Acceleration = (final velocity^2 - initial velocity^2) / 2 * distance

Acceleration = (19.1^2 - 9.2^2) / 2 * 32

Acceleration = (364.81 - 84.64) / 64

Acceleration = 280.17 / 64

Acceleration = 4.3777m/s^2

:)

6 0

3 years ago

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

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

8 0

3 years ago

The accepted value for the density of iron is 7.87 g/cm3. a student records the mass of a 20.00 cm3 block of iron as 153.8 grams

dangina [55]

The density of an object can be calculated using the formula Density = Mass/Volume.

Experimental Density:

Density = 153.8g / 20.00 cm^3
Density = 7.69g/cm^3

Percent error equation:

% Error = | Theoretical Value - Experimental Value|/Theoretical Value * 100
% Error = | 7.87g/cm^3 - 7.69g/cm^3|/7.87g/cm^3 * 100
% Error = 2.29%

Therefore a is the correct answer.

6 0

3 years ago