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geniusboy [140]
4 years ago
12

Two objects of mass m move in opposite directions toward each other. The green object moves at velocity v, and the blue object m

oves at velocity 2v. Which diagram correctly shows the direction and velocity for each of the objects after an elastic collision?
Physics
2 answers:
Kobotan [32]4 years ago
7 0

The answer to this one was C, or the third option.

NeX [460]4 years ago
5 0

. The velocity of a mass attached to a spring is given by v = (1.5 cm/s) sin(ωt + π/2), ..... Which of the following is the motion of objects moving in two dimensions

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A car with a mass of 2200 kg is travelling at a rate of 55 m's. What is the cars momentum? *
Varvara68 [4.7K]
  • Mass=2200kg
  • Velocity=55m/s

\\ \sf\bull\dashrightarrow Momentum=Mass\times Velocity

\\ \sf\bull\dashrightarrow Momentum=2200(55)

\\ \sf\bull\dashrightarrow Momentum=121000kgm/s

7 0
3 years ago
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What is the weight on the surface of Earth of an object of mass 2.00 kilograms?
Anni [7]

Answer:

D

Explanation:

19.6 newtons

A 2.00-kilogram object weighs 19.6 newtons on Earth.

6 0
3 years ago
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A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axles. A light cord wrapped around
Alex_Xolod [135]

Answer:

\alpha =\frac{m*g*R}{I-m*R^2}

a = \frac{m*g*R^2}{I-m*R^2}

T=\frac{I*m*g}{I-m*R^2}

Explanation:

By analyzing the torque on the wheel we get:

T*R=I*\alpha    Solving for T:   T=I/R*\alpha

On the object:

T-m*g = -m*a    Replacing our previous value for T:

I/R*\alpha-m*g = -m*a

The relation between angular and linear acceleration is:

a=\alpha*R

So,

I/R*\alpha-m*g = -m*\alpha*R

Solving for α:

\alpha =\frac{R*m*g}{I+m*R^2}

The linear acceleration will be:

a =\frac{R^2*m*g}{I+m*R^2}

And finally, the tension will be:

T =\frac{I*m*g}{I+m*R^2}

These are the values of all the variables: α, a, T

8 0
3 years ago
The main energy source during exercise of low to moderate intensity is
zhannawk [14.2K]

Answer:

THUS,THE MAJOR SOURCES OF ENERGY DURING EXERCISE ARE CARBOHYDRATES AND FATS.

Explanation:

6 0
3 years ago
A house has well-insulated walls. It contains a volume of 105 m3 of air at 305 K.
REY [17]

Answer: 85.46\ kJ

Explanation:

Given

Volume of air V=105\ m^3

Temperature of air T=305\ K

Increase in temperature \Delta T=0.7^{\circ}C

Specific heat for diatomic gas is C_p=\dfrac{7R}{2}

Energy required to increase the temperature is

\Rightarrow Q=nC_pdT\\\\\Rightarrow Q=n\times \dfrac{7R}{2}\times \Delta T\\\\\Rightarrow Q=\dfrac{7}{2}nR\Delta T\\\\\Rightarrow Q=\dfrac{7}{2}\times \dfrac{PV}{T}\times \Delta T\quad [\text{using PV=nRT}]

Insert the values

\Rightarrow Q=\dfrac{7}{2}\times \dfrac{1.01325\times 10^5\times 105}{305}\times 0.7\\ \text{Assuming air pressure to be atmospheric P=}1.01325\times 10^5\ N/m^2\\\\\Rightarrow Q=0.8546\times 10^5\\\Rightarrow Q=85.46\ kJ

6 0
3 years ago
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