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borishaifa [10]
3 years ago
10

Two identical boxes (equal dimensions and mass) are at rest, the first on a 30-degree incline and the second on a 45-degree incl

ine. Which of the following statements is FALSE?
(A) The frictional forces acting on the two boxes have the same magnitude.
(B) The normal contact force acting on the first box is larger than the normal contact force acting on the second
(C) The gravitational forces acting on the two boxes have the same magnitude.
(D) The net forces acting on the boxes have equal magnitude.
Physics
1 answer:
klemol [59]3 years ago
3 0

Answer:

A)

Explanation:

If both boxes are at rest, this means that the friction force must be equal to the component of the gravity force, trying to accelerate the box down the incline.

If the angle that the incline forms with the horizontal is θ, the component of the weight along the incline is as follows:

Fgp = m*g*sinθ

As sin θ is different for the 30-degree incline than for the 45-degree one, the components of  the weight along the incline, that must be equal to the friction forces (due to the boxes are at rest), can't be equal each other, as masses are the same.

So, the frictional forces acting on the two boxes have NOT the same magnitude, so A) is FALSE.

All the other choices are true.

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What is the amount of thermal energy needed to make 5 kg of ice at - 10 °C to
agasfer [191]

Answer:

The amount of thermal energy needed is 15167500 joules.

Explanation:

By First Law of Thermodynamics, we see that amount of thermal energy (Q), in joules, is equal to the change in internal energy. From statement we understand that change in internal energy consisting in two latent components (U_{l,ice}, U_{l,steam}), in joules, and two sensible component (U_{s,w}), in joules, that is:

Q = U_{l,ice} + U_{s, w} + U_{s,ice} + U_{l,steam} (1)

By definitions of Sensible and Latent Heat, we expanded the formula:

Q = m\cdot (h_{f,w}+h_{v,w}+c_{ice}\cdot \Delta T_{ice}+c_{w}\cdot \Delta T_{w}) (2)

Where:

m - Mass, in kilograms.

h_{f,w} - Latent heat of fussion of water, in joules per kilogram.

h_{v,w} - Latent heat of vaporization of water, in joules per kilogram.

c_{ice} - Specific heat of ice, in joules per kilogram per degree Celsius.

c_{w} - Specific heat of water, in joules per kilogram per degree Celsius.

\Delta T_{ice} - Change in temperature of ice, measured in degrees Celsius.

\Delta T_{w} - Change in temperature of water, measured in degrees Celsius.

If we know that m = 5\,kg, h_{f,w} = 3.34\times 10^{5}\,\frac{J}{kg}, h_{v,w} = 2.26\times 10^{6}\,\frac{J}{kg}, c_{ice} = 2.090\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}, c_{w} = 4.186\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}, \Delta T_{ice} = 10\,^{\circ}C and \Delta T_{w} = 100\,^{\circ}C, then the amount of thermal energy is:

Q = 15167500\,J

The amount of thermal energy needed is 15167500 joules.

7 0
3 years ago
The elementary particle called a muon is unstable and decays in about 2.20μs2.20μs , as observed in its rest frame, into an elec
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Answer:

5.865 μs

Explanation:

t₀ = Time taken to decay a muon = 2.20 μs

c = Speed of Light in vacuum = 3×10⁸ m/s

v = Velocity of muon = 0.927 c

t = Lifetime observed

Time dilation

t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{1-\frac{(0.927c)^2}{c^2}}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{1-0.927^2}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{0.140671}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{0.3750}\\\Rightarrow t=5.865\times 10^{-6}\ seconds

∴Lifetime observed for muons approaching at 0.927 the speed of light is 5.865 μs

3 0
3 years ago
Needing help explaining these!!! (2 pics attached)
Dmitriy789 [7]

#28

Fluid always flow from higher pressure to lower pressure so as we can see the figure liquid is coming out of the spray bottle so it clearly shows that the pressure outside the tube will be lower than the pressure inside the tube.

#29

Momentum is defined as product of mass and its velocity

so here we will have

P = mv

here we have

m = 38 kg

v = 2.2 m/s

so we will have

P = (38 kg)(2.2 m/s)

P = 83.6 kg m/s

6 0
3 years ago
An above ground swimming pool of 30 ft diameter and 5 ft depth is to be filled from a garden hose (smooth interior) of length 10
STALIN [3.7K]

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

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where,

v = flow speed = ?

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\rho = density of water = 1000 kg/m³

Therefore,

v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

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where,

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V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = [\frac{\pi (9.144\ m)^2}{4}][1.524\ m] = 100.1 m³

Therefore,

t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\

<u>t = 18353.5 s = 305.9 min = 5.1 hours</u>

Learn more about dynamic pressure here:

brainly.com/question/13155610?referrer=searchResults

7 0
3 years ago
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