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3 years ago

Liana draws a diagram to show the direction of the electric force acting on a charged particle in a uniform electric field.

Physics

1 answer:

ahrayia [7]3 years ago

7 0

Answer:

changing the force to point right is the correct answer :)

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Flapping flight is very energy intensive. A wind tunnel test

steposvetlana [31]

Answer:

The metabolic power for starting flight=134.8W/kg

Explanation:

We are given that

Mass of starling, m=89 g=89/1000=0.089 kg

1 kg=1000 g

Power, P=12 W

Speed, v=11 m/s

We have to find the metabolic power for starting flight.

We know that

Metabolic power for starting flight= $\frac{P}{m}$

Using the formula

Metabolic power for starting flight= $\frac{12}{0.089}$

Metabolic power for starting flight=134.8W/kg

Hence, the metabolic power for starting flight=134.8W/kg

4 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!<br><br> Which number marks a crest?

lyudmila [28]

number 2 because the curve demstrates the crest GOOD LUCK i hope i got you the correct answer if not sorry

5 0

3 years ago

Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 40.0 s to speed up from rest t

Vinvika [58]

Answer

Time period T = 1.50 s

time t = 40 s

r = 6.2 m

Angular speed ω = 2π/T

= $\dfrac{2\pi }{1.5}$

= 4.189 rad/s

Angular acceleration α = $\dfrac{\omega}{t}$

= $\dfrac{4.189}{40}$

= 0.105 rad/s²

Tangential acceleration a = r α = 6.2 x 0.105 = 0.651 m/s²

b)The maximum speed.

v = 2πr/T

= $\dfrac{2\pi \times 6.2}{1.5}$

= 25.97 m/s

So centripetal acceleration.

a = $\dfrac{v^2}{r}$

= $\dfrac{25.97^2}{6.2}$

= 108.781 m/s^2

= 11.1 g

in combination with the gravitation acceleration.

$a_{total} = \sqrt{(11.1g)^2+g^2}$

$a_{total}= 11.145 g$

6 0

3 years ago

A car company wants to ensure its newest model can stop in less than 450 ft when traveling at 60 mph. If we assume constant dece

seraphim [82]

Answer:

The value of acceleration that accomplishes this is 8.61 ft/s² .

Explanation:

Given;

maximum distance to be traveled by the car when the brake is applied, d = 450 ft

initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s

final velocity of the car when it stops, v = 0

Apply the following kinematic equation to solve for the deceleration of the car.

v² = u² + 2as

0 = 88.02² + (2 x 450)a

-900a = 7747.5204

a = -7747.5204 / 900

a = -8.61 ft/s²

|a| = 8.61 ft/s²

Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .

4 0

2 years ago

Heat energy can be transferred from a source to a receiver by I. radiation. II. convection. III. conduction. A.

nadya68 [22]

It can be transferred by I, II, and III

Hope this helps ! :}

7 0

3 years ago

Read 2 more answers