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2 years ago

11

Students repeat the experiment but replace block X and block Y with block W and block Z , as shown in Figure 3. Block W and bloc

k Z have identical mass M . When the experiment is conducted, the students observe that the blocks do not stick together, as shown in Figure 4. The students predict the speed of block Z after the collision by assuming that the collision is perfectly elastic. They then collect data about the actual speeds of both blocks immediately after the collision, as shown in the table.
Why does the predicted speed of block Z after the collision not agree with the actual speed of block Z after the collision?

1 answer:

Lena [83]2 years ago

8 0

The block Z would be seen in figure 10 when 4 strident turn around

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A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences

NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, $q=3\times 10^{-6}\ C$

Mass of charge particle, $m=2\times 10^{-6}\ C$

Speed, $v=5\times 10^{6}\ m/s$

Acceleration, $a=3\times 10^{4}\ m/s^2$

We need to find the minimum magnetic field that would produce such an acceleration. So,

$ma=qvB\ sin\theta$

For minimum magnetic field,

$ma=qvB$

$B=\dfrac{ma}{qv}$

$B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}$

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0

2 years ago

Sea breezes that occur near the shore are attributed to a difference between land and water with respect to what property?

ddd [48]

Answer:

a. mass density

Explanation:

<em>Land and sea breeze that occur near the shore are due to the variation of mass density of air with change in temperature.</em>

When the air gets heated it becomes rarer in density and thus rises up in the atmosphere and its space is occupied by a cooler and denser air that flows to the place.

<em>During the day the land is warmer than the sea so the sea breeze blows and during the night the water bodies are warmer than the land so the land breeze blows.</em>

7 0

2 years ago

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

In the winter sport of bobsledding, athletes push their sled along a horizontal ice surface and then hop on the sled as it start

Hitman42 [59]

Answer:

$v_f = 16.6 m/s$

Explanation:

As we know by force equation that force along the inclined planed due to gravity is given as

$F_g = mg sin\theta$

so the acceleration due to gravity along the plane is given as

$a = \frac{F_g}{m}$

now we have

$a = g sin\theta$

$a = (9.81 sin4.0)$

$a = 0.68 m/s^2$

now we know that

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 9.2^2 = 2(0.68)(140)$

$v_f = 16.6 m/s$

4 0

2 years ago

A particular material has an index of refraction of 1.25. What percent of the speed of light in a vacuum is the speed of light i

beks73 [17]

Answer:

80% (Eighty percent)

Explanation:

The material has a refractive index (n) of 1.25

Speed of light in a vacuum (c) is 2.99792458 x 10⁸ m/s

We can find the speed of light in the material (v) using the relationship

n = c/v, similarly

v = c/n

therefore v = 2.99792458 x 10⁸ m/s ÷ (1.25) = 239 833 966 m/s

v = 239 833 966 m/s

Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as

(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%

Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)

3 0

2 years ago

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