All electromagnetic waves have the same speed in which of the following?

Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projec

sammy [17]

Answer:

1) The maximum jump height is reached at A. $0.337s$

2) The maximum center of mass height off of the ground is B. $1.64m$

3) The time of flight is C. $0.834s$

4) The distance of jump is B. $7.49m$

Explanation:

First of all we need to decompose velocity in its rectangular components, so

$v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s$

1) We use, $v_{fy}=v_{iy}-gt$ , as we clear it for $t$ and using the fact that $v_{fy}=0$ at max height, we obtain $t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s$

2) We can use the formula $y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ for $t=0.337s$ , so

$y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m$

3) We can use the formula $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find total time of fligth, so $0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66$ , as it is a second-grade polynomial, we find that its positive root is $t=0.834s$

4) Finally, we use $x=v_{x}t=8.05m/s(0.834s)=6.71m$ , as it has an additional displacement of $0.77m$ due the leg extension we obtain,

$x=6.71m+0.77m=7.48m$ , aprox $x=7.49m$

5 0

3 years ago

Four weightlifters (A-D) enter a competition. The mass, distance, and time of their lifts are shown in the table.

siniylev [52]

Let Pa, Pb, Pc, and Pd be the powers delivered by weightlifters A, B, C, and D, respectively.

Use this equation to determine each power value:

P = W÷Δt

P is the power, W is the work done by the weightlifter, and Δt is the elapsed time.

A) Determining Pa:

Pa = W÷Δt

The weightlifter does work to lift the weight up a certain distance. Therefore the work done is equal to the weight's gain in gravitational potential energy. The equation for gravitational PE is

PE = mgh

PE is the potential energy, m is the mass of the weight, g is the acceleration of objects due to earth's gravity, and h is the distance the weight was lifted.

We can equate W = PE = mgh, therefore we can make the following substitution:

Pa = mgh÷Δt

Given values:

m = 100.0kg

g = 9.81m/s²

h = 2.25m

Δt = 0.151s

Plug in the values and solve for Pa

Pa = 100.0×9.81×2.25÷0.151

Pa = 14600W (watt is the SI derived unit of power)

B) Determining Pb:

Let us use our new equation derived in part A to solve for Pb:

Pb = mgh÷Δt

Given values:

m = 150.0kg

g = 9.81m/s²

h = 1.76m

Δt = 0.052s

Plug in the values and solve for Pb

Pb = 150.0×9.81×1.76÷0.052

Pb = 49800W

C) Determining Pc:

Pc = mgh÷Δt

Given values:

m = 200.0kg

g = 9.81m/s²

h = 1.50m

Δt = 0.217s

Plug in the values and solve for Pc

Pc = 200.0×9.81×1.50÷0.217

Pc = 13600W

D) Determining Pd:

Pd = mgh÷Δt

Given values:

m = 250.0kg

g = 9.81m/s²

h = 1.25m

Δt = 0.206s

Plug in the values and solve for Pd

Pd = 250.0×9.81×1.25÷0.206

Pd = 14900W

Compare the following power values:

Pa = 14600W, Pb = 49800W, Pc = 13600W, Pd = 14900W

Pc is the lowest value.

Therefore, weightlifter C delivers the least power.

7 0

3 years ago

A moving roller coaster speeds up with constant acceleration for 2.3s until it reaches a velocity of 35m/s. During this time, th

nata0808 [166]

Hope this answer helps: option B

Explanation:

9 0

3 years ago

What are the principals of electric arc welding

Wewaii [24]

Principles of arc welding. Arc welding is a welding process, in which heat is generated by an electric arc struck between an electrode and the work piece. Electric arc is luminouselectrical discharge between two electrodes through ionized gas.

7 0

3 years ago

Please can some one explaine fore me What is work done

olga nikolaevna [1]

In Physics, 'work' has a very clear definition:

It's (strength of a force) times (distance through which the force acts).

'Work' has the units of Energy.

If you push against a shopping cart with 30 newtons of force, and
you keep pushing while the cart moves 4 meters, then you have
done (30 x 4) = 120 newton-meters of work = 120 "Joules".

5 0

3 years ago