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2 years ago

Will mark brainlest helpp!!!!!!

Physics

2 answers:

cestrela7 [59]2 years ago

7 0

initial velocity=u=0m/s
Final velocity=60km/h=v

Convert to m/s

$\\ \Large\sf\longmapsto v=60\times \dfrac{5}{18}=16.6m/s$

Time=t=5min=5×60=300s

We know

$\boxed{\Large{\sf Acceleration=\dfrac{v-u}{t}}}$

$\\ \Large\sf\longmapsto Acceleration=\dfrac{16.6-0}{300}$

$\\ \Large\sf\longmapsto Acceleration=\dfrac{16.6}{300}$

$\\ \Large\sf\longmapsto Acceleration=0.05m/s^2$

Distance=s

Using second equation of kinematics

$\boxed{\Large{\sf s=ut+\dfrac{1}{2}at^2}}$

$\\ \Large\sf\longmapsto s=0(300)+\dfrac{1}{2}(0.05)(300)^2$

$\\ \Large\sf\longmapsto s=0.05\times 150^2$

$\\ \Large\sf\longmapsto s=0.05\times 22500$

$\\ \Large\sf\longmapsto s=1125m$

Luda [366]2 years ago

6 0

Answer:

I don't know if it is correct or not.

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An archer puts a 0.30 kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m.a. Assuming

Vlad [161]

Answer:

Explanation:

Given

mass of archer $m=0.3\ kg$

Average force $F_{avg}=201\ N$

extension in arrow $x=1.3\ m$

Work done to stretch the bow with arrow

$W=F\cdot x$

$W=201\times 1.3=261.3\ m$

This work done is converted into kinetic Energy of arrow

$W=\frac{1}{2}mv^2$

where v= velocity of arrow

$261.3=\frac{1}{2}\times 0.3\times v^2$

$v=\sqrt{1742}$

$v=41.73\ m/s$

(b)if arrow is thrown vertically upward then this energy is converted to Potential energy

$W=mgh$

$261.3=0.3\times 9.8\times h$

$h=\frac{261.3}{0.3\times 9.8}$

$h=88.87\ m$

4 0

3 years ago

Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m^2 o

VARVARA [1.3K]

The rate of heat loss by radiation is equal to <u>-207.5kW</u>

Why?

To calculate the heat loss rate (or heat transfer rate) by radiation, from the given situation, we can use the following formula:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

Where,

E, is the emissivity of the body.

A, is the area of the body.

T, are the temperatures.

S, is the Stefan-Boltzmann constant, which is equal to:

$5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }$

Now, before substitute the given information, we must remember that the given formula works with absolute temperatures (Kelvin), so, we need to convert the given values of temperature from Celsius degrees to Kelvin.

We know that:

$K=Celsius+273.15$

So, converting we have:

$T_{1}=1110\°C+273.15=1383.15K\\\\T_{2}=36.2\°C+273.15=309.35K$

Therefore, substituting the given information and calculating, we have:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

$HeatLossRate=1*5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }*1m^{2} *((309.35K)^{4} -(1383.15})^{4} )\\\\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(95697.42K^{4} -3.66x10^{12}K^{4})\\ \\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(-3.66x10^{12} K^{4})=-207522W=-207.5kW$

Hence, we have that the rate of heat loss is equal to -207.5kW.

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3 years ago

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Answer:

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2 years ago

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2 years ago

Will mark brainlest helpp!!!!!!​

Will mark brainlest helpp!!!!!!