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slava [35]
3 years ago
10

How much pressure is applied to the ground by a 78 kg man who standing on square stilts that measure 0.04 m on each edge ? Answe

r in units of Pa.
Physics
1 answer:
nadya68 [22]3 years ago
5 0

Answer:

19500Pa

Explanation:

pressure=force/area

force=78*10=780N

Pressure=780/0.04

=19500Pa

You might be interested in
A thin block of soft wood with a mass of 0.078 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is
natali 33 [55]

Answer:

Final speed of the bullet is 228.3 m/s

Explanation:

As we know that there is no external force on the system of wooden block and the bullet

so we can say momentum of the system is conserved here

so here we can say

P_i = P_f

m_1v_1 = m_1v_{1f} + m_2v_{2f}

4.67\times 10^{-3}(613) = 4.67 \times 10^{-3}v_{1f} + (0.078)(23)

so we will have

2.86 = 4.67\times 10^{-3} v_{1f} + 1.794

v_{1f} = 228.3 m/s

7 0
3 years ago
A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with incompressib
Natasha_Volkova [10]

Answer:

F = 1958.4 N

Explanation:

By volume conservation of the fluid on both sides we can say that volume of fluid displaced on the side of the car must be equal to the volume of fluid on the other side

so we have

L_1A_1 = L_2A_2

1.20(\pi 18^2) = L_2(\pi 5^2)

L_2 = 15.55 m

so the car will lift upwards by distance 1.2 m and the other side will go down by distance 15.55 m

So here the net pressure on the smaller area is given as

P = P_{atm} + \frac{12,000}{\pi (0.18)^2} + \rho g (1.2 + 15.55)

excess pressure exerted on the smaller area is given as

P_{ex} = \frac{12000}{\pi (0.18)^2} + 800(9.81)(16.75)

P_{ex} = 2.49\times 10^5 Pascal

now the force required on the other side is given as

F = P_ex (area)

F = (2.49 \times 10^5)(\pi (0.05)^2)

F = 1958.4 N

3 0
3 years ago
What is the highest energy photon that can be absorbed by a ground-state hydrogen atom without causing ionization?
lakkis [162]

Highest energy photon absorbed: 2.18\cdot 10^{-18}J

Explanation:

An atom is said to be (positively) ionised when it absorbs a photon, and as a consequence, an electron becomes energetic enough to escape the atom, leaving an excess of positive charge behind.

In order for the electron to escape, the energy of the absorbed photon must be exactly equal to the (negative) energy of the level in which the electron lies.

For an hydrogen atom, the energy levels are given by

E_n = -13.6 \frac{1}{n^2}

where this energy is measured in electronvolts, and n is the number of the energy level.

Since the energy is negative, this means that the electron which requires most energy is the one lying in the ground state (n=1). Therefore, for an electron in the ground state, the most energy that can be absorbed from the incoming photon is

E_1 = -13.6 eV

Converting into Joules, this is equal to

E_1 = 13.6 \cdot 1.6\cdot 10^{-19}=2.18\cdot 10^{-18}J

Learn more about hydrogen atom:

brainly.com/question/2757829

#LearnwithBrainly

3 0
3 years ago
One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i
kati45 [8]

Answer:

The force exerted on an electron is 7.2\times10^{-18}\ N

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance  = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}

E_{1}=1.0183\times10^{3}\ N/C

Using formula of electric field again

E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}

E_{2}=-1.064\times10^{3}\ N/C

We need to calculate the resultant electric field

Using formula of electric field

E=E_{1}+E_{2}

Put the value into the formula

E=1.0183\times10^{3}-1.064\times10^{3}

E=-0.045\times10^{3}\ N/C

We need to calculate the force exerted on an electron

Using formula of electric field

E = \dfrac{F}{q}

F=E\times q

Put the value into the formula

F=-0.045\times10^{3}\times(-1.6\times10^{-19})

F=7.2\times10^{-18}\ N

Hence, The force exerted on an electron is 7.2\times10^{-18}\ N

8 0
3 years ago
Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?
Alexandra [31]

11) 1.04\cdot 10^7 J

12) 1.04\cdot 10^7 J

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

PE=mgh

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

g=9.8 m/s^2 (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

PE=(8325)(9.8)(127)=1.04\cdot 10^7 J

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

KE_t + PE_t = KE_b + PE_b

where

KE_t is the kinetic energy at the top

PE_t is the potential energy at the top

KE_b is the kinetic energy at the bottom

PE_b is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

KE_t=0

Also, at the bottom the height is zero, so the potential energy is zero

PE_b=0

Therefore, we find:

KE_b=PE_t=1.04\cdot 10^7 J

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

KE=1.04\cdot 10^7 J

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

KE_1 + PE_1 = KE_2 + PE_2

where

KE_1 is the kinetic energy at the top of the 1st hill

PE_1 is the potential energy at the top of the 1st hill

KE_2 is the kinetic energy at the top of the 2nd hill

PE_2 is the potential energy at the top of the 2nd hill

From the previous questions, we know that

KE_1=0

and

PE_1=1.04\cdot 10^7 J

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J

So, the kinetic energy at the second hill is

KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J

And so, the speed is

v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s

4 0
2 years ago
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