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3 years ago

Rank the work done on the charged particles from highest to lowest

Physics

1 answer:

Artist 52 [7]3 years ago

4 0

Answer:

correct order of work done is given as

2 > 3 > 4 > 1

Explanation:

When positive charge is dragged opposite to electric field then in that case work done is positive while if negative charge is dragged opposite to electric field then the work done is negative

So here we will have

positive work for charge 2, 3 and 4 while for negative charge particle 1 the work done is negative

now here we also know that work done is product of force and displacement

so here work depends on displacement of charge particle

maximum displacement is force particle 2

minimum is for particle 4

so correct order of work done is given as

2 > 3 > 4 > 1

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Volcanoes change the earth by

SOVA2 [1]

I believe the answer is b) slowly heating the surface

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3 years ago

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A wrench 0.500 m long is applied to a nut with a force of 80.0 N. Because of the cramped space, the force must be exerted upward

riadik2000 [5.3K]

Answer:

Torque, $\tau=34.6\ N.m$

Explanation:

It is given that,

Length of the wrench, l = 0.5 m

Force acting on the wrench, F = 80 N

The force is acting upward at an angle of 60.0° with respect to a line from the bolt through the end of the wrench. We need to find the torque is applied to the nut. We know that torque acting on an object is equal to the cross product of force and distance. It is given by :

$\tau=Fr\ sin\theta$

$\tau=80\times 0.5\ sin(60)$

$\tau=34.6\ N.m$

So, the torque is applied to the nut is 34.6 N.m. Hence, this is the required solution.

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3 years ago

In which of the following elevator situations would the acceleration be positive. Select TWO answers

kap26 [50]

Both options 5 and 6

Explanation:

Let us consider option 5,

In option 5 body is moving up with initial velocity lower than that of final velocity which gets accelerated. Therefore the acceleration is positive in this case.

Let us consider option 6,

In option 6 body is moving down with initial velocity lower than that of final velocity which gets accelerated. Therefore the acceleration is positive in this case.

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3 years ago

What determines the path that an object in projectile motion follows

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3 years ago

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An aircraft is in level flight at 225 km/hr through air at standard conditions. The lift coefficient at this speed is 0.45 and t

mojhsa [17]

Answer:

- the effective lift area for the aircraft is 8.30 m²

- the required engine thrust is 1275 N

- required power is 79.7 kW

Explanation:

Given the data in the question;

Speed V = 225 km/hr = 62.5 m/s

The lift coefficient CL = 0.45

drag coefficient CD = 0.065

mass = 900 kg

g = 9.81 m/s²

a) the effective lift area for the aircraft

we know that for a steady level flight, weight = lift and thrust = drag

Using the equation for the lift force

F $_L$ = C $_L$ $\frac{1}{2}$ ρV²A = W

we substitute

0.45 × $\frac{1}{2}$ × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )

1081.05 × A = 8829

A = 8829 / 1081.05

A = 8.30 m²

Therefore, the effective lift area for the aircraft is 8.30 m²

b) the required engine thrust and power to maintain level flight.

we use the expression for drag force

F $_D$ = T = C $_D$ $\frac{1}{2}$ ρV²A

we substitute

= 0.065 × $\frac{1}{2}$ × 1.21 × ( 62.5 )² × 8.30

T = 1275 N

Since drag and thrust force are the same,

Therefore, the required engine thrust is 1275 N

Power required;

P = TV

p = 1275 × 62.5

p = 79687.5 W

p = ( 79687.5 / 1000 )kW

p = 79.7 kW

Therefore, required power is 79.7 kW

8 0

3 years ago