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3 years ago

What is the difference between thrust and applied force?

Physics

1 answer:

Tamiku [17]3 years ago

7 0

Answer:

Thrust acts on the accelerated object in the direction opposite to the applied force hence it accelerates the object in the direction opposite to the applied force. ... Its magnitude is equal to that of applied force. It always increases the velocity of the object.

Explanation:

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Andrew gathered several different species of seed plants. Which three characteristics do all the plants most likely share? have

BigorU [14]

Answer:

Explanation:

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2 years ago

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A 22-turn circular coil of wire has diameter 1.02 m. It is placed with its axis along the direction of the Earth's magnetic fiel

ollegr [7]

Answer:

ξ = 0.00845020162 V or 8.4 mV

Explanation:

Magnetic flux measures the total magnetic field that passes through a known area. Magnetic flux describe the effect of magnetic field in a given area. Mathematically,

magnetic flux (Ф) = BA cos ∅

where

A = test area

B = magnetic field

before the flip

Ф = Bπr²N

N = number of turn

magnitude of induced emf = N |ΔФ/Δt|

ξ = 2Nπr²B/dt

ξ = 2 × 22 × π × (1.02/2)² × 0.000047/0.2

ξ = 44 × π × 0.51² × 0.000047/0.2

ξ = 44 × π × 0.2601 × 0.000047/0.2

ξ = 0.0005378868 × 3.142/0.2

ξ = 0.00169004032/0.2

ξ = 0.00845020162 V or 8.4 mV

8 0

3 years ago

S To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volum

erastovalidia [21]

To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volume V the ratio of the sphere will be $\frac{4.83598}{c^{\frac{1}{3} } }$ .

We know that the surface area and volume of the sphere is given by:

$A=4 \pi r^{2}\\V=\frac{4}{3} \pi r^{3}$

Therefore, the ratio between the surface area and the volume for the sphere will be:

$\frac{A}{V}=\frac{4 \pi r^{2}\\}{\frac{4}{3} \pi r^{3}}=\frac{3}{r}$

Equating the volume to the constant c, we will find the value of $r$ .

$V=c=\frac{4}{3} \pi r^{3}\\r= (\frac{3c}{4\pi} )^{\frac{1}{3} }$

Substituting the value of r in the ration between surface area and volume, we get:

$\frac{A}{V}=\frac{3}{ (\frac{3c}{4\pi} )^{\frac{1}{3} }}$

Calculating the constants, we get:

$\frac{4.83598}{c^{\frac{1}{3} } }$

Hence, the ration between surface area and volume is $\frac{4.83598}{c^{\frac{1}{3} } }$

To learn more about surface area and volume of sphere, refer to:

brainly.com/question/4387241

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1 year ago

A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch

Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

$x=V_{o}cos \theta t$ ----------------------(1)

$y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}$ ---------(2)

$V=V_{o}-gt$ ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

$V_{o}$ is the golf ball's initial velocity

$\theta=0\°$ is the angle (it was a horizontal shot)

t is the time

y is the final height of the ball

$y_{o}$ is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

$t=\sqrt{\frac{2 y_{o}}{g}}$

$t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}$

$t=1.597$ s

Substituting (6) in (1):

$67.1 =V_{o} cos(0\°) 1.597$ -------------------(4)

Step 2: Finding $V_{o}$ :

From equation(4)

$67.1 =V_{o}(1) 1.597$

$V_0 = \frac{6.71}{1.597}$

$V_{o}=42.01$ m/s (8)

Substituting $V_{o}$ in (3):

$V=42.01 -(9.8)(1.597)$

v =42 .01 - 15.3566

V=26.359 m/s

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3 years ago

In this reaction diagram which part represents the doffrence in energy between the reactants and the products?

Annette [7]

Answer:

The correct answer is - option C. G.

Explanation:

In this reaction diagram, there is a representation of the reaction profile. The reaction profile shows the change that takes place during a reaction in the energy of reactants or substrate and products. In this profile, activation energy looks like a hump in the line, and the minimum energy required to initiate the reaction.

The overall energy of the reaction, including or excluding activation energy depends on the nature of the reaction if it is exothermic or endothermic. and products are represented by the G which shows the difference between the energy of the reactants and products.

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3 years ago

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