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monitta
3 years ago
9

Please help I’m a senior trying to graduate Work as area under curve

Physics
1 answer:
Stells [14]3 years ago
8 0
I need a picture to help you
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A 5.58 kg object with a speed of 35 m/s strikes a steel plate at an angle of 45.0 degrees with the normal to the plate, and rebo
MAVERICK [17]

Answer:

magnitude of vector is 276.19 kg m/s

Explanation:

The initial momentum is vector of magnitude

5.58 \times 35 = 195.3 (kg m/s)  And driven in a coherent manner with initial vector.

same magnitude is momentum after the impact, but it is oriented perpendicularly to initial momentum vector.

So, you have 2 momentum vector of  specified magnitude perpendicular to one another.

 The contrast between such two vectors is a right angle triangle hypotenuse of 195.3 sides

magnitude of vector is  \sqrt{ 195.3^2 + 195.3^2} = 276.19

8 0
3 years ago
What are the two major topics studied by a physicist
Alexxx [7]
Math and the study of motion
6 0
3 years ago
Which two types of waves require matter in order to travel
Anton [14]

sound \: wave \:  \\ light \: wave

4 0
2 years ago
The student applies a net force of 4.5 N to the 1.5 kg textbook.
Kitty [74]

Answer:

3 m/s²

Explanation:

acceleration = Force / mass;  4.5/1.5 = 3

4 0
3 years ago
For a particular reaction, Δ H ∘ = − 93.8 kJ and Δ S ∘ = − 156.1 J/K. Assuming these values change very little with temperature,
svetoff [14.1K]

To solve this problem it is necessary to apply the concepts related to Gibbs free energy and spontaneity

At constant temperature and pressure, the change in Gibbs free energy is defined as

\Delta G = \Delta H - T\Delta S

Where,

H = Entalpy

T = Temperature

S = Entropy

When the temperature is less than that number it is negative meaning it is a spontaneous reaction. \Delta  G is also always 0 when using single element reactions. In numerical that implies \Delta G = 0

At the equation then,

\Delta G = \Delta H - T\Delta S

0 = \Delta H - T\Delta S

\Delta H = T\Delta S

T = \frac{\Delta H}{\Delta S}

T = \frac{-93.8kJ}{-156.1J/K}

T = \frac{-93.8*10^3J}{-156.1J/K}

T = 600.89K}

Therefore the temperature changes the reaction from non-spontaneous to spontaneous is 600.89K

5 0
3 years ago
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