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skad [1K]
3 years ago
8

Please help me with this question​

Physics
1 answer:
topjm [15]3 years ago
7 0

Answer:

Answer is option d

Explanation:

,Mark as brainlist

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A 1.0 kg block is attached to an unstretched
KonstantinChe [14]

Answer:

Change in  potential energy of the block-spring-Earth

system between Figure 1 and Figure 2 = 1 Nm.

Explanation:

Here, spring constant, k  = 50 N/m.

given block comes down eventually 0.2 m below.

here, g = 10 m/s.

let block be at a height h above the ground in figure 1.

⇒In figure 2, potential energy of the block-spring-Earth

system = m×g×(h - 0.2) + 1/2× k × x². where, x = change in spring length.

⇒ Change in  potential energy of the block-spring-Earth

system between Figure 1 and Figure 2 = (m×g×(h - 0.2)) - (1/2× k × x²)

              =  (1×10×0.2) - (1/2×50×0.2×0.2) = 1 Nm.

4 0
3 years ago
What affects the amount of solar energy the planet receives?
solmaris [256]
The sun is facing a certain side of the earth
4 0
3 years ago
Read 2 more answers
what frequency of light must an electron in hydrogen absorb to jump from the n=2 state to the n=5 state? (a) 2.86 Hz (b) 4.08 Hz
Ostrovityanka [42]

Answer:

(c) 6.91x10^14 Hz

Explanation:

Find the level energy of n=2 and n=5, using  the formula:

E = -E_0/n^2

where  E_0=13.6eV

E_2 =\frac{-13.6}{2^2}=-3.4eV

E_5 =\frac{-13.6}{5^2}=-0.544eV

To jump from n=2 to n=5 the electron absorbs a photon with energy equal to (-0.544) - (-3.4)=2.856eV, using the next formula to find specific wavelength \lambda to that energy

E = hc/\lambda

Where c is the speed of light (c=3 \times10^8m/s) and h is Planck's constant (h=4.14\times10^{-15}eVs). Solve for \lambda:

E = hc/\lambda\\\lambda E = hc\\\lambda = \frac{hc}{E} \\\lambda = \frac{(4.14\times10^{-15})(3 \times10^8)}{2.856}=4.35\times10^{-7}m

The frequency of this wavelength is calculated with this formula:

f=\frac{c}{\lambda}

f=\frac{3\times10^8}{4.3487\times10^{-7}} =6.89\times10^{14}Hz\approx6.9\times10^{14}Hz

8 0
3 years ago
Y UD Eda
PIT_PIT [208]

We have to find acceleration first.

  • Initial velocity=u=44m/s
  • Final velocity=v=22m/s
  • Time=11s=t

\boxed{\sf Acceleration=\dfrac{v-u}{t}}

\\ \sf\longmapsto Acceleration=\dfrac{22-44}{11}

\\ \sf\longmapsto Acceleration=\dfrac{-22}{11}

\\ \sf\longmapsto Acceleration=-2m/s^2

  • Now
  • Distance=s

Using second equation of kinematics

\boxed{\sf s=ut+\dfrac{1}{2}at^2}

\\ \sf\longmapsto s=44(11)+\dfrac{1}{2}(-2)(11)^2

\\ \sf\longmapsto s=484+(-1)(121)

\\ \sf\longmapsto s=484-(121)

\\ \sf\longmapsto s=363m

6 0
3 years ago
Read 2 more answers
Bacteria move back and forth by using their flagella (structures that look like little tails). Speeds of up to 50 μm/s ⎛ ⎝50×10−
Rzqust [24]

Answer:

Explained

Explanation:

Displacement is the change in the position of an object with reference to a starting point. for displacement to occur the position of the object must change.

Here the bacteria although moving but it is moving back and forth, meaning its initial and final positions are the same and hence no displacement. Whereas distance is the total distance traveled no matter in what direction. Hence, The total distance traveled by a bacterium is large for its size, while its displacement is small.

4 0
3 years ago
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