An object moves from point A to point C along the rectangle shown in the figure below.

According to ohm's law what is the resistance of a light bulb if the applied voltage is 21 volts and the current is 3 amps?

Nady [450]

Resistance = voltage/current = 21/3 = 7Ω

6 0

3 years ago

A 0.0750kg teddy bear is dropped from a deck that is 3.00m above the ground. What will be the velocity of the teddy bear as it s

FrozenT [24]

The velocity of the teddy bear as it strikes the ground is 7.67 m/s.

<h3>Velocity of the teddy when it strikes the ground</h3>

The velocity of the teddy when it strikes the ground is calculated from principle of conservation of energy as shown below.

K.E(bottom) = P.E(top)

¹/₂mv² = mgh

v² = 2gh

v = √2gh

where;

h is height of fall of the teddy
g is acceleration due to gravity

v = √(2 x 9.8 x 3)

v = 7.67 m/s

Thus, the velocity of the teddy bear as it strikes the ground is 7.67 m/s.

Learn more about conservation of energy here: brainly.com/question/166559

#SPJ1

4 0

1 year ago

A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top

Anton [14]

Answer:

$y_{max} = 0.829\,m$

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

$v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}$

$v = 2.913\,\frac{m}{s}$

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

$(3\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m) + \frac{1}{2}\cdot (3\,kg)\cdot (2.913\,\frac{m}{s} )^{2} = (3\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s) ^{2}$

After some algebraic handling, a second-order polynomial is formed:

$12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} - (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s$

$1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0$

The roots of the polynomial are, respectively:

$\Delta s_{1} \approx 0.128\,m$

$\Delta s_{2} \approx -0.099\,m$

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

$\Delta s \approx 0.128\,m$

The maximum height that the block reaches after rebound is:

$(3\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s)^{2} = (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot y_{max}$

$y_{max} = 0.829\,m$

4 0

3 years ago

Read 2 more answers

What is energy at motion called

salantis [7]

Energy at motion is: Kinetic energy

7 0

3 years ago

Read 2 more answers

What are three things you already know about the game of baseball?

castortr0y [4]

Answer:

The Positions are, Pitcher, Catcher, First base, second base, third base, shortstop, left field, Center field, and right field. Some pitches that they throw are, 4-seam fastball, Curveball, 2-seam fastball, Slider, Knuckleball, 12-6 Curve, Knucklecurve, Slurve, Changeup, Splitter, Cutter, Split-finger fastball, Sinker, Palm ball, Eephus pitch, Screwball, Cut Fastball, Circle Changup, and Forkball. I think thats all the piches i know.

6 0

2 years ago