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3 years ago

Define velocity is the question

Physics

2 answers:

kherson [118]3 years ago

5 0

$\huge\boxed{ \sf\purple{αɳʂɯҽɾ \huge\redღ︎︎}}$

$\sf{Velocity \: is \: the \: speed \: of \: something \: in \: a \: given \: direction.}$

Oxana [17]3 years ago

4 0

Answer:

The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of an object's speed and direction of motion (e.g. 60 km/h to the north).

Explanation:

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What does the principle of conservation of matter say

Vikentia [17]

Answer:

it says that matter cannot be created or destroyed so the number of atoms is the same after chemical reactions

8 0

4 years ago

A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping. Even without knowing how bi

MariettaO [177]

Answer:

3801.13 N

Explanation:

Pressure exerted on a surface is equivalent to applied force divided by the cross sectional area. Then, the applied force will be equal to the product of the pressure exerted and the cross sectional area.

Where given:

Atmospheric pressure (P1) = 1.013*10^5 Pa

T1 = 20+273.15 = 293.15 K

P2 = ?

T2 = 120+273.15 = 393.15 K

Using the gas equation: P1/T1 = P2/T2

Therefore, P2 = P1*T2/T1 = 1.013*10^5 *393.15/293.15 = 13.6*10^4 Pa

The net pressure = P2 - P1 = 13.6*10^4 - 1.013*10^5 = 34.6 kPa

The net force $F_{120} = net pressure * area$

Area = 0.11 m^2

Thus:

The net force $F_{120} = 34555.7*0.11$ = 3801.13 N

6 0

3 years ago

HELP NOW PLEASE ANSWER THE QUESTION THE PICTURE IS ATTACHED BELOW

Andrej [43]

Answer:

Explanation:

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7 0

3 years ago

A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.

Alexxandr [17]

Answer:

Kinetic energy of the projectile at the vertex of the trajectory: $900\; {\rm J}$ .

Work done when firing this projectile: $2500\; {\rm J}$ .

Explanation:

Since the drag on this projectile is negligible, the horizontal velocity $v_{x}$ of this projectile would stay the same (at $30\; {\rm m\cdot s^{-1}}$ ) throughout the flight.

The vertical velocity $v_{y}$ of this projectile would be $0\; {\rm m\cdot s^{-1}}$ at the vertex (highest point) of its trajectory. (Otherwise, if $v_{y} > 0$ , this projectile would continue moving up and reach an even higher point. If $v_{y} < 0$ , the projectile would be moving downwards, meaning that its previous location was higher than the current one.)

Overall, the velocity of this projectile would be $v = 30\; {\rm m\cdot s^{-1}}\!$ when it is at the top of the trajectory. The kinetic energy $\text{KE}$ of this projectile (mass $m = 2.0\; {\rm kg}$ ) at the vertex of its trajectory would be:

$\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (30\; {\rm m\cdot s^{-1}})^{2} \\ &= 900\; {\rm J} \end{aligned}$ .

Apply the Pythagorean Theorem to find the initial speed of this projectile:

$\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \left(\sqrt{900 + 1600}\right)\; {\rm m\cdot s^{-1}} \\ &= 50\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Hence, the initial kinetic energy $\text{KE}$ of this projectile would be:

$\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (50\; {\rm m\cdot s^{-1}})^{2} \\ &=2500\; {\rm J} \end{aligned}$ .

All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be $2500\; {\rm J}$ .

7 0

2 years ago

A uniform electric field exists everywhere in the x, y plane. This electric field has a magnitude of 4500 N/C and is directed in

Anton [14]

Answer with Explanation:

We are given that

$E=4500 N/C$