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vovikov84 [41]
3 years ago
7

11)

Physics
1 answer:
Vikki [24]3 years ago
6 0

Answer:

salt w nulas...........

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While standing on a balcony a child drops a penny. The penny lands on the ground floor 1.5 s later. How fast was the penny trave
sveticcg [70]

Answer:

14.7 m/s.

Explanation:

From the question given above, the following data were obtained:

Time (t) = 1.5 s

Acceleration due to gravity (g) = 9.8 m/s².

Height = 11.025 m

Final velocity (v) = 0 m/s

Initial velocity (u) =?

We, can obtain the initial velocity of the penny as follow:

H = ½(v + u) t

11.025 = ½ (0 + u) × 1.5

11.025 = ½ × u × 1.5

11.025 = u × 0.75

Divide both side by 0.75

u = 11.025/0.75

u = 14.7 m/s

Therefore, the penny was travelling at 14.7 m/s before hitting the ground.

8 0
3 years ago
Consider the numbers 23.68 and 4.12. The sum of these numbers has ____ significant figures, and the product of these numbers has
damaskus [11]

Answer:multiplying will give us 7 significant figures and addition will give us 3 significant figures

Explanation:

After multiplying the two numbers they resulting value will give a value in its 4 decimal places because both given values are in 2 decimal places. The 4 dp is gotten by the addition of the decimal places of both given numbers (2+2) and

The result of its addition will give us a value in its 1dp and 3 significant figures since the addition of 23.68 and 4.12 will give us 27.8

7 0
3 years ago
2. Sara y Antonio son mellizos. Cuando nacieron, Sara pesaba 600 gramos más que Antonio. Sus pesos ya se han igualado, gracias a
valkas [14]

Answer:

8,25 creo

Explanation:

nose no estoy segura

7 0
3 years ago
Learning Task 1: Define or describe the following words relat are not familiar with the word, you may ask assistance from y your
Tems11 [23]

Answer:

1.) TABLE = The table is 2.74 m (9.0 ft) long, 1.525 m (5.0 ft) wide, and 76 cm (2.5 ft) high with any continuous material so long as the table yields a uniform bounce of about 23 cm (9.1 in) when a standard ball is dropped onto it from a height of 30 cm (11.8 in), or about 77%.

2.) TENNIS = Tennis is a racket sport that can be played individually against a single opponent (singles) or between two teams of two players each (doubles). Each player uses a tennis racket that is strung with cord to strike a hollow rubber ball covered with felt over or around a net and into the opponent's court.

3.) PING-PONG = Table tennis, also known as ping-pong and whiff-whaff, is a sport in which two or four players hit a lightweight ball, also known as the ping-pong ball, back and forth across a table using small rackets. ... Spinning the ball alters its trajectory and limits an opponent's options, giving the hitter a great advantage.

4.) NET = This is stretched across the centter of the table by a cord attached to a post at either end. It measures 6ft long and the ball must pass over it for a rally to continue.

5.) BALL = The ball, which is spherical and hollow, was once made of white celluloid. Since 1969 a plastic similar to celluloid has been used. The ball, which may be coloured white, yellow, or orange, weighs about 0.09 ounce (2.7 grams) and has a diameter of about 1.6 inches (4 cm).

6.) RACKET = A table tennis racket is made up of two distinct parts - a wooden blade which incorporates the handle together with table tennis rubbers affixed to each side of the blade using water-based glue.

Explanation:

All my answers are

about the tools of the

game of tennis

3 0
2 years ago
A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.
bija089 [108]

a)

i) Potential for r < a: V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

ii) Potential for a < r < b:  V(r)=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

iii) Potential for r > b: V(r)=0

b) Potential difference between the two cylinders: V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c) Electric field between the two cylinders: E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

Explanation:

a)

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

E=\frac{\lambda}{2\pi \epsilon_0 r}

where

\lambda is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

E=0

So the potential where the electric field is zero is constant:

V=const.

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density +\lambda and an equal negative charge density -\lambda. Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)

However, we know that the potential at b is zero, so

V(r)=V(b)=0

ii) The electric field in the region a < r < b instead it is given only by the positive charge +\lambda distributed over the surface of the inner cylinder of radius a, therefore it is

E=\frac{\lambda}{2\pi r \epsilon_0}

And so the potential in this region is given by:

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r} (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

And so, for r<a,

V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

b)

Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.

We have:

- Potential at the surface of the inner cylinder:

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

- Potential at the surface of the outer cylinder:

V(b)=0

Therefore, the potential difference is simply equal to

V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c)

Here we want to find the magnitude of the electric field between the two cylinders.

The expression for the electric potential between the cylinders is

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

The electric field is just the derivative of the electric potential:

E=-\frac{dV}{dr}

so we can find it by integrating the expression for the electric potential. We find:

E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

So, this is the expression of the electric field between the two cylinders.

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
3 years ago
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