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2 years ago

Which type(s) of electromagnetic radiation do human bodies emit? Which type(s) can our senses detect?

1 answer:

6 0

Our bodies emit heat, and nerve endings in our skin can detect it.
Our eyes can detect visible light, but our bodies don't emit that.

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A 24.7-g bullet is fired from a rifle. It takes 2.73 × 10-3 s for the bullet to travel the length of the barrel, and it exits th

Gala2k [10]

Answer:

$F_a_v_g=7093333.33N*s$

Explanation:

The impulse or average force in classical mechanics is the variation in the linear momentum that a physical object experiences in a closed system. It is defined by the following equation:

$F_a_v_g=m*\frac{\Delta v}{\Delta t}=m*\frac{v_2-v_1}{t_2-t_1}$

Where:

$m=mass\hspace{3}of\hspace{3}the\hspace{3}object$

$v_2=final\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}at\hspace{3}the\hspace{3}end\hspace{3}of\hspace{3}the\hspace{3}time\hspace{3}interval$

$v_1=initial\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}when\hspace{3}the\hspace{3}time\hspace{3}interval\hspace{3}begins.$

$t_2=final\hspace{3}time$

$t_1=initial\hspace{3}time$

Asumming v1=0 and t1=0:

$F_a_v_g=m* \frac{v_2}{t_2} =(24.7)*\frac{784}{2.73*10^{*3} } =7093333.333N*s$

8 0

3 years ago

There is a force between two charges; if the distance between the two charges is doubled, by what factor does the force between

Luba_88 [7]

It will double it and yes it will change them.

4 0

3 years ago

Read 2 more answers

A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi

nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

$W = F * d$

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

$d=h*N$ , for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

$V_f^2-V_i^2 = 2a\Delta X$

Where,

$V_f =$ Final velocity

$V_i =$ Initial Velocity

a = Acceleration

$\Delta X =$ Displacement

PART A) For the particular case of work we know then that,

$W = F*d$

$W = m*g*(h*N)$

$W = 50*9.8*(0.3*30)$

$W = 4.41kJ$

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

$V_f^2-V_i^2 = 2a\Delta X$

Here,

$V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow$ 3 steps in one second

$v_f = 0$

Replacing,

$V_f^2-V_i^2 = 2a\Delta X$

$0-0.9^2=2a(30*0.3)$

Re-arrange for a,

$a = -\frac{0.9^2}{2*30*0.3}$

$a = -45*10^{-3}m/s^2$

At this point we can calculate the time, which is,

$t = \frac{\Delta V}{a}$

$t = \frac{0-0.9}{-45*10^{-3}}$

$t = 20s$

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

6 0

3 years ago

A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho

fiasKO [112]

Answer:

$v_f=8.17\frac{m}{s}$

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

$W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m$

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

$W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$