A boy weighing 50 kg is running at 2m/s. What is its kinetic energy? * 10 points

As speed (velocity) increases, potential energy increases true or false

Lyrx [107]

Answer:

true

Explanation:

3 0

2 years ago

A 20.0 kg crate sits at rest at the bottom of a 15.0-m-long ramp that is inclined at 34.0° above horizontal. A constant horizont

ankoles [38]

Answer:

987 joules, 3.01s

Explanation:

(A)

from the attached diagram

net force, Fnet, pulling the crate up the ramp is given by

Fnet = FcosФ - WsinФ - Fr

where FcosФ is the component of horizontal force 290N resolved parallel to the plane

WsinФ = mgsinФ = component of the weight of the crate resolved parallel to the plane

Fr = constant opposing frictional force

Fnet = 290cos34⁰ - 20 × 9.8 × sin34° - 65

Fnet = 240.421 - 109.602 - 65

Fnet = 65.82N

Work done on the crate up the ramp, W, is given by

W = Fnet × d (distance up the plane)

W = 65.819 × 15

W = 987.285 joules

W = 987 joules (to 3 significant Figures)

(B)

to calculate the time of travel up the ramp

we use the equation of motion

$s = ut + \frac{1}{2}at^{2}$

where s = distance up the plane, 15m

u = Initial velocity of the crate, which is 0 for a body that is initially at rest

a = acceleration up the plane, given by

$a = \frac{Fnet}{m}$

where m = mass of the crate, 20 kg

now, $a = \frac{65.819}{20} \\a = 3.291\frac{m^{2} }{s}$

from, $s = ut + \frac{1}{2}at^{2}$

$15 = 0*t + \frac{1}{2}* 3.291 * t^{2}$

15 = 0 + 1.645 $t^{2}$

15 = 1.645 $t^{2}$

$t = \sqrt{\frac{15}{1.645} }$

$t = 3.019$

t = 3.01s (to 3 sig fig)

7 0

3 years ago

In an electric shaver, the blade moves back and forth over a distance of 2.0 mm in simple harmonic motion, with frequency 124 Hz

bagirrra123 [75]

Answer:

(a) A = 1 mm

(b) $V_{max}=0.77872 m/s$

(c) $a_{max}=606.4 m/s^{2}/tex]Explanation:Distance moved back and forth = 2 mm Frequency, f = 124 HzSo, amplitude is the half of the distance traveled back and forth. (a) So, amplitude, A = 1 mm(b) Angular frequency, ω = 2 π f = 2 x 3.14 x 124 = 778.72 rad/s The formula for the maximum speed is given by [tex]V_{max}=\omega \times A$

$V_{max}=778.72 \times 0.001$

$V_{max}=0.77872 m/s$

(c) The formula for the maximum acceleration is given by

$a_{max}=\omega ^{2}A$

$a_{max}=778.72 ^{2}\times 0.001$

[tex]a_{max}=606.4 m/s^{2}/tex]

8 0

3 years ago

For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo

pickupchik [31]

The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.

In that particular situation, you can prove it like this: 

initial velocity is Vo 
launch angle is α 

initial vertical velocity is 
Vv = Vo×sin(α) 

horizontal velocity is 
Vh = Vo×cos(α) 

total time in the air is the the time it needs to fall back to a height of 0 m, so 
d = v×t + a×t²/2 
where 
d = distance = 0 m 
v = initial vertical velocity = Vv = Vo×sin(α) 
t = time = ? 
a = acceleration by gravity = g (= -9.8 m/s²) 
so 
0 = Vo×sin(α)×t + g×t²/2 
0 = (Vo×sin(α) + g×t/2)×t 
t = 0 (obviously, the projectile is at height 0 m at time = 0s) 
or 
Vo×sin(α) + g×t/2 = 0 
t = -2×Vo×sin(α)/g 

Now look at the horizontal range. 
r = v × t 
where 
r = horizontal range = ? 
v = horizontal velocity = Vh = Vo×cos(α) 
t = time = -2×Vo×sin(α)/g 
so 
r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) 
r = -(Vo)²×sin(2α)/g 

To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. 

dr/dα = d[-(Vo)²×sin(2α)/g] / dα 
dr/dα = -(Vo)²/g × d[sin(2α)] / dα 
dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα 
dr/dα = -2 × (Vo)² × cos(2α) / g 

Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when 
cos(2α) = 0 
2α = 90° 
α = 45°

4 0

2 years ago

What is a positive impact of artificial fertilizers

dybincka [34]

The top benefit of pesticides is their effectiveness against pests that would otherwise decimate crops large and small.

5 0

3 years ago