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3 years ago

On April 26, 1939, Great Britain did this in response to Hitler's aggressive moves?

1 answer:

8 0

On April 26, 1939, Great Britain re-introduced conscription to Hitler's aggressive moves. So for the given question the correct option is option "b". Conscription actually means the compulsory drafting or enlistment of people in the government service. IN case of Great Britain in the year 1939 the compulsory drafting was made in the military to increase the number of soldiers. This was done to fight with Hitler and stop him from conquering great Britain. In some countries conscription is still present, but it dates back to a long time. China is a country that has conscription.

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a moving cart hits a spring, traveling at 5.0 m/s at the time of contact. at this instant the cart is motionless by how much is

viva [34]

This state of motionlessness occurs because all of the kinetic energy in the car is absorbed by the spring in the form of elastic potential energy. The mathematical representation is:
1/2 mv² = 1/2 kx²
25m = kx², where m is the mass of the cart, k is the spring constant and x is the spring's extension.

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3 years ago

How to atoms behave in non-magnetic items?

Anastaziya [24]

Answer:

By altering the quantum interactions of the electrons in the atoms of a metal's atoms, scientists from the University of Leeds have generated magnetism in metals that aren’t normally magnetic.

Explanation:

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2 years ago

(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r

Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface $U_1=\frac{GM_em}{R_e}$

Gravitational Potential Energy at height h is $U_2=\frac{GM_em}{R_e+h}$

Energy required to lift the satellite $E_1=U_1-U_2$

$E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}$

Now Energy required to orbit around the earth

$E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}$

$\Delta E=E_1-E_2$

$\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}$

$E_1=E_2$ (given)

$\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0$

$\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0$

$h=\frac{R_e}{2}$

$h=3.19\times 10^6\ m$

(b)For greater height $E_1$ is greater than $E_2$

thus energy to lift the satellite is more than orbiting around earth

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3 years ago

Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

labwork [276]

Answer: $\tau=1.03\times 10^{-4}\ N-m$

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

$\tau=NIAB\ \sin\theta$