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3 years ago

Give an example of a positive slope

Physics

2 answers:

Mekhanik [1.2K]3 years ago

6 0

Answer:

uu yt? gcc5feesserhujnggbgxf

Explanation:

Aleks04 [339]3 years ago

3 0

The bikers moved along the x-axis in a positive direction, they also moved along the y-axis in a positive direction. Hope this helped! Ccredits: from internet source but not plagiarism because I changed it up a bit

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The two pucks of equal mass did not move linearly (they came to a stop) after the collision due to the conservation of linear mo

weeeeeb [17]

Compared to the pucks given, the pair of pucks will rotate at the same rate.

Answer: Option A

<u>Explanation:</u>

The law of conservation of the angular momentum expresses that when no outer torque follows upon an article, no difference in angular momentum will happen. At the point when an item is turning in a shut framework and no outside torques are applied to it, it will have no change in angular momentum.

The conservation of the angular momentum clarifies the angular quickening of an ice skater as she brings her arms and legs near the vertical rotate of revolution. In the event, that the net torque is zero, at that point angular momentum is steady or saved.

By twice the mass yet keeping the speeds unaltered, also twice the angular momentum's to the two-puck framework. Be that as it may, we likewise double the moment of inertia. Since $L=I \times \omega$ , the turning rate of the two-puck framework must stay unaltered.

4 0

3 years ago

Please HELP!!

Rudik [331]

Answer:

I think is 2.

Explanation:

(The entire range of wavelengths or frequencies of electromagnetic radiation extending from gamma rays to the longest radio waves and including visible light)

7 0

3 years ago

A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$

Therefore

$V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437$

Charge, Q is given by CV hence for the first capacitor charge will be $Q_1=C_1V$

Here, $Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C$

8 0

3 years ago

If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int

vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

$\Delta U = Q-W$

where

$\Delta U$ is the change in internal energy of the system

$Q$ is the heat absorbed by the system

$W$ is the work done by the system on the surrounding

In this problem, the work done by the system is

$W=-225 kcal=-941.4 kJ$

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

$Q=-5 \cdot 10^2 kJ=-500 kJ$

with a negative sign as well because it is released by the system.

Therefore, by using the initial equation, we find

$\Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ$

8 0

3 years ago

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

7 0

3 years ago