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3 years ago

Look at the circuit diagram. Which of these components is part of the circuit?

Physics

1 answer:

tangare [24]3 years ago

4 0

I think the answer is c AC power source
Hope this help you?

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A car is traveling at 21m/s. It accelerates at 1.4m/s2 for 11s. How fast is the car moving after the acceleration?

nata0808 [166]

Answer:

v=u+at

v=21+1.4(11)

v=21+15.4

v=36.4

3 0

3 years ago

You throw a glob of putty straight up toward the ceiling, which is 3.50 mm above the point where the putty leaves your hand. The

LuckyWell [14K]

Answer: $4.65\ m/s$

Explanation:

Given

Distance putty has to travel is 3.5 m

The initial speed of putty is 9.50 m/s

Using equation of motion to determine the velocity of putty just before it hits ceiling

$v^2-u^2=2as$

$\Rightarrow v^2-(9.5)^2=2(-9.8)(3.5)\\\\\Rightarrow v^2=9.5^2-68.6\\\Rightarrow v=\sqrt{90.25-68.6}\\\Rightarrow v=4.65\ m/s$

So, the velocity of putty just before hitting is $4.65\ m/s$

5 0

3 years ago

A neutron star and a black hole are 3.34 x 1012 m from each other at a certain point in their orbit. The neutron star has a mass

m_a_m_a [10]

Answer:

F=1.65 x 10²⁶ N

Explanation:

Given that

Distance ,R= 3.34 x 10¹² m

Mass m₁= 2.78 x 10³⁰ kg

Mass ,m₂= 9.94 x 10³⁰ kg

we know that gravitational force F given as

$F=G\dfrac{m_1m_2}{R^2}$

G=Constant

G=6.67 x 10⁻¹¹ Nm²/kg²

Now by putting the values

$F=6.67\times 10^{-11}\times \dfrac{2.78\times 10^{30}\times 9.94\times 10^{30}}{(3.34\times 10^{12})^2}\ N$

F=1.65 x 10²⁶ N

Therefore the force between these two mass will be 1.65 x 10²⁶ N.

5 0

3 years ago

How to make a tangram with paper?

sattari [20]

Just follow these simple steps:
Fold a rectangular piece of paper so that a square is formed. ...
Cut the square into two triangles.
Take one triangle and fold it in half. ...
Take the other triangle and crease it in the middle. ...
Fold the trapezoid in half and fold again. ...
Fold the remaining small trapezoid and cut it in two.

3 0

3 years ago

Consider electrons of kinetic energy 6.0 eV and 600 keV. For each electron, find the de Broglie wavelength, particle speed, phas

irinina [24]

Answer:

For 6.0 eV

0.5 nm, 1.45*10^6 m/s, 6.17*10^10 m/s, 1.45*10^6 m/s

For 600 eV

1.26*10^-3 nm, 2.66*10^8 m/s, 3.37*10^8 m/s, 2.66*10^8 m/s

Explanation:

See attachment for calculation

5 0

3 years ago