Find the kinetic energy of a 0.1 kg toy truck moving at a speed of 1.1 m/s

an object weighing 15 newtons is lifted from the ground to a height of 0.22 meter what is the increase in the object's gravitati

kicyunya [14]

GPE= weight•height= 15 N• 0.22meter= 3.3 Joules
I hope this helps ~~Charlotte~~

5 0

3 years ago

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A surfer is moving at a constant velocity of 5.2 m/s north relative to a wave which is pushing him west at a constant velocity o

Vika [28.1K]

The speed and distances are directly proportional. Use ratios to solve for vertical y-distance. The ratio of x-distance west to y-distance north equals the x-velocity to y-velocity.

x/y = vx/vy
41/y = 8.6/5.2
41/y = 1.65
41/1.65 = y
24.8 m = y

5 0

2 years ago

Determine the wavelength of incident electromagnetic radiation required to cause an electron transition from the n = 5 to the n

ollegr [7]

The correct answer is: wavelength = 4562 nm

Explanation:

Rydberg's formula is given as:
$\frac{1}{\lambda} = R[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ]$ --- (1)

Where
R = Rydberg's constant = 1.096 * 10^7 per meter
$n_1$ = 5
$n_2$ = 7

λ = Wavelength

Plug in the values in (1):

(1)=> $\frac{1}{\lambda} = (1.096 * 10^7)[ \frac{1}{5^2} - \frac{1}{7^2} ]$

$\frac{1}{\lambda} = (1.096 * 10^7)[ 0.04 - 0.020 ] \\ \lambda = \frac{1}{(1.096 * 10^7)[0.020 ]} \\ \lambda = 4562 nm$

8 0

3 years ago

A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-mhigh rise and angle is 35 degrees.Find her final speed at th

Shalnov [3]

Wt. = Fg = m*g = 60kg * 9.8N/kg=588 N.=
Wt. of skier. 

Fp=588*sin35 = 337 N.=Force parallel to 
incline. 
Fv = 588*cos35 = 482 N. = Force perpendicular to incline. 

Fk = u*Fv = 0.08 * 482 = 38.5 N. = Force 
of kinetic friction. 
d =h/sinA = 2.5/sin35 = 4.36 m. 

Ek + Ep = Ekmax - Fk*d 
Ek = Ekmax-Ep-Fk*d 
Ek=0.5*60*12^2-588*2.5-38.5*4.36=2682 J. 
Ek = 0.5m*V^2 = 2682 J. 
30*V^2 = 2682 
V^2 = 89.4 
V = 9.5 m/s = Final velocity.

4 0

3 years ago

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330 grams of boiling water (temperature 100°C, specific heat capacity 4.2 J/K/gram) are poured into an aluminum pan whose mass i

Alexus [3.1K]

Answer:

T = 74°C

Explanation:

Given Mw = mass of water = 330g, Ma = mass of aluminium = 840g

Cw = 4.2gJ/g°C = specific heat capacity of water and Ca = 0.9J/g°C = specific heat capacity of aluminium

Initial temperature of water = 100°C.

Initial temperature of aluminium = 29°C

When the boiling water is poured into the aluminum pan, heat is exchanged and after a short time the water and aluminum pan both come to thermal equilibrium at a common temperature T.

Heat lost by water equal to the heat gained by aluminium pan.

Mw × Cw×(100 –T) = Ma × Ca × (T–29)

330×4.2×(100– T) = 890×0.9×(T–29)

1386(100 – T) = 801(T –29)

1386/801(100 – T) = T – 29

1.73(100 – T) = T – 29

173 –1.73T = T –29

173+29 = T + 1.73T

202 = 2.73T

T = 202/2.73

T = 74°C

6 0

3 years ago