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Tanzania [10]
3 years ago
6

Sorry that i ask for too much but I'm horrible at this ⭐-_-

Physics
2 answers:
Evgesh-ka [11]3 years ago
6 0

Answer:

1. A

2. C

3. D

4.A

Explanation:

salantis [7]3 years ago
6 0
1 A
2 C
3 D
4 A hope this helps
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What is the frequency of light with a wavelength of 7.9 x 10^-9 m? ( the speed of light is 3.00 x 10^8)
nlexa [21]
Using the formula v=f times lambada
then v=the speed of light.
and f=what’s we’re looking for
and lambada=the wavelength.

so then you sub what you have (v and lambada) in the formula.
then multiply the frequency(f) by the given wavelength and then solve for f

3 0
3 years ago
on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker
Aleksandr-060686 [28]
B4 the tackle: 

<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>

<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>


<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>

<span>The vector triangle is right angled: </span>

<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>

<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>

<span>v(f) = 5.6 m/s (to 2 sig figs) </span>


<span>direction of v(f) is the same as the direction of the final momentum </span>

<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>


<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>




<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
4 0
3 years ago
In a transformer, energy is carried from the primary coil to the secondary coil by:________
likoan [24]

In a transformer, energy is carried from the primary coil to the secondary coil by magnetic field in the iron core.

To find the answer, we have to know more about the transformer.

<h3>How transformer works?</h3>
  • An item utilized in the transfer of electric energy is a transformer.
  • AC current is used for transmission.
  • It is frequently used to modify the supply voltage between circuits without altering the AC frequency.
  • The fundamentals of mutual and electromagnetic induction govern how the transformer operates.
  • Magnetic field through the primary coil changes when primary coil current varies. the iron core of the secondary coil likewise has a magnetic field.
  • EMF is therefore generated in the secondary coil.

Thus, we can conclude that, in a transformer, energy is carried from the primary coil to the secondary coil by magnetic field in the iron core.

Learn more about the transformer here:

brainly.com/question/26787198

#SPJ4

5 0
2 years ago
What is the definition of differentiation geology?
UkoKoshka [18]
Any process in which a mixture of materials separates out partially
5 0
4 years ago
A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o
inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) =  \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_i^2) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2  ) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 -  \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 -  \frac{3}{4}v_f^2\\\\

h = \frac{3}{4g}(v_1^2 -v_f^2)

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0
3 years ago
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