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Answer:

a) the markers are 29.76 m far apart in the laboratory

b) According to B, the markers are 23.21 m far apart in the laboratory

Explanation:

Given that;

speed of particle = 0.624c = (0.624 × 3×10⁸m/s)

{ where c is the speed of light}    

lifetime of the particle t = 159ns  = 1.59 × 10⁻⁷

a) Observer A places markers in the laboratory at the locations where the particle is produced and where it decays. How far apart are those markers in the laboratory

lets apply the formula distance = velocity × time

lets substitute

distance = (0.624×3×10⁸ m/s)  × (1.59 × 10⁻⁷ s)

distance = 29.76 m

Therefore, the markers are 29.76 m far apart in the laboratory

b) Observer B, who observes the particle to be at rest and measures its lifetime to be 124ns. According to B, how far apart are the two markers in the laboratory?

lifetime of the particle t = 124ns  = 1.24 × 10⁻⁷

distance = (0.624×3×10⁸ m/s)  × (1.24 × 10⁻⁷ s)

distance = 23.21 m

Therefore,  According to B, the markers are 23.21 m far apart in the laboratory

Answer:I think sprinting

Explanation:

Answer:

Assumption: there is no object between this point charge and the observer.

Explanation:

The electric field of a point charge is inversely proportional to the square of the distance from that point charge.

Let [tex]k[/tex] denote Coulomb's constant ([tex]k \approx 8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-1}[/tex].) Let the magnitude of that point charge be [tex]q[/tex]. At a distance of [tex]r[/tex] from this charge, the electric field due to this charge would be:

[tex]\displaystyle E = \frac{k \cdot q}{r^{2}}[/tex].

Convert the magnitude of the point charge in this question to standard units:

[tex]q = 1\; \rm nC = 10^{-9}\; \rm C[/tex].

Apply that equation to find the magnitude of the electric field due to this point charge:

[tex]r = 1\; \rm m[/tex]:

[tex]\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(1\; \rm m)^{2}} \\ &\approx 9.0\; \rm N \cdot C^{-1}\end{aligned}[/tex].

[tex]r = 2\; \rm m[/tex]:

[tex]\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(2\; \rm m)^{2}} \\ &\approx 2.2\; \rm N \cdot C^{-1}\end{aligned}[/tex].

[tex]r = 3\; \rm m[/tex]:

[tex]\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(3\; \rm m)^{2}} \\ &\approx 1.0\; \rm N \cdot C^{-1}\end{aligned}[/tex].

The direction of the electric field at a point is the same as the direction of a force from this field onto a positive point charge at this point.

Because the [tex](+1\; \rm nC)[/tex] point charge here is positive, the electric field of this charge would repel other positive point charges. Hence, the electric field around this [tex](+1\; \rm nC)\![/tex] point charge at any point in the field would point away from this charge.

Answer:

Explanation:

Let us calculate the average velocity .

Average velocity = total distance / total time

Time taken to drive first 15 mi

= 15 / 60 = .25 h

Time taken to drive next 15 mi

= 15 / 30 = .5 h

Total time = .25 + .5 = .75 h

Total distance = 30 mi

average speed = 30 / .75

= 40 mi / h

So average speed is 40 mph which is less than given velocity of 45 mph .

Hence if she travels at 45 mph , she will take less time .

Answer:

Factors Affecting the Strength of the Magnetic Field of an Electromagnet: Factors that affect the strength of electromagnets are the nature of the core material, strength of the current passing through the core, the number of turns of wire on the core and the shape and size of the core

Explanation:

Follow me for more

The following factors affect the strength of an electromagnet; number of turns, wire size, current and the presence of iron core.

This is a type of magnet  in which the magnetic field is produced by an electric current.

The following factors affect the strength of an electromagnet;

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Answer:

It is kinetic energy.

Explanation:

The kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate the volleyball from rest to its stated velocity. Having gained this energy from the player spiking it during its acceleration, the body maintains this kinetic energy unless its speed changes.

Answer:

True

Explanation:

Friction occurs because no surface is perfectly smooth. Even surfaces that look smooth to the unaided eye make look rough or bumpy when viewed under a microscope. You know that friction produces heat. That’s why rubbing your hands together makes them warmer. But do you know why? Friction causes the molecules on rubbing surfaces to move faster, so they have more energy. This gives them a higher temperature, and they feel warmer. Heat from friction can be useful. It not only warms your hands.

