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3 years ago

How do air particles move above a candle flame

Physics

2 answers:

stiv31 [10]3 years ago

3 0

Answer: .

Explanation:

Sav [38]3 years ago

3 0

It circulates. As warm air moves up, cooler air rushes in at the bottom of the flame. Once the cooler air is heated, it then rises up (like before) and is replaced by cooler air at the base of the flame.

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Do you have to take physics to graduate high school?

Lina20 [59]

Usually, you have to take two years of physical sciences with lab. It doesn't have to be physics, it can be chemistry or biology.

6 0

3 years ago

If a force does a positive amount of work on an object, does the object’s speed increase, decrease, or remain the same? Justify

iris [78.8K]

Answer:

see that there is no dependence on speed, so the work remains constant

Explanation:

Work is defined by the expression

W = F. d

where the boldface indicates vectors, this equation can be written in scalar form

W = f d cos θ

where θ is the angle between force and displacement.

We see that there is no dependence on speed, so the work remains constant

The power is

P = W / t

P = f d / t

p = F v

we see that the power is the one that depends on the speed of the body

6 0

3 years ago

How far (in feet) could a pitcher throw a baseball on flat, level ground if he can throw it at 100mph? (Neglect wind drag and th

ehidna [41]

Answer:

R = 668.19 ft

Explanation:

given,

speed of the ball thrown by the pitcher = 100 mph

to travel maximum distance θ = 45°

distance traveled by the ball = ?

using formula

1 mph = 0.44704 m/s

100 mph = 44.704 m/s

$R = \dfrac{u^2sin2\theta}{g}$

$R = \dfrac{44.704^2sin2\times 45}{9.81}$

R = 203.71 m

1 m = 3.28 ft

R = 203.71 × 3.28

R = 668.19 ft

hence, ball will go at a distance of 668.19 ft when pitcher throw it at 100 mph.

6 0

3 years ago

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

$0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s$

So now we can find the magnitude of the initial velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s$

4 0

4 years ago

Consider an electrical transformer that has 10 loops on its primary coil and 20 loops on its secondary coil. What is the current

rusak2 [61]

Answer:

C. 2.5 A

Explanation:

Transformer: A transformer is an electromechanical device that is used to change the voltage of an alternating current.

The current and the number of loops in a transformer is related as shown below

Ns/Np = Ip/Is........................... Equation 1

Where Ns = Secondary loop, Np = primary loop, Ip = primary current, Is = secondary current.

Making Is the subject of the equation

Is = NpIp/Ns........................ Equation 2

Given: Np = 10 loops, Ns = 20 loops, Ip = 5.0 A.

Substitute into equation 2

Is = (10×5.0)/20

Is = 50/20

Is = 2.5 A.

Hence the current in the primary coil = 2.5 A.

The right option is C. 2.5 A

6 0

3 years ago