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2 years ago

Juan whose weight is 500 N is standing on the ground. The force the ground exerts on

Physics

2 answers:

Anna71 [15]2 years ago

8 0

Answer:

more than 500 n i think the answer will

kari74 [83]2 years ago

7 0

Answer:

Equal to 500 N

Explanation:

The force the ground exerts on him is Normal force.

Since, no other vertical forces acting on him.

Therefore,

The force the ground exerts on him is 500 N.

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In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is

Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

$\theta$ = Angle = 20°

y = -20 cm

Velocity components

$u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s$

$u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s$

Acceleration components

$a_x=0$

$a_y=-9.81\ m/s^2$

$y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0$

$t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629$

Time taken is 0.26088 seconds

$x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m$

The distance the beetle travels on the ground is 0.3677181864 m

6 0

3 years ago

Explain how Copernicus concluded that stars were farther away than planets. Draw a diagram showing how this principle applies to

Rainbow [258]

<span>Copernicus decided this with more of an educated guess than anything. For example is when your standing right next to a plane it's huge Right? Well when it's flying it looks really small. He used the same reasoning for stars. Since it looks small it must be farther away.</span>

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2 years ago

Determine the gravitational potential energy, in kJ, of 3 m3 of liquid water at an elevation of 40 m above the surface of Earth.

melomori [17]

Explanation:

We will calculate the gravitational potential energy as follows.

$P.E_{1} = mgz_{1}$

$P.E_{1} = (\rho V)gz_{1}$

= $1000 kg/m^{3} \times 3 m^{3} \times 9.7 \times 40 m$

= 1164000 J

or, = 1164 kJ (as 1 kJ = 1000 J)

Now, we will calculate the change in potential energy as follows.

$\Delta P.E = mg(z_{2} - z_{1})$

= $\rho \times V \times g (z_{2} - z_{1})$

= $1000 \times 3 \times 9.7 (10 - 40)m$

= -873000 J

or, = -873 kJ

Thus, we can conclude that change in gravitational potential energy is -873 kJ.

4 0

2 years ago

Ay tawaging bayani sa makabagong panahon?

agasfer [191]

Answer:

Explanation: Bagong bayani ang turing sa ating mga Overseas Filipino. Workers (OFWs) sapagkat ang kanilang pagpapagod at pakikipagsapalaran sa ibayong dagat ay hindi lamang ... kasigurahan ang pagkukunan ni Lito sa panahon ng kanyang.

7 0

2 years ago

Write the terms involved in s=ut+1\2

cricket20 [7]

s = displacement; u = initial velocity; t = time of motion

7 0

2 years ago