QUICK! If i have a 95.6% in my as my grade in that class what is 30% off of that grade?

What would be the Elastic Potential Energy (EPE) stored in a spring with a constant k = 200 N/m that is pulled to stretch 0.45m?

harkovskaia [24]

Answer:

During the summer, oxygen is supplied by surface winds. M/NPT 16 to 26 0.14 to 0.45 m 1/ 77.80 $ 28 Phone: 418 81-1 Fax: 418 81-2882 SZ5-2025 SZ22-200 SZ2-250 SZ24-00 12.5 x 60 12.5 x 60 12.5 x 60 8 x 10 10 x 15 10 x 20 15 x HI-950 can take measures between 0.1 C to 999.9 C/ F. K, J & T-type: The HI-9551

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3 0

3 years ago

Which situation is work not being done? A) A bookcase is slid across carpeting. B) A stack of books is carried at waist level ac

Nastasia [14]

Answer:

B) A stack of books is carried at waist level across a room

Explanation:

Work is defined as:

$W=Fd cos \theta$

where

F is the force applied

d is the displacement of the object

$\theta$ is the angle between the direction of the force and of the displacement

From the formula, we see that the work done is zero when the force and the displacement are perpendicular to each other. Let's now analyze each situation:

A) A bookcase is slid across carpeting. --> work is done, because the force that pushes the bookcase is in the same direction of the displacement

B) A stack of books is carried at waist level across a room. --> no work is done, because the force to carry the book is vertical, while the displacement of the books is horizontal

C) A chair is lifted vertically with respect to the floor. --> work is done, because the force that lifts the chair is vertical, and the displacement is vertical as well

D) A table is dropped onto the ground. --> work is done, because the force of gravity (that makes the table falling down) is vertical and the displacement of the table is also vertical.

5 0

3 years ago

An object accelerating may be changing in what two ways

Oksana_A [137]

Velocity and direction

8 0

3 years ago

To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0129-kg bullet i

ipn [44]

Answer:

t=0.42s

Explanation:

Here you have an inelastic collision. By the conservation of the momentum you have:

$m_1v_1+m_2v_2=(m_1+m_2)v$

m1: mass of the bullet

m2: wooden block mass

v1: velocity of the bullet

v2: velocity of the wooden block

v: velocity of bullet and wooden block after the collision.

By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:

$m_1(-v_1)+m_2v_2=(m1+m2)(-v2)$

Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:

$-(0.0129kg)(767m/s)+(1.17kg)v_2=(1.1829kg)(-v_2)\\\\v_2=4.20m/s$

Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:

$v_2=v_o+gt\\\\t=\frac{v_2-v_o}{g}=\frac{4.2m/s-0m/s}{9.8m/s^2}=0.42s$