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2 years ago

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You are on an elevator that is plummeting toward the ground at 30 m/s and you jump up right before it hits the ground. you jump

upward with a velocity of 5 m/s. at what speed do you hit the ground?

1 answer:

sashaice [31]2 years ago

8 0

<span>Elevator speed = 30 m/s my jump speed = 5 m/s Since the Elevator and Jumping action are of opposite direction, the speed should be subtracted to arrive at the right speed. Actual speed = (Elevator speed - My Jump Speed) = ( 30 - 5 ) m/s = 25 m/s So at the speed of 25 m/s I would hit the ground.</span>

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The potential (relative to infinity) at the midpoint of a square is 3.0 V when a point charge of +Q is located at one of the cor

Sveta_85 [38]

Answer:

d) 12 V

Explanation:

Due to the symmetry of the problem, the potential (relative to infinity) at the midpoint of the square, is the same for all charges, provided they be of the same magnitude and sign, and be located at one of the corners of the square.

We can apply the superposition principle (as the potential is linear with the charge) and calculating the total potential due to the 4 charges, just adding the potential due to any of them:

V = V(Q₁) + V(Q₂) +V(Q₃) + V(Q₄) = 4* 3.0 V = 12. 0 V

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2 years ago

Батискаф витримує тиск 60 МПа. Чи можна провести дослідження

Elanso [62]

1) Yes

2) $6.34\cdot 10^9 N$

Explanation:

1)

To solve this part, we have to calculate the pressure at the depth of the batyscaphe, and compare it with the maximum pressure that it can withstand.

The pressure exerted by a column of fluid of height h is:

$p=p_0+\rho g h$

where

$p_0 = 101,300 Pa$ is the atmospheric pressure

$\rho$ is the fluid density

$g=10 m/s^2$ is the acceleration due to gravity

h is the height of the column of fluid

Here we have:

$\rho=1030 kg/m^3$ is the sea water density

h = 5440 m is the depth at which the bathyscaphe is located

Therefore, the pressure on it is

$p=101,300+(1030)(10)(5440)=56.1\cdot 10^6 Pa = 56.1 MPa$

Since the maximum pressure it can withstand is 60 MPa, then yes, the bathyscaphe can withstand it.

2)

Here we want to find the force exerted on the bathyscaphe.

The relationship between force and pressure on a surface is:

$p=\frac{F}{A}$

where

p is hte pressure

F is the force

A is the area of the surface

Here we have:

$p=56.1\cdot 10^6 Pa$ is the pressure exerted

The bathyscaphe has a spherical surface of radius

r = 3 m

So its surface is:

$A=4\pi r^2$

Therefore, we can find the force exerted on it by re-arranging the previous equation:

$F=pA=4\pi pr^2 = 4\pi (56.1\cdot 10^6)(3)^2=6.34\cdot 10^9 N$

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2 years ago

A ray of light crosses a boundary between two transparent materials. The medium the ray enters has a larger index of refraction.

Margarita [4]

Answer:

True, True, False, False, False, False.

Explanation:

The refraction index of a material is given by the formula n=c/v, where c is the speed of light in vacuum and v the speed of light in the material. If a ray of light crosses a boundary between two transparent materials and the medium the ray enters has a larger index of refraction it means that in this new medium the speed of light is smaller than on the other one, and then its wavelength is also reduced since f must remain the same (and $\lambda=v/f$ ), otherwise there is a discontinuity on number of vibrations per second, which cannot happen. So we know that:

1) The wavelength of the light decreases as it enters into the medium with the greater index of refraction. True.

2) The frequency of the light remains constant as it transitions between materials. True.

3) The speed of the light remains constant as it transitions between materials. False.

4) The speed of the light increases as it enters the medium with the greater index of refraction. False.

5) The frequency of the light decreases as it enters into the medium with the greater index of refraction. False.

6) The wavelength of the light remains constant as it transitions between materials. False.

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2 years ago

16. A 7500 kg 18-wheeler traveling at 20 m/s exits onto the runaway truck ramp on the freeway.

miskamm [114]

Answer:

<em>765,000 Joule</em>

Explanation:

<u>Principle of Conservation of Energy </u>

The total energy in an isolated system cannot be created or destroyed, but transformed. Moving objects have kinetic energy, objects placed in some height above a reference level have gravitational potential energy. When they change their motion variables, one energy converts into the other, but if the numbers don't fit, we know there was some other type of energy acting into the system. The most common reason for energy 'losses' is the thermal energy, produced when objects move in rough surfaces or take friction from the air.

The 7,500 kg truck is originally traveling at 20 m/s to a certain height we'll set to 0. Thus, its total energy is

$\displaystyle E_1=\frac{mv^2}{2}$

$\displaystyle E_1=\frac{7,500\ 20^2}{2}$

$E_1=1,500,000\ Joule$

When it comes to a stop, its speed is 0 and its height is 10 m higher than before. It means all the kinetic energy was transformed into other types of energy. The gravitational potential energy is

$U=mgh=(7,500)(9.8)(10)=735,000\ Joule$

Since this number is not equal to the previous value of the energy, the difference is due to thermal energy dissipated by friction

$E_t=1,500,000\ Joule-735,000\ Joule=765,000\ Joule$

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2 years ago

When a 75.0-kg man slowly adds his weight to a vertical spring attached to the ceiling, he reaches equilibrium when the spring i

Sergio039 [100]

Answer:

1) $k=11.319kN/m$

2)displacement $=13.02cm$

3) $k_{eq}=5.65kN/m$

Explanation:

At equilibrium position the weight of the man should be balanced by force in the spring

thus we have at equilibrium

$kx=mg\\\\k=\frac{mg}{x}$

Applying values we get

$k=\frac{75\times 9.81}{0.065}\\\\k=11.319kN/m$

2)

When we add another identical spring we get an equivalent spring with spring constant as

$\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}$

Applying values we get

$\frac{1}{k_{eq}}=\frac{1}{11.319}+\frac{1}{11.319}\\\\k_{eq}=5.65kN/m$

Thus at equilibrium we have

$x_{2}k_{eq}=mg\\\\x_{2}=\frac{mg}{k_{eq}}\\\\x_{2}=\frac{75\times 9.81}{5.65}\times 10^{-3}=13.02cm$

3) Equivalent spring constant will be as calculated earlier $k_{eq}=5.65kN/m$

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2 years ago