This is an example of a ____________ resource using ______

Andre, whose mass is 77.1 kg, sits on a diving board above a dunk tank. If he is sitting 0.90 m above the water, what is his tot

Ganezh [65]

Answer:

680 J

Explanation:

Mechanical energy = potential energy + kinetic energy

ME = PE + KE

ME = mgh + ½ mv²

ME = (77.1 kg) (9.8 m/s²) (0.90 m) + ½ (77.1 kg) (0 m/s)²

ME = 680 J

8 0

4 years ago

Read 2 more answers

Compare the force of air resistance and the force of gravity on an object falling at its terminal velocity.

Sholpan [36]

The viscous force on an object moving through air is proportional to its velocity.
The only forces acting on an object when falling are air resistance and its weight itself. The weight acts vertically downwards whereas air resistance acts vertically upward.
Let F be the viscous force due to air molecules, B be buoyant force due to air and W be the weight of falling object. Initially, the velocity of falling object and hence the viscous force F is zero and the object is accelerated due to force
(W-B). Because of the acceleration the velocity increases and accordingly the viscous force also increases. At a certain instant, the viscous force becomes equal to W-B. The net force then becomes zero and the object falls with constant velocity. This constant velocity is called terminal velocity.
Thus at terminal velocity, air resistance and force of gravity becomes equal.

7 0

4 years ago

James threw a ball vertically upward with a velocity of 41.67ms-1 and after 2 second David threw a ball vertically upward with a

Reptile [31]

Answer:

When have passed 3.9[s], since James threw the ball.

Explanation:

First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.

We'll use the kinematics equations to find these two unknowns.

$y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can find the velocity after 2 seconds.

$v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

$y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\$

As we can see the equation is based on Time (t).

Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

$y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]$

Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.

Time = 2 + 1.9 = 3.9[s]

4 0

3 years ago

only a relatively small part of the electromagnetic spectrum is visible. what determines which bands of the electromagnetic spec

swat32

Answer:

hsnzbssj

Explanation:

<h2>skibbbbbbbbbbbbibi</h2>

3 0

3 years ago

Radar uses radio waves of a wavelength of 2.9 m . The time interval for one radiation pulse is 100 times larger than the time of

Mandarinka [93]

Answer:

145 m

Explanation:

Given:

Wavelength (λ) = 2.9 m

we know,

c = f × λ

where,

c = speed of light ; 3.0 x 10⁸ m/s

f = frequency

thus,

$f=\frac{c}{\lambda}$

substituting the values in the equation we get,

$f=\frac{3.0\times 10^8 m/s}{2.9m}$

f = 1.03 x 10⁸Hz

Now,

The time period (T) = $\frac{1}{f}$

or

T = $\frac{1}{1.03\times 10^8}$ = 9.6 x 10⁻⁹ seconds

thus,

the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s

Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s

Now,

For radar to detect the object the pulse must hit the object and come back to the detector.

Hence, the shortest distance will be half the distance travelled by the pulse back and forth.

Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}

Thus, the minimum distance to target = $\frac{290}{2}$ = 145 m

6 0

3 years ago

This is an example of a ____________ resource using _______.

This is an example of a ______ resource using _.