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3 years ago

Look at the blue line in the graph.

Physics

1 answer:

Tamiku [17]3 years ago

4 0

Answer:

it is uneven

Explanation:

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A student weighing 700 N climbs at constant speed to the top of an 8 m vertical rope in 10 s. The average power expended by the

nasty-shy [4]

Answer:

The average power the student expended to overcome gravity is 560W, Watts is a units of work it is Joules /time or kg* $m^{2}$ / $s^{3}$

Explanation:

Weight = 700N

$F= m*g*h \\m*g= 700N\\F= 700N*h\\F=700N*8m\\F= 5600 N*m \\F=5600 J$

The power is the work in (Joules) or (N*m) in a determinate time (s) to get Watts (W) units for work

$Work= \frac{5600 J}{10s} = \frac{5600 \frac{kg*m^{2} }{s^{2} } }{10 s} \\Work= 560\frac{kg*m^{2} }{s^{3} } \\Work =560 W$

3 0

3 years ago

if a runners power is 400 watts as she runs, how much chemical energy does she convert into other forms in 10 minutes

AnnZ [28]

Answer:

Energy converted = $240000\,Joules = 240\, kJoules$

Explanation:

Recall that Power is the rate at which energy is transferred therefore defined by the mathematical formula: $Power\,=\,\frac{Energy\,transferred}{time}$

Since the information on the power of the runner is given, as well as the time the energy conversion takes place, we can then use this equation to find how much energy is been converted. Notice that we just need to change the given time *10 minutes) into the appropriate units (seconds)to get the answer in SI units of energy (Joules). The conversion of 10 minutes into seconds is done by multiplying : 10 minutes * 60 seconds/minute = 600 seconds.

We use this then to find the energy converted by the runner:

$Power\,=\,\frac{Energy\,transferred}{time}\\400 \,W = \frac{E}{600\,sec} \\400 \,W * 600\,sec=E\\E=240000\,Joules = 240\, kJoules$

3 0

3 years ago

ln the constellation of Orion, you can see a group of stars and other objects that appear in the shape of a sword. Ln the middle

bagirrra123 [75]

Answer:

Nebula

Explanation:

Given that in the constellation of Orion, you can see a group of stars and other objects that appear in the shape of a sword. Ln the middle of the sword, a bright "fuzzy star" appears. Astronomers looking at this object through telescopes refer to it as a "stellar nursery." Another name for this object is called NEBULA.

The space where new stars are forming anew is known as nebulae

7 0

4 years ago

A 82-kg fisherman in a 112-kg boat throws a package of mass m = 15 kg horizontally toward the right with a speed of vi = 4.8 m/s

abruzzese [7]

Answer:

0.37 m/s to the left

Explanation:

Momentum is conserved. Initial momentum = final momentum.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

Initially, both the fisherman/boat and the package are at rest.

0 = m₁ v₁ + m₂ v₂

Plugging in values and solving:

0 = (82 kg + 112 kg) v + (15 kg) (4.8 m/s)

v = -0.37 m/s

The boat's velocity is 0.37 m/s to the left.

8 0

3 years ago

A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS

Flura [38]

Answer:

(a) the electrical power generated for still summer day is 1013.032 W

(b)the electrical power generated for a breezy winter day is 1270.763 W

Explanation:

Given;

Area of panel = 2 m × 4 m, = 8m²

solar flux GS = 700 W/m²

absorptivity of the panel, αS = 0.83

efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp

panel emissivity , ε = 0.90

Apply energy balance equation to determine he electrical power generated;

transferred energy + generated energy = 0

(radiation + convection) + generated energy = 0

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0$

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s$

(a) the electrical power generated for still summer day

$T_s = T_{\infty} = 35 ^oC = 308 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W$

(b)the electrical power generated for a breezy winter day

$T_s = T_{\infty} = -15 ^oC = 258 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W$

3 0

3 years ago