No spacecraft has been built yet that was able to absorb harmful
radiations in space, change weather conditions on Earth, or destroy
meteors and comets which might strike Earth.
We should continue to send robotic spacecrafts into space
because they help discard some myths about objects in space.
In other words, they help us learn things that we never knew before.
Answer:
<h3>JAWAB SECEPATNYA pliss</h3><h3 /><h3>Anda memiliki rangkaian paralel 10 volt, dengan 2 resistor di atasnya. Berapakah tegangan pada</h3><h3>resistor pertama? Di seberang kedua?</h3><h3 /><h3>(saya akan menandai tercerdas tolong bantu)</h3>
Explanation:
Hukum Ohm
= tegangan
= kuat arus
= ketahanan
Kalau kamu mau mencari tegangan listrik, kamu gunakan rumus V = I.R. Kalau ternyata kamu perlu mencari kuat arus listrik, maka gunakan rumus I = V/R. Nah, kalau yang kamu cari adalah hambatan listrik, maka gunakan rumus R = V/I.
Answer:
70 cm
Explanation:
0.5 kg at 20 cm
0.3 kg at 60 cm
x = Distance of the third 0.6 kg mass
Meter stick hanging at 50 cm
Torque about the support point is given by (torque is conserved)
The position of the third mass of 0.6 kg is at 20+50 = 70 cm
<span>Thermocline is a layer between
warm water from the ocean’s surface and cool water from below the ocean. In here,
the temperature decreases rapidly from the warmer layer to the colder layer. A thermocline forms due to the heat of the sun
heating the ocean’s surface. Because of the difference in density between warm
and cooler ocean water, cooler ocean water sinks and warmer ocean water floats.
This is caused due to the heat and mass transfer between particles of the
ocean. The answer is letter C. The sun’s radiation does not extend below a
certain depth; therefore, deeper ocean water is colder than surface water.</span>
Explanation:
The 11Ω, 22Ω, and 33Ω resistors are in parallel. That combination is in series with the 4Ω and 10Ω resistors.
The net resistance is:
R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)
R = 20Ω
Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:
V = IR
120 V = I (20Ω)
I = 6 A
So the voltage drops are:
V = (4Ω) (6A) = 24 V
V = (10Ω) (6A) = 60 V
That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:
V = 120 V − 24 V − 60 V
V = 36 V
So the currents are:
I = 36 V / 11 Ω = 3.27 A
I = 36 V / 22 Ω = 1.64 A
I = 36 V / 33 Ω = 1.09 A
If we wanted to, we could also show this using Kirchhoff's laws.