Explanation:

Q= mc∆T

∆T= 5-24=- 19

Q= 0.5*4186*-19

Q= -39767 J

negative sign show heat releases

Answer:

Velocity = -20.45 i + 35.54 j km/h

Explanation:

given data

speed v = 41.0 km/h

direction = 60.08 north of west

solution

first we get here Speed in west direction that is express as

Speed in west direction = v × cos(60.08)     ..........................1

Speed in west direction = 41 × cos(60.08)

Speed in west direction = 20.45 km/h

now we get Speed in north direction that is

Speed in north direction = v × sin(60.08)        ............................2

Speed in north direction = 41 × sin(60.08)

Speed in north direction = 35.54 km/h

so Velocity will be here

Velocity = -20.45 i + 35.54 j km/h

Answer:

iron/copper

iron faster because it in the core so iron is the answer with would be 15C because its the hottest in the core and is very quick there to being moved by the core so iron being 15C is the answer.

Explanation:

Answer:

1.092 cycles

Explanation:

We need to convert the arc length of 6.86 radians into cycles.

We know that,

1 radian = 0.1592 cycles

To convert 6.86 radians to cycles, we use unitary method.

6.86 radian = (0.1592 × 6.86) cycles

= 1.092 cycles

Hence, there are 1.092 cycles in 6.86 radians. Hence, this is the required solution.

Answer:

it will take 4 seconds to travel 2 meters.

Answer:

change is imperceptible

          w_f = 7.272  10⁻⁵ rad / s

Explanation:

For this exercise we can use the conservation of angular momentum.

Initial. Before the crash

        L₀ = I w₀

final. After the crash

         L_f = I w_f + p r

where the moment is

        p = mv

As the system is formed by the two bodies, the forces during the impact are internal, therefore the angular momentum is conserved

        L₀ = L_f

        I w₀ = I w_f + m v r

        w_f = w₀ - [tex]m \frac{ v \ r }{I}[/tex]

We can approximate the Earth to a sphere, so its angular momentum is

       I = 2/5 M r²

we substitute

         w_f = w₀ - [tex]\frac{5 \ m \ v}{2 \ M \ r}[/tex]

We can find the angular velocity of the Earth with the duration of a spin which is the period of one day

        w₀ = 2π / T

        T = 24 h (3600 s / 1h) = 86 400 s

         w₀ = 2π / 86400

         w₀ = 7.272 10⁻⁵ rad / s

let's calculate

         w_f = 7.27 10⁻⁵ - [tex]\frac{5 \ 250 \ 10^{13} \ 33.0 \ 10^{3} }{ 2 \ 5.98 \ 10^{24} \ 6.37 10^{6} }[/tex]

         w_f = 7.272 10⁻⁵ - 1.0829 10⁻¹³

         w_f = 7.27199999 10⁻⁵

this change is imperceptible

          w_f = 7.272  10⁻⁵ rad / s

Answer:

i dont know!

Explanation:

Answer:

Explanation:

Given that:

The area of the metal tube = 1.60 cm² = 1.60 × 10⁻⁴ m²

Length of cube L = 0.150 m

The exerted force by water = 1.00 × 10³ N

At the top of the cube;

The force on water is F = pgL

F = 9.8 m/s² × (1000 kg/m³) (0.150 m)(0.150 m)²

F = 33.075 N

∴

[tex]pgh = \dfrac{F}{A}[/tex]

[tex]h = \dfrac{F}{Apg}[/tex]

[tex]h = \dfrac{33.075}{(1.60 \times 10^{-4} \ m^2) \times (1000 \ kg/m^3)(9.8)}[/tex]

h = 21.09 m

The volume of the water V = L³

V = (0.150 m)³

V = 0.003375 m³

weight of the water w = p(V + Ah)

w = 1000 (0.003375 + (1.60 × 10⁻⁴ × 21.09))

w = 6.75 kg

Answer:

Potential difference [tex]V=1.6\times 10^{-10}\ V[/tex]

Explanation:

Capacitance of capacitor, C = 0.9 MF

Charge, [tex]Q=1.4\times 10^{-4}\ C[/tex]

We need to find the amount of potential difference. We know that,

[tex]Q=CV[/tex]

V is potential difference

So,

[tex]V=\dfrac{Q}{C}\\\\V=\dfrac{1.4\times 10^{-4}}{0.9\times 10^6}\\\\V=1.6\times 10^{-10}\ V[/tex]

So, the required potential difference is [tex]1.6\times 10^{-10}\ V[/tex].

