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Vadim26 [7]
3 years ago
7

Help Please!!

Physics
1 answer:
Setler [38]3 years ago
3 0
A is the answer to the question
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a cricket stamp

Explanation:

4 0
3 years ago
What is a tool that measures the size of a force
MakcuM [25]
<span>this is a dynamometer</span>
3 0
3 years ago
A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The
Yuki888 [10]

Answer:

Approximately 0.560\; {\rm m}, assuming that:

  • the height of 3.21\; {\rm m} refers to the distance between the clay and the top of the uncompressed spring.
  • air resistance on the clay sphere is negligible,
  • the gravitational field strength is g = 9.81\; {\rm m\cdot s^{-2}}, and
  • the clay sphere did not deform.

Explanation:

Notations:

  • Let k denote the spring constant of the spring.
  • Let m denote the mass of the clay sphere.
  • Let v denote the initial speed of the spring.
  • Let g denote the gravitational field strength.
  • Let h denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let x denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of x, the elastic potential energy \text{PE}_{\text{spring}} in this spring would be:

\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}.

The initial kinetic energy \text{KE} of the clay sphere was:

\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}.

When the spring is at the maximum compression:

  • The clay sphere would be right on top of the spring.
  • The top of the spring would be below the original position (when the spring was uncompressed) by x.
  • The initial position of the clay sphere, however, is above the original position of the top of the spring by h = 3.21\; {\rm m}.

Thus, the initial position of the clay sphere (h = 3.21\; {\rm m} above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by (h + x).

The gravitational potential energy involved would be:

\text{GPE} = m\, g\, (h + x).

No mechanical energy would be lost under the assumptions listed above. Thus:

\text{PE}_\text{spring} = \text{KE} + \text{GPE}.

\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x).

Rearrange this equation to obtain a quadratic equation about the only unknown, x:

\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0.

Substitute in k = 1570\; {\rm N \cdot m^{-1}}, m = 4.87\; {\rm kg}, v = 5.21\; {\rm m\cdot s^{-1}}, g = 9.81\; {\rm m \cdot s^{-2}}, and h = 3.21\; {\rm m}. Let the unit of x be meters.

785\, x^{2} - 47.775\, x - 219.453 \approx 0 (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that x \ge 0:

\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}.

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately 0.560\; {\rm m}.

5 0
2 years ago
If a body having mass 40kg started moving initially with rest and it takes a velocity of 20m/sec in time 4 seconds. Find the val
baherus [9]

{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Mass \ of \ the \ body \ (m) = 40 \ kg}

\:\:\:\:\bullet\:\:\:\sf{Final \ velocity \ of \ the \  body \ (v) = 20 \ m/s}

\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ the \  body \ (u) = 0}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Force \ exerted \ by \ the \  body \ ( F)}

\\

{\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\

☯ <u>Using 1st equation of motion </u>

\\

\dashrightarrow\:\: \sf{v = u + at}

\\

\dashrightarrow\:\: \sf{20 = 0 + a(4)}

\\

\dashrightarrow\:\: \sf{20 = 4a}

\\

\dashrightarrow\:\: \sf{\dfrac{\cancel{20}}{\cancel{4}} = a}

\\

\dashrightarrow\:\: \sf{a = 5}

\\

☯ <u>Now, Finding the force exerted </u>

\\

\dashrightarrow\:\: \sf{F = ma}

\\

\dashrightarrow\:\: \sf{F = 40 \times 5}

\\

\dashrightarrow\:\: \sf{F = 200 \ N}

\\

☯ <u>Hence</u>, \\

\:\:\:\:\star\:\:\:\sf{The \ force \ exerted \ by \ the \ body \ is \ 200N}

8 0
3 years ago
A horse does 910 J of work in 380 seconds while pulling a wagon. What is the power output of the horse? Round your answer to two
Alekssandra [29.7K]
2.39 Watts roughly since watts is joules per second it’s just 910j/380s
3 0
3 years ago
Read 2 more answers
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