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3 years ago

How much must you raise the

Physics

1 answer:

Oksana_A [137]3 years ago

3 0

Answer:

4.83 °C

Explanation:

001 = λ*ΔT

ΔT = 10^-3*10^6/207 = 4.83°C

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How does the resultant displacement change as the angle between two vectors increases from 90° to 120°?

user100 [1]

The resultant displacement between the two vectors will increase.

The resultant of the two vectors is given by parallelogram law of vectors.

The parallelogram law of vector addition states that if two vectors are represented in magnitude and direction by the adjacent sides of a parallelogram, the diagonal of the parallelogram drawn from the point of intersection of the vectors represents the resultant vector in magnitude and direction.

The resultant of these vectors, say vector A, and B, is given as;

$R^2 = a^2 + b^2 -2ab.Cos (\theta )$

When;

θ = 90°

$R^2 = a^2 + b^2$

When;

θ = 120°

$R^2 = a^2 + b^2 + ab$

Thus, the resultant displacement between the two vectors will increase.

Learn more here: brainly.com/question/20885836

8 0

3 years ago

Positive charge Q is distributed uniformly along the x-axis fromx=0 to x=a. A positive point charge q is located on the positive

dybincka [34]

Answer:

a. b- x= y

dx = -dy

b. F = $\frac{-kQqi}{r (a+r)}$

c. F = $\frac{-kQqi}{r^{2} }$

Explanation:

a. x components:

$dE = \frac{kdq}{(a+r-x^{2}) } \\$

= $\frac{kQdx}{(a(a+r-x)^2}$

Integrating and solving gives:

b- x= y

dx = -dy

b. the force is given by the equation derived from (a.):

F = $\frac{-kQqi}{r (a+r)}$

c. Given that r>>a, the expression becomes:

F = $\frac{-kQqi}{r^{2} }$

Explanation:

When the size of the charge distribution is less than the distance to the deviation point of the charge then the charge distribution would produce the same effect such as a linear charge.

6 0

3 years ago

A soccer player takes a corner kick, lofting a stationary ball 33.0° above the horizon at 15.0 m/s. If the soccer ball has a mas

Alexxandr [17]

Explanation:

It is given that,

Mass of the soccer ball, m = 0.425 kg

Speed of the ball, u = 15 m/s

Angle with horizontal, $\theta=33^{\circ}$

Time for which the player's foot is in contact with it, $\Delta t = 5.1\times 10^{-2}\ s$

Part A,

The x component of the soccer ball's change in momentum is given by :

$\Delta p_x=mv\ cos\theta$

$\Delta p_x=0.425\times 15\ cos(33)$

$p_x=5.34\ kg-m/s$

The y component of the soccer ball's change in momentum is given by :

$\Delta p_y=mv\ sin\theta$

$\Delta p_y=0.425\times 15\ sin(33)$

$p_y=3.47\ kg-m/s$

Hence, this is the required solution.

3 0

3 years ago

Any two application of gravity

Brrunno [24]

Answer:

Well the definition of an application is the act of putting to a special use or purpose so lam assuming that you want specific uses that scientists make of gravity in their work.

Well our first application has helped us to send satellites around the solar system with what Nasa calls gravity assist. Using a particular planets gravity to slingshot a satellite to another destination. Look it up.

The next application much simpler but here on Earth. There are many hydro-electric power stations in use all over the world. Water is stored at a high level and released falling 100s of metres to a turbine where it generates electricity.

Hope that helps.

Explanation:

5 0

3 years ago

Which term describes the quantity of energy transferred from a warmer object to a cooler object?

Mars2501 [29]

Answer: B

Explanation:

6 0

3 years ago