Calculate the time taken by the light to pass through a nucleus of diameter 1.56 10 -16 m. (speed of light is 3 10 8 m/s)

A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.

Pie

#1

Volume of lead = 100 cm^3

density of lead = 11.34 g/cm^3

mass of the lead piece = density * volume

$m = 100 * 11.34 = 1134 g$

$m = 1.134 kg$

so its weight in air will be given as

$W = mg = 1.134* 9.8 = 11.11 N$

now the buoyant force on the lead is given by

$F_B = W - F_{net}$

$F_B = 11.11 - 11 = 0.11 N$

now as we know that

$F_B = \rho V g$

$0.11 = 1000* V * 9.8$

so by solving it we got

V = 11.22 cm^3

(ii) this volume of water will weigh same as the buoyant force so it is 0.11 N

(iii) Buoyant force = 0.11 N

(iv)since the density of lead block is more than density of water so it will sink inside the water

#2

buoyant force on the lead block is balancing the weight of it

$F_B = W$

$\rho V g = W$

$13* 10^3 * V * 9.8 = 11.11$

$V = 87.2 cm^3$

(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight = 11.11 N

(iii) Buoyant force = 11.11 N

(iv) since the density of lead is less than the density of mercury so it will float inside mercury

#3

Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid

3 0

1 year ago

A hydrogen discharge lamp emits light with two prominent wavelengths: 656 nm (red) and 486 nm (blue). The light enters a flint-g

mezya [45]

Answer:

The angle between the red and blue light is 1.7°.

Explanation:

Given that,

Wavelength of red = 656 nm

Wavelength of blue = 486 nm

Angle = 37°

Suppose we need to find the angle between the red and blue light as it leaves the prism

$n_{r}=1.572$

$n_{b}=1.587$

We need to calculate the angle for red wavelength

Using Snell's law,

$n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}$

Put the value into the formula

$1.572\sin37=1\times\sin\theta_{r}$

$\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})$

$\theta_{r}=71.0^{\circ}$

We need to calculate the angle for blue wavelength

Using Snell's law,

$n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}$

Put the value into the formula

$1.587\sin37=1\times\sin\theta_{b}$

$\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})$

$\theta_{b}=72.7^{\circ}$

We need to calculate the angle between the red and blue light

Using formula of angle

$\Delta \theta=\theta_{b}-\theta_{r}$

Put the value into the formula

$\Delta \theta=72.7-71.0$

$\Delta \theta=1.7^{\circ}$

Hence, The angle between the red and blue light is 1.7°.

8 0

1 year ago

A truck is traveling down a road with a 4-percent grade at a speed of 75 mi/h when its brakes are applied to slow it down to 22.

kvasek [131]

Answer:

3.964 s

Explanation:

Metric unit conversion:

1 miles = 1.6 km = 1600 m.

1 hour = 60 minutes = 3600 seconds

75 mph = 75 * 1600 / 3600 = 33.3 m/s

22.5 mph = 22.5 * 1600/3600 = 10 m/s

Let g = 9.81 m/s2

Friction is the product of coefficient and normal force, which equals to the gravity

$F_f = \mu N = \mu mg$

The deceleration caused by friction is friction divided by mass according to Newton 2nd law.

$a = F_f / m = \mu mg / m = \mu g = 0.6 *9.81 = 5.886 m/s^2$

So the time required to decelerate from 33.3 m/s to 10 m/s so the wheels don't slide, with the rate of 5.886 m/s2 is

$t = \frac{\Delta v}{a} = \frac{33.3 - 10}{5.886} = 3.964 s$

3 0

1 year ago

A piston–cylinder device contains 0.15 kg of air initially at 2 MPa and 350°C. The air is first expanded isothermally to 500 kPa

Paraphin [41]

Answer:

<h2>jeusYgwyhedswusjsj</h2>

Explanation:

sjauajshsuyeueueheysb

3 0

1 year ago

A 4.0-mF capacitor initially charged to 50 V and a 6.0-mF capacitor charged to 30 V are connected to each other with the positiv

Juli2301 [7.4K]

Answer:

The final charge on the 6.0 mF capacitor would be 12 mC

Explanation:

The initial charge on 4 mF capacitor = 4 mf x 50 V = 200 mC

The initial Charge on 6 mF capacitor = 6 mf x 30 V =180 mC

Since the negative ends are joined together the total charge on both capacity would be;

q = $q_{1} -q_{2}$

q = 200 - 180

q = 20 mC

In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage

q = (4 x V) + (6 x V)

20 = 10 V

V = 2 V

For the final charge on 6.0 mF;

q = CV

q = 6.0 mF x 2 V

q = 12 mC

Therefore the final charge on the 6.0 mF capacitor would be 12 mC

5 0

2 years ago

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