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3 years ago

7. A spring balance reads force in Newton. The scale is 20 cm long and reads from 0 to

Physics

1 answer:

GarryVolchara [31]3 years ago

7 0

Answer:

The potential energy when it reads 40 N is $PE = 5.33 \ J$

Explanation:

From the question we are told that

The lowest reading of the spring balance is 0 N and this is at 0 cm = 0 m

The height reading of the spring balance is 60 N and this is at 20 cm = 0.20 m

Generally the length corresponding to the reading of 40 N is mathematically represented as

$d = \frac{40 * 0.20 }{60 }$

=> $d = 0.133 \ m$

Generally the potential energy is mathematically represented as

$PE = m * g * d$

Here

$F = mg = 40 N$

$PE = 40 * 0.133$

=> $PE = 5.33 \ J$

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A single loop of current is immersed in an externally applied uniform magnetic field of 3 Tesla oriented in the positive y direc

vlabodo [156]

Answer:

mu=12Tm^2

Explanation:

the magnetic moment mu of a single loop is given by:

$\mu = I A B$

where I is the current, B is the magnetic field and A is the area of the loop. By replacing we obtain:

$\mu=(0.5A)(4m*2m)(3T)=12Tm^2$

hope this helps!!

6 0

3 years ago

A 22-turn circular coil of wire has diameter 1.02 m. It is placed with its axis along the direction of the Earth's magnetic fiel

ollegr [7]

Answer:

ξ = 0.00845020162 V or 8.4 mV

Explanation:

Magnetic flux measures the total magnetic field that passes through a known area. Magnetic flux describe the effect of magnetic field in a given area. Mathematically,

magnetic flux (Ф) = BA cos ∅

where

A = test area

B = magnetic field

before the flip

Ф = Bπr²N

N = number of turn

magnitude of induced emf = N |ΔФ/Δt|

ξ = 2Nπr²B/dt

ξ = 2 × 22 × π × (1.02/2)² × 0.000047/0.2

ξ = 44 × π × 0.51² × 0.000047/0.2

ξ = 44 × π × 0.2601 × 0.000047/0.2

ξ = 0.0005378868 × 3.142/0.2

ξ = 0.00169004032/0.2

ξ = 0.00845020162 V or 8.4 mV

8 0

3 years ago

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$