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4 years ago

What materials in a torch are conductors or insulators?

Physics

2 answers:

Oduvanchick [21]4 years ago

7 0

The outer portion is insulator like the headlight, handheld. The battery end portions & switch assembly are conductors.

balu736 [363]4 years ago

3 0

<span>The outer portion is insulator like the headlight, handheld. The battery end portions & switch assembly are conductors.
thank you!</span>

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A cannon is fired from a castle wall at some unknown height above the ground. The cannonball leaves the cannon with speed 30.0m/

Sauron [17]

Answer:

Part a)

$t = 3.85 s$

Part b)

$h = 72.67 m$

Part C)

$v_x = 25.98 m/s$

$v_y = 0$

Part d)

In horizontal direction velocity will remain constant

$v_x = 30 cos30 = 25.98 m/s$

in vertical direction we have

$v_y = -22.77 m/s$

Explanation:

Part a)

Horizontal speed of the cannon

$v = 30.0 m/s$

angle of projection

$\theta = 30^o$

now we have

horizontal speed = $v_x = vcos30 = 30 cos30 =25.98 m/s$

vertical speed = $v_y = vsin30 = 30 sin30 = 15 m/s$

now the time taken by it to cover the distance 100 m from the wall

$x = v_x t$

$100 = 25.98 t$

$t = 3.85 s$

Part b)

Since it hits the ground in the same time

so the height of the castle is given as

$h = \frac{1}{2}gt^2$

$h = \frac{1}{2}(9.81)(3.85^2)$

$h = 72.67 m$

Part C)

At highest point of the projection

the vertical component of the velocity will become zero

so we will have

$v_x = 25.98 m/s$

$v_y = 0$

Part d)

In horizontal direction velocity will remain constant

so we have

$v_x = 30 cos30 = 25.98 m/s$

in vertical direction we have

$v_y = v_i + at$

$v_y = 15 - 9.81(3.85)$

$v_y = -22.77 m/s$

Part e)

8 0

3 years ago

Read 2 more answers

Using the picture below, what is the displacement of the triangle?

zavuch27 [327]

Answer:

i think it is 40 kilometers in the positive direction... if not im sorry

Explanation:

7 0

3 years ago

In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm a

Assoli18 [71]

Answer:

(d) 113.19 J

(e) 113.19 J

(f) 10061 N/m

Explanation:

15 cm = 0.15 m

Let g = 9.8 m/s2

$E_p = mgx = 7*9.8*0.15 = 10.29 J$

(d) By using law of energy conservation with potential energy reference being 0 at the maximum compression point. As the ball falls and come to a stop at the compression point, its potential energy is transferred to elastic energy, which is the work that the mattress does on the ball:

$E_p = E_e$

$E_e = mgh$

where h = 1.5 + 0.15 = 1.65 m is the vertical distance that it falls.

$E_e = 7*9.8*1.65 = 113.19 J$

(e) Before the compression, the potential energy of the mattress is 0. After the compression, the potential energy is 113.19J. So it has increased by 113.19J due to the potential energy transferred from the falling ball.

(f) $E_e = 113.19 = kx^2/2$

$k0.15^2/2 = 113.19$

$k = 10061 N/m$

8 0

3 years ago

Read 2 more answers

A student has to work the following problem: A block is being pulled along at constant speed on a horizontal surface a distance

brilliants [131]

Answer:

The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.

Explanation:

Since block moves with constant speed

So, frictional force

f = FCosq

Work done by friction

W = - fd

W = - fd Cos q

The answer may be greater or less than - fdSinq. It depends on the value of q which is less than, or equal to 45°.

6 0

3 years ago

A particle of mass 10 g and charge 80 mC moves through a uniform magnetic ﬁeld,in a region where the free-fall acceleration is .

Alex777 [14]

Answer:

$B=1.223\frac{T\cdot m}{s}\frac{1}{v}$

Explanation:

According to the first Newton's law, when a particle moves with constant velocity, the sum of forces on it is zero. So, we have:

$F_m=F_g$

Here $F_m=qvBsin\theta$ is the magnetic force and $F_g=mg$ is the gravitational force. The velocity of the particle is perpendicular to the magnetic field, so $\theta=90^\circ$ . Replacing and solving for B:

$qvBsin(90^\circ)=mg\\B=\frac{mg}{qv}\\B=\frac{mg}{q}\frac{1}{v}\\B=\frac{(10*10^{-3}kg)(9.8\frac{m}{s^2})}{80*10^{-3}C}\frac{1}{v}\\B=1.223\frac{T\cdot m}{s}\frac{1}{v}$

4 0

3 years ago