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Dovator [93]
3 years ago
5

When was the first electric typewriter produced and by what company

Physics
1 answer:
LiRa [457]3 years ago
6 0
In 1920, after returning from Army service, he produced a successful model and in 1923 turned it over to the Northeast Electric Company of Rochester for development.
You might be interested in
A portable music player, operating with four 1.5 V cells connected in series, provides a resistance of 15 000 Ω. What amount of
nata0808 [166]

Answer:

Explanation:

Given that,

A portable music player is operating with 4 cell batteries connected in series, and each cell has a P.D of 1.5V.

Then,

Total potential difference is

P.D_total = V1 + V2 + V3 + V4

P.D_total = 1.5 + 1.5 + 1.5 + 1.4

P.D_total = 6V.

The music player provides a resistance of 15,000Ω

R = 15,000Ω

We want to find the current (I) flowing through the music player?

Using ohms law

V = IR

Where

V is the potential difference

I is the current

R is the resistance

Therefore,

I = V/R

I = 6 / 15,000

I = 4 × 10^-4 A

I = 0.4 × 10^-3 A

I = 0.4 mA.

So, 0.4mA is passing through the music player

7 0
3 years ago
HELP PLZZZ!!!! Hurry
Mashcka [7]

Answer:

3. When the number of turns, N is doubled, the strength of the electromagnet is also doubled

4. Doubling the voltage, doubles the strength of the electromagnet

5. The number of paper clips a 7.5 V battery would pick is approximately 28 paper clips

The number of paper clips a 7.5 V battery would pick is 59 paperclips

6. For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is approximately 7 paperclips

For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is 16 paperclips

Explanation:

3. The Magnetomotive Force, MMF = The Number of Turns on the Coil, N × The Current I Flowing in the Coil, I

∴ MMF = N × I

When the number of turns, N is doubled, the magnetomotive force, MMF is also doubled, and the strength of the electromagnet is doubled

4. Given that the voltage, V applied to the coil = The current, I flowing × The resistance, R of the coil, we have

V = I × R

Therefore, for a given constant resistance in the coil, doubling the voltage, doubles the current and therefore doubles the strength of the electromagnet

5. The average slope for the 25-coil electromagnet = (23 - 12)/(6 - 3) = 3.\bar 6

The number of paper clips a 7.5 V battery would pick = 12 + (7.5 - 3) × 11/3 = 28.5 paperclips  ≈ 28 paper clips

The average slope for the 50-coil electromagnet = (48 - 26)/(6 - 3) = 7.\bar 3

The number of paper clips a 7.5 V battery would pick = 26 + (7.5 - 3) × 22/3 = 59 paperclips

6. The slope calculated from a start point of approximately 0.4 V, is given as follows;

The slope for the 25-coil electromagnet = (12 - 6)/(3 - 0.4) = 30/13

Therefore, for the 25-coil electromagnet,  the average number of paper clips a 1 V battery would pick = 6 + (1 - 0.4) × 30/13) = 96/13 ≈ 7 paperclips

The slope for the 50-coil electromagnet = (26 - 13)/(3 - 0.4) = 5

Therefore, for the 50-coil electromagnet,  the average number of paper clips a 1 V battery would pick = 13 + (1 - 0.4) × 5 = 16 paperclips

8 0
3 years ago
Read 2 more answers
Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m

The beetle can jump to a height of <u>2.1 m</u>



7 0
3 years ago
Which answer is right? thanks if u help me! :)
enyata [817]

Answer:7.5 x 10* 14

Explanation:

7 0
3 years ago
Read 2 more answers
If a nucleus decays by gamma decay to a daughter nucleus, which of the following statements about this decay are correct? (There
GarryVolchara [31]

Answer: Option (b) is the correct answer.

Explanation:

A gamma particle is basically a photon of electromagnetic radiation with a short wavelength.

Symbol of a gamma particle is ^{0}_{0}\gamma. Hence, charge on a gamma particle is also 0.

For example, ^{234}_{91}Pa \rightarrow ^{234}_{91}Pa + ^{0}_{0}\gamma + Energy

So, when a nucleus decays by gamma decay to a daughter nucleus then there will occur no change in the number of protons and neutrons of the parent atom but there will be loss of energy as a nuclear reaction has occurred.

Thus, we can conclude that the statement daughter nucleus has the same number of nucleons as the original nucleus., is correct about if  a nucleus decays by gamma decay to a daughter nucleus.

5 0
3 years ago
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