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Soloha48 [4]
3 years ago
12

15. Using 3 – 4 sentences, explain, in your own words, two ways you could cause an electrical current using only a wire and a ho

rseshoe magnet.
Physics
1 answer:
never [62]3 years ago
5 0
An electric current will start to form when a part of the wire is within a magnetic field. Using the magnet you could put the whole wire into the magnetic field of the magnet and move it across the magnet so that a current starts to run.

Another way you could do it is to make the wire so that it looks like a coil and stick the magnet in the middle of the coil of wire. This will cause an induced current in the coil of wire.
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Explain nuclear reactions, nuclear fission and nuclear fusion.
SVETLANKA909090 [29]

Answer:

Nuclear reaction is a reaction in which nucleus of the atom undergoes a change.

Nuclear fusion is a process which involves the conversion of two small nuclei to form a heavy nuclei along with release of energy.

Example : 4_1^1\textrm{H}\rightarrow _2^4\textrm{He}+2_{+1}^0\textrm{e}+\text{energy}

Nuclear fission is a process which involves the conversion of a heavier nuclei into two or more small and stable nuclei along with the release of energy.

Example: _{92}^{235}\textrm{U}+_0^1\textrm{n}\rightarrow _{56}^{143}Ba+_{36}^{90}Kr+3_0^1\textrm{n}

8 0
3 years ago
What’s the force of a pitching machine on a baseball? Baseballs pitched by a machine have a horizontal velocity of 30 meters/sec
Leya [2.2K]
The acceleration of the baseball is:
a= \frac{v_f-v_i}{\Delta t}
where v_f and v_i are the final and initial speed of the ball, and \Delta t is the time interval in which the force acted.

Replacing the numbers, we get
a= \frac{30 m/s-0m/s}{0.5 s}=60 m/s^2
And at this point, we can use Newton's second law F=ma to find the value of the force of the pitching machine:
F=ma=(0.15 kg)(60 m/s^2)=9 N
5 0
3 years ago
The volume of vessel is 6 litres. convert it into ml​
dalvyx [7]
6000 ml
There are 1000ml in one liter
5 0
3 years ago
12. Suppose a person uses a mechanical iack to lift half the weight of a car with a
Likurg_2 [28]

a) The force that must be applied is 73.5 N

b) The actual efficiency is 82 %

Explanation:

a)

Since the jack is 100% efficient, all the input work is converted into output work. So we can write:

W_{in} = W_{out}\\F_{in} d_{in} = F_{out} d_{out}

where:

F_{in} is the input force applied on the jack

d_{in} is the arm of the input force

F_{out} is the output force applied on the jack

d_{out} is the arm of the output force

Here we have:

F_{out}=\frac{mg}{2}= \frac{(1200 kg)(9.8 m/s^2)}{2}=5880 N is the output force (half the weight of the car)

d_{in} = 40.0 cm = 0.40 m is the arm of the input force

d_{out} = 5.0 mm = 0.005 m is the arm of the output force

Solving for F_{in}, we find the force that must be applied in input to lift the car:

F_{in} = \frac{F_{out}d_{out}}{d_{in}}=\frac{(5880)(0.005)}{0.40}=73.5 N

b)

The efficiency of the jack is given by the ratio between the output work and the input work:

\eta = \frac{W_{out}}{W_{in}}=\frac{F_{out}d_{out}}{F_{in}d_{in}}

where we have:

F_{out}=5880 N is the output force (half the weight of the car)

d_{in} = 40.0 cm = 0.40 m is the arm of the input force

d_{out} = 5.0 mm = 0.005 m is the arm of the output force

Here we are told that the input force this time is

F_{in}=90.0N

Substituting into the equation, we find the new efficiency of the jack:

\eta = \frac{(5880)(0.005)}{(90.0)(0.40)}=0.82

Which means an efficiency of 82%.

Learn more about levers:

brainly.com/question/5352966

#LearnwithBrainly

8 0
4 years ago
An infinitely long line of charge has a linear charge density of 7.50×10^−12 C/m . A proton is at distance 14.5 cm from the line
Nata [24]

Answer:

10.22 cm

Explanation:

linear charge density, λ = 7.5 x 10^-12 C/m

distance from line, r = 14.5 cm = 0.145 m

initial speed, u = 3000 m/s

final speed, v = 0 m/s

charge on proton, q = 1.6 x 10^-19 C

mass of proton, m = 1.67 x 10^-27 kg

Let the closest distance of proton is r'.

The potential due t a line charge at a distance r' is given  by

V=-2K\lambda ln\left (\frac{r'}{r}  \right )

where, K = 9 x 10^9 Nm^2/C^2

W = q V

By use of work energy theorem

Work = change in kinetic energy

qV = 0.5m(u^{2}-v^{2})

By substituting the values, we get

V=\frac{mu^{2}}{2q}

-2K\lambda ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{2q}

- ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{4Kq\lambda }

- ln\left ( \frac{r'}{r} \right )=\frac{1.67 \times 10^{-27}\times 3000\times 3000}{4\times 9\times 10^{9}\times 1.6\times 10^{-19}\times 7.5\times 10^{-12} }

- ln\left ( \frac{r'}{r} \right )=0.35

\frac{r'}{r} =e^{-0.35}

\frac{r'}{r} =0.7047

r' = 14.5 x 0.7047 = 10.22 cm

5 0
4 years ago
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