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3 years ago

I’ll mark brainliest just answer ASAP

Physics

1 answer:

True [87]3 years ago

6 0

C.) During a Spring Tide

This is because there is a greater difference in the high and low tides, causing the water to flow longer through the turbine

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You are on a rollercoaster going around a curve. Your speed is a constant 40 miles per hour. Are you accelerating?

vodomira [7]

No, you are at a constant rate which means that you are always at 40mph

8 0

4 years ago

What happens when an object is moved against gravity, such as rolling a toy car up a ramp?

BlackZzzverrR [31]

Answer:

it goes up until we help it to but the moment we stop support it gets affected by gravity and goes back

Explanation:

3 0

3 years ago

A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part

mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

$v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t$

$v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}$

At <em>t</em> = 5.0 s, the particle has velocity

$v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}$

$v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}$

The speed at this time is the magnitude of the velocity :

$\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}$

The direction of motion at this time is the angle $\theta$ that the velocity vector makes with the positive <em>x</em>-axis, such that

$\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}$

4 0

3 years ago

A planet of mass m moves around the Sun of mass M in an elliptical orbit. The maximum and minimum distance of the planet from th

zzz [600]

Answer:

the relation between the time period of the planet is

T = 2π √[( r1 + r2 )³ / 8GM ]

Explanation:

Given the data i the question;

mass of sun = M

minimum and maximum distance = r1 and r2 respectively

Now, using Kepler's third law,

" the square of period T of any planet is proportional to the cube of average distance "

T² ∝ R³

average distance a = ( r1 + r2 ) / 2

we know that

T² = 4π²a³ / GM

T² = 4π² [( ( r1 + r2 ) / 2 )³ / GM ]

T² = 4π² [( ( r1 + r2 )³ / 8 ) / GM ]

T² = 4π² [( r1 + r2 )³ / 8GM ]

T = √[ 4π² [( r1 + r2 )³ / 8GM ] ]

T = 2π √[( r1 + r2 )³ / 8GM ]

Therefore, the relation between the time period of the planet is

T = 2π √[( r1 + r2 )³ / 8GM ]

5 0

3 years ago

Explain thermodynamics?

Maurinko [17]

Explanation:

the branch of physical science that deals with the relations between heat and other forms of energy (such as mechanical, electrical, or chemical energy), and, by extension, of the relationships between all forms of energy

7 0

4 years ago

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