Answer:

A.

Explanation:

A large fraction of astronomical gamma rays are screened by Earth's atmosphere. Gamma rays are ionizing radiation and are thus biologically hazardous. Due to their high penetration power, they can damage bone marrow and internal organs.

The extremely high energy of gamma rays allows them to penetrate just about anything. They can even pass through bones and teeth. This makes gamma rays very dangerous. They can destroy living cells, produce gene mutations, and cause cancer.

25.10 or 24.6m/s both answers are correct

Explanation:

firstly , divide the multiply with 2 and divide the result by acceleration due to gravity which is 9.8m/s or 10m/s

123*2=246

246/9.8 or 10

= 25.10 or 24.6m/s

Answer:

south

Explanation:

That depends on the circle's radius, which you neglected to mention.

If the radius is 1 meter, then an arc of 3.05 degrees is 5.32 centimeters long.

If the radius is 1 kilometer, then an arc of 3.05 degrees is 53.2 meters long.

Answer:

convergence and trench

Explanation:

LOL

Answer:

the mass of the mud sitting above a 3.5 m2 area of the valley floor is 8509273.5 kg

Explanation:

Given the data in the question;

first we compute the total volume of mud

V = ( 1.9×1000)m × (0.55×1000)m × (2.4×1000)m

V = 2.508 × 10⁹ m³

same volume of mud spread uniformly over 1.4 × 1.4 km² area

Hence,

V = (1.4×1000)m × (1.4×1000)m × depth

2.508 × 10⁹ = 1400m × 1400m × depth

2.508 × 10⁹ m³ = 1960000m² × depth

depth = 2.508 × 10⁹ m³ / 1960000 m²

depth = 1279.59 m

so volume of the mud sitting above 3.5 m² area of the valley floor will be;

⇒ 3.5 m² × 1279.59 m

⇒ 4478.565 m³

so mass will be

m = 4478.565 × 1900 kg

m = 8509273.5 kg

Therefore, the mass of the mud sitting above a 3.5 m2 area of the valley floor is 8509273.5 kg

Answer: 4R

Explanation:R=pL/A

R second wire=p2L/1/2A

=4R

Answer:

h = 18.4 cm

Explanation:

Given that,

The speed of the bob at the bottom of the swing is 1.9m/s.

We need to find the initial height of the bob. Let it is h.

We can find it using the conservation of energy i.e.

[tex]mgh=\dfrac{1}{2}mv^2[/tex]

Where

v is speed of the bob

So,

[tex]h=\dfrac{v^2}{2g}\\\\h=\dfrac{(1.9)^2}{2\times 9.8}\\\\h= 0.184\ m[/tex]

or

h = 18.4 cm

So, the initial height of the bob is 18.4 cm.

The initial height of bob will be "18.4 cm".

Speed would be the scalar quantity that describes "how quickly an attribute moves." This could sometimes be defined as this same rate whereby an object travels a given distance.

According to the question,

The speed of the bob = 1.9 m/s

Let,

The initial height be "h".

By using the conservation of energy, we get

→ mgh = [tex]\frac{1}{2}[/tex]mv²

or,

→      h = [tex]\frac{v^2}{2g}[/tex]

By substituting the values, we get

          = [tex]\frac{(1.9)^2}{2\times 9.8}[/tex]

          = 0.184 m or,

          = 18.4 cm

Thus the above answer is correct.

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Answer:

The answer is B

Explanation:

keep it up

Answer:

4.9 x 10-23 N

Explanation:

I took the test

Answer:

100

w=f*s

20*5=100....

100 Joules of work is done on the cart.

Work done by a force is defined as the product of the displacement and the component of the applied force on the object in the direction of displacement. When we push a block with some force, the body moves with some acceleration, so it is called work done.

Work done is expressed as W=Fd and its unit is joules which can be defined as the amount of work done by a force in Newton is applied to an object, as a result of which it is displaced in meter.

For above given information,

Force= 20 N

Distance= 5 m

So, work done= 20*5= 100 Joules

Thus, 100 Joules of work is done on the cart.

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Answer: look at the file