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1 month ago

A student with a mass of 66.0 kg climbs a staircase in 44.0 s. If the distance between the base and the top of the staircase is

14.0 m, how much power will the student deliver by climbing the stairs

Physics

1 answer:

serg [3.5K]1 month ago

4 0

$power = 205.8 \: watt \\ solution \\ mass = 66 \: kg \\ time = 44 \: sec \\ distance = 14 \\ now \\ power = \frac{w}{t} \\ \: \: \: \: \: \: \: \: = \frac{f \times d}{t} \\ \: \: \: \: \: = \frac{m \times g \times d}{t} \\ \: \: \: \: \: \: = \frac{66 \times 9.8 \times 14}{44} \\ \: \: \: \: = \frac{9055.2}{44} \\ \: \: \: \: \: = 205.8 \: watt \\ hope \: it \: helps$

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What will be the final temperature if a 4.00 g silver ring at 41.0◦C if it gives off 18.0 J of heat to the surroundings? The spe

kicyunya [3294]

Final temperature to determine: Given the following details, the calculations proceed as follows: Mass of the silver ring is m = 4 g, initial temperature is presented, and the heat released is Q = -18 J (indicating heat loss). The specific heat of silver is considered next to find the final temperature.

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1 month ago

James Cameron piloted a submersible craft to the bottom of the Challenger Deep, the deepest point on the ocean's floor, 11,000 m

serg [3582]

Answer:

$4.1\cdot 10^8 N$

Explanation:

To begin with, we must determine the pressure acting on the sphere, which is calculated using:

$p=p_0 + \rho g h$

where

$p_0 =1.01\cdot 10^5 Pa$ denotes the atmospheric pressure

$\rho = 1000 kg/m^3$ represents the density of the water

$g=9.8 m/s^2$ signifies the acceleration due to gravity

$h=11,000 m$ indicates the depth

By substituting these values,

$p=1.01\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(11,000 m)=1.08\cdot 10^8 Pa$

The sphere's radius is calculated as r = d/2 = 1.1 m/2 = 0.55 m

Thus, the sphere's total surface area can be expressed as

$A=4 \pi r^2 = 4 \pi (0.55 m)^2=3.8 m^2$

Consequently, the inward force acting on the sphere equals

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8 0

1 month ago

Read 2 more answers

A nerve signal is transmitted through a neuron when an excess of Na+ ions suddenly enters the axon, a long cylindrical part of t

Yuliya22 [3333]

Answer:

1.32.225 N/C, moving away from the point charge

2. 8.972*10^-12 C

3. the field is oriented away from the axon

Explanation:

The calculation for the electric field is illustrated below:

E = k*|q|/r²

Where:

E = electric field; k = 8.98755*10⁹ N*m²/C²; r = distance separating the field being measured from the point charge = 0.05 m; q = point charge

For a length of 0.100 m of the axon, the value of q is calculated as:

q = (5.6*10¹¹)*(+e)*(0.001)

+e = charge of an electron = 1.60217*10^-19 C

Therefore:

q = (5.6*10¹¹)*(1.60217*10^-19)*(0.0001) = 8.972*10^-12 C

Consequently:

E = (8.98755*10⁹)*(8.972*10^-12)/0.05² = 32.255 N/C

A positive point charge produces an electric field that radiates outward, while a negative point charge creates an electric field directed inward.

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2 months ago

A Chevrolet Corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. What

kicyunya [3294]

Response:

83.1946504051 m

Rationale:

u = Starting velocity = $60\ mph=\dfrac{60\times 1609.34}{3600}=26.82233\ m/s$

s = Distance traveled = $123\ ft=\dfrac{123}{3.281}=37.4885705578\ m$

$\theta$ = Incline = $26^{\circ}$

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-26.82233^2}{2\times 37.4885705578}\\\Rightarrow a=-9.5954230306\ m/s^2$

Friction coefficient

$\mu=-\dfrac{a}{g}\\\Rightarrow \mu=\dfrac{9.5954230306}{9.81}\\\Rightarrow \mu=0.978126710561$

$mg sin\theta - u mg cos\theta = ma\\\Rightarrow a=9.81(sin26-0.978126710561cos26)\\\Rightarrow a=-4.32382\ m/s^2$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-26.82233^2}{2\times -4.32382}\\\Rightarrow s=83.1946504051\ m$

The calculated stopping distance is 83.1946504051 m

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2 months ago

50 POINTS! A Boy throws a ball horizontally a distance of 22m downrange from the top of a tower that is 20.0m tall. What is his

Softa [3030]

At time $t$ , the ball's horizontal and vertical velocities can be represented as

$v_x=v_{xi}$

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The ball travels a distance of 22 m horizontally from the throw point, thus

$22\,\mathrm m=v_{xi}t$

With this, we determine that the time for the ball to reach the ground is

$t=\dfrac{22\,\rm m}{v_{xi}}$

When it touches down, $y=0$ and

$0=20.0\,\mathrm m-\dfrac{9.8\frac{\rm m}{\mathrm s^2}}2\left(\dfrac{22\,\rm m}{v_{xi}}\right)^2$

$\implies v_i=v_{xi}=11\dfrac{\rm m}{\rm s}$

7 0

1 month ago