at very high temperatures.
C. Nuclear fission produces too little energy for practical
applications.
O D. Nuclear fission is used to power submarines and aircraft carriers.
Answer:
D. Nuclear fission is used to power submarines and aircraft carriers.
Explanation:
A and B describe nuclear fusion. C doesn't apply to nuclear fusion or nuclear fission.
A 5-kg projectile is fired over level ground with a velocity of 200 m/s at an angle of 25
above the horizontal. Just before it hits the ground its speed is 150 m/s. Over the entire trip the
change in the thermal energy of the projectile and air is:
Answer: 43.8 kJ
Explanation:
Given;
mass of the object, m = 5kg
initial velocity of the projectile, v₁ = 200 m/s
final velocity of the projectile, v₂ = 150 m/s
To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.
Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²
KE = Δ¹/₂mv² = ¹/₂m(v₁²-v₂²)
KE = ¹/₂ × 5(200²-150²) = 2.5(17500) = 43750 J = 43.8 kJ
Therefore, the change in thermal energy of the projectile is 43.8 kJ
The mass of Earth is 5.972 x 1024 kg and its orbital radius is an average of 1.496 x 1011 m. Calculate its linear momentum, given the period of one rotation is 3.15 x 107 s
Answer: its linear momentum is 1.78 × 10²⁹ kg.m/s
Explanation:
Given that;
mass of Earth m = 5.972 x 10²⁴ kg
radius r = 1.496 x 10¹¹ m
period t = 3.15 x 10⁷ s
now we know that Earth rotates in a circular path so the distance travelled per rotation is;
d = 2πr we substitute
d = 2π × 1.496 x 10¹¹ m
= 9.4 × 10¹¹ m
Now formula for speed v is;
v = d/t
we substitute
v = 9.4 × 10¹¹ m / 3.15 x 10⁷ s
v = 2.98 × 10⁴ m/s
now we determine the linear momentum p
linear momentum p = mv
we substitute
p = (5.972 x 10²⁴ kg) × (2.98 × 10⁴ m/s)
p = 1.78 × 10²⁹ kg.m/s
Therefore its linear momentum is 1.78 × 10²⁹ kg.m/s
The linear momentum of earth is 17.8 * 10²⁸ kgm/s
The orbital radius (r) of earth is 1.496 x 10¹¹ m. Hence the distance covered by one rotation is:
Distance = 2πr = 2π(1.496 x 10¹¹) m
The period of one rotation is 3.15 x 10⁷ s.
The velocity of earth (v) = distance/time = 2π(1.496 x 10¹¹) m/ 3.15 x 10⁷ s = 298840 m/s
Linear momentum = mass * velocity = 5.972 x 10²⁴ kg * 298840 m/s = 17.8 * 10²⁸ kgm/s
Therefore the linear momentum of earth is 17.8 * 10²⁸ kgm/s
Find out more at: https://brainly.com/question/12194595
g A frequently quoted rule of thumb in aircraft design is that wings should produce about 1000 N of lift per square meter of wing. (The fact that a wing has a top and bottom surface does not double its area.) a)At takeoff the aircraft travels at 61.1 m/s, so that the air speed relative to the bottom of the wing is 61.1 m/s. Given the sea level density of air to be 1.29 kg/m3, how fast (in m/s) must it move over the upper surface to create the ideal lift
Answer:
The value is [tex]u = 72.69 \ m/s[/tex]
Explanation:
From the question we are told that
The amount of force a square meter of an aircraft wing should produce is [tex]F = 1000 \ N[/tex]
The air speed relative to the bottom of the wing is [tex]v = 61.1 \ m/s[/tex]
The air level density of air is [tex]\rho_s = 1.29\ kg/m^3[/tex]
Gnerally this force per square meter of an aircraft wing is mathematically represented as
[tex]F = \frac{1}{2} * \rho_s * A * [ u^2 - v^2 ][/tex]
Here u is the speed air need to go over the top surface to create the ideal lift
A is the area of a square meter i.e [tex]A = 1 \ m^2[/tex]
So
[tex]1000 = \frac{1}{2} * 1.29 * 1 * [ u^2 - 61.1 ^2 ][/tex]
=> [tex]u = 72.69 \ m/s[/tex]
If you place 48 cm object 15 m in front of a convex mirror of focal length 5 m, what is the magnification of the image?
Answer:
m = 0.25
Explanation:
Given that,
Object distance, u = -15cm
Height of the object, h = 48
Focal length, f = cm
We need to find the magnification of the image.
Let v is the image distance. Using mirror's equation.
[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\=\dfrac{1}{5}-\dfrac{1}{(-15)}\\\\=3.75\ cm[/tex]
Magnification,
[tex]m=\dfrac{-v}{u}\\\\=\dfrac{-3.75}{-15}\\\\=0.25[/tex]
Hence, the magnification of the image is 0.25.
A 3 Kg exercise ball is held 2m above the ground. What is the gravitational potential energy?
Answer:
58.8
Explanation:
we should apply formula
m*g*h
3*9.81*2
Which of the following types of energy are present at some point in the energy transfer process in a nuclear power plan? Select all that apply.
Heat energy
Geothermal energy
Tidal energy
Hydroelectric energy
Nuclear energy
Electrical energy
Solar Energy
Solar Energy is the answer to the question tell me if i`m wrong
9. Who was the FIRST person to propose that the continents might fit together
Answer:
Alfred Wegener
Explanation:
A 0.0600-kilogram ball traveling at 60.0 meters per second hits a concrete wall. What speed must a 0.0100-kilogram bullet have in order to hit the wall with the same magnitude of momentum as the ball?
Answer:
the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s
Explanation:
The computation of the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is as follows:
The ball momentum is
[tex]\Delta Q = mv\\\\\Delta Q = 6 \times 1^-2 \times 60\\\\\Delta Q = 3.6 kg \times m/s[/tex]
As there is a similar momentum
So,
[tex]\Delta Q = MV\\\\3.6 = 10^-2V\\\\V = 360 m/s[/tex]
Hence, the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s