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3 years ago

What is this feature?

Physics

1 answer:

Viktor [21]3 years ago

3 0

Answer: c Delta

Explanation:

A delta is a landform created by deposition of sediment that is carried by a river as the flow leaves its mouth and enters slower-moving or stagnant water. This occurs where a river enters an ocean, sea, estuary, lake, reservoir, or (more rarely) another river that cannot carry away the supplied sediment.

Hope this helps!!

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The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 54/(7 + x2 + y2), where T is measured in °C and x,

ANTONII [103]

Answer:

dt/dx = -0.373702

dt/dy = -1.121107

Explanation:

Given data

T(x, y) = 54/(7 + x² + y²)

to find out

rate of change of temperature with respect to distance

solution

we know function

T(x, y) = 54 /( 7 + x² + y²)

so derivative it x and y direction i.e

dt/dx = -54× 2x / (7 +x² + y²)² .........................1

dt/dy = -54× 2y / (7 + x² + y²)² .........................2

now put the value point (1,3) as x = 1 and y = 3 in equation 1 and 2

dt/dx = -54× 2(1) / (7 +(1)² + (3)²)²

dt/dx = -0.373702

and

dt/dy = -54× 2(3) / (7 + (1)² + (3)²)²

dt/dy = -1.121107

7 0

3 years ago

What tool should a scientist use to measure an object in milligrams

Luden [163]

A scale can be used to measure miligrams,centigrams,grams and small numbers

8 0

3 years ago

Compare and contrast four types <br> of friction

malfutka [58]

Some of these frictions depend on the Pressure, temperature of atmosphere.

Static Friction: This is the friction force when two objects in contact are not moving relative to each other. This friction is higher than kinetic friction.

Kinetic or Dynamic friction: this the friction force opposing the motion of objects, when two objects in contact are in motion relative to each other. It is less than the static friction. The two surfaces are rubbing against each other as they move.

Rolling friction: This is the friction when two objects are in contact and one object is rolling over the other - like a wheel on a road. The point of contact appears as stationary. The rolling friction is very less compared to static friction & dynamic friction.

Lubricated friction: this is the friction between two solid surfaces in contact with a layer of lubricant fluid flowing in between them. This friction is the least.

Fluid friction - viscosity : this is friction between two adjacent layers that are moving relative to each other at different speeds in a fluid. This is not high.

Internal friction: when an object is compressed and forced to deform, like in a piece of rubber, there is friction between the layers, that opposes this deformation.

Skin friction is the friction that opposes movement of a fluid across a solid surface. This is also called drag. When a coin is dropped in water, there is a friction called drag on the coin. Same is the case when a ball is thrown, a drag is experienced by the ball due to the drag of air.

6 0

3 years ago

Can some one please help!!!!!!<br> ASAP pleaseeeeeeee❗️❗️❗️❗️❗️❗️❗️❗️

Thepotemich [5.8K]

a) The launch velocity of the rocket is 5.48 m/s

b) The maximum height is 1.53 m

Explanation:

We can solve this part by applying the law of conservation of energy, by considering the kinetic energy and the elastic potential energy only, since there is no change in gravitational potential energy and no friction is involved.

The total energy when the spring is compressed is:

$E=KE_i + PE_{si}$

with

$KE_i = 0$ (initial kinetic energy is zero)

$PE_{si} = \frac{1}{2}kx^2$ is the elastic potential energy stored in the spring, with

k = 450 N/m (spring constant)

x = 0.10 m (compression of the spring)

The total energy when the spring is relased is:

$E=KE_f + PE_{sf}$

with

$KE_f = \frac{1}{2}mv^2$ (final kinetic energy), with

m = 0.15 kg (mass of the rocket)

v = velocity of launch of the rocket

$PE_{sf} = 0$ (elastic potential energy is zero when the spring is released)

Combining the two equations we get

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

And solving for v,

$v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(450)(0.10)^2}{0.15}}=5.48 m/s$

In this part instead we consider only the kinetic energy and the gravitational potential energy, since the spring is at rest so its energy is now zero.

The total energy at the launch is:

$E=KE_i + PE_{gi}$