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2 years ago

! ' 30 !! !

Physics

1 answer:

ELEN [110]2 years ago

7 0

Answer:

Hey

Unlike visible light, X-rays can go through you without getting absorbed. But some of it gets absorbed but most of it passes through, this tiny bit that does get absorbed shows you the image of the baby.

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A physicist hangs a 150-g object on a spring whose spring constant is a value of 13.22 Newtons/meter and has a spring force of 2

Nataly_w [17]

Answer:

so the answer is this because the answer is that

Explanation:

and the reason why the answer is this and that is because the answer is that

4 0

2 years ago

A beam of light has a wavelength of 4.5 x10^-7 meter in a vacuum. the frequency of this light is

valkas [14]

The basic relationship between frequency and wavelength for light (which is an electromagnetic wave) is
$c= f \lambda$
where c is the speed of light, f the frequency and $\lambda$ the wavelength of the wave.
Using $\lambda=4.5 \cdot 10^{-7} m$ and $c=3 \cdot 10^8 m/s$ , we can find the value of the frequency:
$f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{4.5 \cdot 10^{-7} m}=6.7 \cdot 10^{14} Hz$

3 0

3 years ago

An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and stic

Vika [28.1K]

Answer:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

Given that,

Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

$R =u\cos\theta t$

Put the value into the formula

$\dfrac{230}{6} = u\cos\theta$

$u\cos\theta=38.33$ .....(I)

We need to calculate the height

Using vertical component

$H=u\sin\theta t-\dfrac{1}{2}gt^2$

Put the value in the equation

$16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2$

$u\sin\theta=\dfrac{16+9.8\times18}{6}$

$u\sin\theta=32.06$ .....(II)

Dividing equation (II) and (I)

$\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}$

$\tan\theta=0.8364$

$\theta=\tan^{-1}0.8364$

$\theta=39.90^{\circ}$

(a). We need to calculate the initial speed

Using equation (I)

$u\cos\theta\times t=38.33$

Put the value into the formula

$u =\dfrac{230}{6\times\cos39.90}$

$u=49.96\ m/s$

(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

5 0

3 years ago

A jet aircraft is traveling at 262 m/s in hor-

NeTakaya

Solution :

Speed of the air craft, $S_a$ = 262 m/s

Fuel burns at the rate of, $S_b$ = 3.92 kg/s

Rate at which the engine takes in air, $S_{air}$ = 85.9 kg/s

Speed of the exhaust gas that are ejected relative to the aircraft, $S_{exh}$ =921 m/s

Therefore, the upward thrust of the jet engine is given by

$F=S_{air}(S_{exh}-S_a)+(S_b \times S_{exh})$

F = 85.9(921 - 262) + (3.92 x 921)

= 4862635.79 + 3610.32

= $4.8 \times 10^6 \ N$

Therefore thrust of the jet engine is $4.8 \times 10^6 \ N$ .

3 0

2 years ago

A circular loop of wire with radius 2.00 cm and resistance 0.600 Ω is in a region of a spatially uniform magnetic field B⃗ that

Anika [276]

Answer:

Incomplete questions

Let assume we are asked to find

Calculate the induced emf in the coil at any time, let say t=2

And induced current

Explanation:

Flux is given as

Φ=NAB

Where

N is number of turn, N=1

A=area=πr²

Since r=2cm=0.02

A=π(0.02)²=0.001257m²

B=magnetic field

B(t)=Bo•e−t/τ,

Where Bo=3T

τ=0.5s

B(t)=3e(−t/0.5)

B(t)=3exp(-2t)

Therefore

Φ=NAB

Φ=0.001257×3•exp(-2t)

Φ=0.00377exp(-2t)

Now,

Induce emf is given as

E= - dΦ/dt

E= - 0.00377×-2 exp(-2t)

E=0.00754exp(-2t)

At t=2

E=0.00754exp(-4)

E=0.000138V

E=0.138mV

b. Induce current

From ohms laws

V=iR

Given that R=0.6Ω

i=V/R

i=0.000138/0.6

i=0.00023A

i=0.23mA

5 0

2 years ago