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3 years ago

How long will it take an object traveling at 7 m/s to reach a distance of 26 meters

Physics

1 answer:

Blababa [14]3 years ago

7 0

Answer:

3 hours 71 minutes

Explanation:

As because Speed= distance/time taken

so time taken= 3.71

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Solenoid 2 has twice the radius and six times the number of turns per unit length as solenoid 1. The ratio of the magnetic field

Xelga [282]

Answer:

Explanation:

The magnetic field inside a solenoid is given by the following formula:

$B = \mu_{0}nI$

where,

B = Magnetic Field Inside Solenoid

μ₀ = permittivity of free space

n = No. of turns per unit length

I = Current Passing through Solenoid

For Solenoid 1:

$B_{1} = \mu_{0}n_{1}I ------------------- equation 1$

For Solenoid 2:

n₂ = 6n₁

Therefore,

$B_{1} = \mu_{0}n_{2}I\\B_{1} = 6\mu_{0}n_{1}I ----------------- equation 2$

Diving equation 1 and equation 2:

$\frac{B_{2}}{B_{1}} = \frac{6\mu_{0}nI}{\mu_{0}nI}\\\\\frac{B_{2}}{B_{1}} = 6$

Hence, the correct option is:

8 0

3 years ago

Calculate the induced electric field (in V/m) in a 40-turn coil with a diameter of 18 cm that is placed in a spatially uniform m

kozerog [31]

Answer:

ε = 6.617 V

Explanation:

We are given;

Number of turns; N = 40 turns

Diameter;D = 18cm = 0.18m

magnetic field; B = 0.65 T

Time;t = 0.1 s

The formula for the induced electric field(E.M.F) is given by;

ε = |-NAB/t|

A is area

ε is induced electric field

While N,B and t remain as earlier described.

Area = π(d²/4) = π(0.18²/4) = 0.02545

Thus;

ε = |-40 × 0.02545 × 0.65/0.1|

ε = 6.617 V

(we ignore the negative sign because we have to take the absolute value)

6 0

3 years ago

A person can jump a maximum horizontal distance (by using a 45◦ projectile angle) of 5 m on Earth. The acceleration of gravity i

snow_lady [41]

Answer:30 m

Explanation:

Given

Maximum Horizontal distance is 5 m on earth

launching angle $=45^{\circ}$

Acceleration due to gravity on earth is $9.8 m/s^2$

Acceleration due to gravity on moon is $\frac{9.8}{6}=1.63 m/s^2$

Range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

$R_{earth}=\frac{u^2\sin 2\theta }{g}=5$ ----1

$R_{moon}=\frac{u^2\sin 2\theta }{\frac{g}{6}}$ -----2

Divide 1 & 2

$\frac{5}{R_{moon}}=\frac{1}{6}$

$R_{moon}=30 m$

4 0

3 years ago

1.A river flowing steadily at a rate of 240 m3/s is considered for hydroelectric power generation. It is determined that a dam c

SIZIF [17.4K]

Answer:

the power that can be generated by the river is 117.6 MW

Explanation:

Given that;

Volume flow rate of river v = 240 m³/s

Height above the lake surface a h = 50 m

Amount of power can be generated from this river water after the dam is filled = ?

Now the collected water in the dam contains potential energy which is used for the power generation,

hence, total mechanical energy is due to potential energy alone.

$E_{mech}$ = m(gh)

first we determine the mass flow rate of the fluid m

m = p×v

where p is density ( 1000 kg/m³

so we substitute

m = 1000kg/m³ × 240 m³/s

m = 240000 kg/s

so we plug in our values into ( $E_{mech}$ = m(gh) kJ/kg )

$E_{mech}$ = 240000 × 9.8 × 50

$E_{mech}$ = 117600000 W

$E_{mech}$ = 117.6 MW

Therefore, the power that can be generated by the river is 117.6 MW

4 0

3 years ago

Using moving water to produce electricity is an example of _____.

denis-greek [22]

Using moving water to produce electricity is an example of changing one form of energy into another form of energy.

8 0

4 years ago