Ask question

Ask question

All categories

3 years ago

12

Which spacecraft did john glenn pilot when he became the first man to orbit the earth in 1962?

1 answer:

Makovka662 [10]3 years ago

7 0

Spacecraft used is "Friendship 7". Hope it helps.

You might be interested in

Angular Velocity The Moon rotates once on its axis in 27.3 days. Its radius is 27.3 days

dolphi86 [110]

B is the answer I’m sure

5 0

3 years ago

Considerable scientific work is currently under way to determine whether weak oscillating magnetic fields such as those found ne

VLD [36.1K]

Answer:

The the maximum emf is $2.52\times10^{-11}\ V$

Explanation:

Given that,

Magnetic field $B = 1.40\times10^{-3}\ T$

Frequency = 60 Hz

Diameter = 7.8 μm

We need to calculate the maximum emf

Using formula of emf

$\epsilon=NBA\omega$

Where, N = number of turns

B= magnetic field

A = area

Put the value in to the formula

$\epsilon=1\times1.40\times10^{-3}\times\pi\times(\dfrac{7.8\times10^{-6}}{2})^2\times2\times\pi\times60$

$\epsilon=2.52\times10^{-11}\ V$

Hence, The the maximum emf is $2.52\times10^{-11}\ V$

5 0

3 years ago

An object is released from rest high above the surface of the earth. When it has fallen halfway to the surface, its kinetic ener

aksik [14]

Answer:

There will be an increase in the kinetic energy

Explanation:

A falling object converts the gravitational potential energy to the kinetic energy. The potential energy is then converted to kinetic energy followed by the conversation:

$E_{p} = E_{k}$

where Ep and Ek are potential and kinetic energies respectively.

This potential energy is then converted to kinetic energy. Halfway, the kinetic energy is equal to KE1.

However, the kinetic energy is given by the equation:

$KE = \frac{1}{2}mv^{2}$

As the velocity increases, the kinetic energy increases. Hence KE2 will be greater than KE1

4 0

4 years ago

2. One mole of a monatomic ideal gas undergoes a reversible expansion at constant pressure, during which the entropy of the gas

Anna35 [415]

Answer:

The initial and final temperatures of the gas is 300 K and 600 K.

Explanation:

Given that,

Entropy of the gas = 14.41 J/K

Absorb gas = 6236 J

We know that,

$ds=\dfrac{dQ}{dt}$

At constant pressure,

$dQ=C_{p}dt$

$\Delta s=\int_{T_{1}}^{T_{2}}{\dfrac{C_{p}dT}{T}}$

$\Delta s=C_{p}ln\dfrac{T_{2}}{T_{1}}$

Put the value into the formula

$14.41=2.5\times8.3144(ln\dfrac{T_{2}}{T_{1}})$

$\dfrac{14.41}{2.5\times8.3144}=ln\dfrac{T_{2}}{T_{1}}$

$0.693=ln\dfrac{T_{2}}{T_{1}}$

$ln2=ln\dfrac{T_{2}}{T_{1}}$

$T_{2}=2T_{1}$ ...(I)

We need to calculate the initial and final temperatures of the gas

Using formula of energy

$\Delta Q=C_{p}\Delta T$

Put the value into the formula

$6236=2.5\times8.3144(T_{2}-T_{1})$

$6236=20.786(T_{2}-T_{1})$

$T_{2}-T_{1}=\dfrac{6236}{20.786}$

$T_{2}-T_{1}=300$

Put the value of T₂

$2T_{1}-T_{1}=300$

$T_{1}=300\ K$

Put the value of T₁ in equation (I)

$T_{2}=2\times300$

$T_{2}=600\ K$

Hence, The initial and final temperatures of the gas is 300 K and 600 K.

8 0

3 years ago

Two forces P and Q act on an object.<br>Which diagram shows the resultant of these two forces?

denis-greek [22]

The Answer: diagram C

7 0

4 years ago