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cluponka [151]
3 years ago
15

A gardener pushes her 90 kg wheelbarrow in a straight line out of her garage and across the yard in 18 s. If her average velocit

y is 1.5 m/s, what is her displacement?
A. 0.8 km
B. 12 km toward the garden
C. 18 m toward the garage
D. 27 m away from the garage
Physics
1 answer:
Hunter-Best [27]3 years ago
3 0
The answer is D, because displacement equals average velocity times time. 1.5 m/s times 18s equals 27 meters.
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The phrase "light year" is a <u><em>distance</em></u> ... it's the distance that light travels through vacuum in one year.

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2 years ago
Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum
Alina [70]

Answer:

  • Uncertainty in position (∆x) = 24 × 10⁻¹⁵ m
  • Uncertainty in momentum (∆P) = ?
  • Planck's constant (h) = 6.26 × 10⁻³⁴ Js

\longrightarrow \:  \:  \sf\Delta x .\Delta p =  \dfrac{h}{4\pi}

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34}} {4 \times  \frac{22}{7} }

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34}} { \frac{88}{7} }

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34} \times 7} { 8 }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34} } { 8  \times 24 \times  {10}^{ - 15} }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34} } { 192 \times  {10}^{ - 15} }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34}  \times  {10}^{15} } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ -19}   } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{4382 \times  {10}^{ - 2}  \times  {10}^{ -19}   } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{4382 \times  {10}^{ - 21}   } { 192}

\longrightarrow \:  \:  \sf\Delta p = 22.822\times  {10}^{ - 21}

\longrightarrow \:  \:  \sf\Delta p = 2.2822 \times  {10}^{1} \times  {10}^{ - 21}

\longrightarrow \:  \: \underline{ \boxed{ \red{  \bf\Delta p = 2.2822 \times  {10}^{ - 20}  \:  kg/ms}}}

4 0
2 years ago
A block with a mass of 0.28 kg is attached to a horizontal spring. The block is pulled back from its equilibrium position until
krek1111 [17]

Answer:

Explanation:

For spring

n=\sqrt{\frac{k}{m} }

where n is frequency of oscillation and k is force constant and m is mass

Putting the values

1.2=\sqrt{\frac{k}{.28} }

k = .4032 N/m

F= k x

where F is force , k is force constant and x is extension

Putting the given values

1 = .4032 x

x = 2.48 m

8 0
3 years ago
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