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3 years ago

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so no body wants to help me:((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((

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1 answer:

larisa [96]3 years ago

8 0

Yes i will help!!!!!!!!!!!!!!!!!!!!!

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The force required to stretch a Hooke’s-law spring varies from 0 N to 63.5 N as we stretch the spring by moving one end 5.31 cm

Alika [10]

Answer:

Force constant will be 1195.85 N/m

Work done will be 1.6859 J

Explanation:

We have given the force, F = 63.5 N

Spring is stretched by 5.31 cm

So x = 0.0531 m

Force is given , F = 63.5 N

We know that force is given by $F=kx$

So $63.5=k\times 0.0531$

k = 1195.85 N/m

Now we have to find the work done

We know that work done is given by

$W=\frac{1}{2}kx^2=\frac{1}{2}\times 1195.85\times 0.0531^2=1.6859J$

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3 years ago

If the pitched ball was traveling 77 mph before stanton's bat hit it and 120 mph after his bat hit it, by what amount did the sp

Mice21 [21]

here we will use the concept of Newton's III law

as per Newton's III law the impulse given to the ball is same as the impulse lost by the bat

So here we will say

impulse gain by the ball = impulse lost by the bat

$m_1(v_f - v_i) = m_2(\Delta v)$

given that

$m_1 = 5 ounce$

$m_2 = 32 ounce$

For ball the change in speed will be

$v_f - v_i = (120 - 77)mph$

now from above equation

$5\times (120 - 77) = 32 \times \Delta v$

$\Delta v = 6.72 mph$

so speed of bat will decrease by 6.72 mph

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3 years ago

calculate the spring constant if a weight of 250N is added to a spring which increases in length by 20cm

ZanzabumX [31]

Since, F = k . ∆x

Therefore, k = F / ∆x = 250 / 0.2 = 1250 N/m

(ps: convert 20 cm into 0.2 m)

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Answer:

Only object a and b???????

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14. This illustration shows the path that a ball follows when it is thrown into the air. The speed of the ball's motion changes

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Ignore this comment sorry

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