If you wanted to learn more about ancient scientific knowledge, whose life

inysia [295]

The energy transformations that occur as you coast down long hill on a bicycle, including the brakes to make the bike stop at the bottom, is that at the top of the hill you have high GPE AND LOW KE, on your way down you have HIGH KE AND LOW GPE, and at the bottom you have thermal energy due to the stop of the brakes.

the law of conversation of energy and describe the energy transformations that occur as you coast down a long hill on a bicycle and then apply the brakes to make the bike stop at the bottom.

3 0

3 years ago

A 5kg cart moving to the right with a velocity of 16 m/s collides with a concrete wall and

Kryger [21]

Explanation:

mass, m = 5kg

initial velocity, u = 16m/s

final velocuty, v = -22m/s

change in momentum, ∆p = ?

∆p = m (v-u)

5(-22-16)

5(38)

∆p = 190kgm/s

check the calculations!

8 0

3 years ago

A beam of alpha particles ( q = +2e, mass = 6.64 x 10-27 kg) is accelerated from rest through a potential difference of 1.8 kV.

Mrrafil [7]

Answer:

The magnetic field required required for the beam not to be deflected is $B = 0.0036T$

Explanation:

From the question we are told that

The charge on the particle is $q = +2e$

The mass of the particle is $m = 6.64 *10^{-27} kg$

The potential difference is $V_a = 1.8 kV = 1.8 *10^{3} V$

The potential difference between the two parallel plate is $V_b = 120 V$

The separation between the plate is $d = 8 mm = \frac{8}{1000} = 8*10^{-3}m$

The Kinetic energy experienced by the beam before entering the region of the parallel plate is equivalent to the potential energy of the beam after the region having a potential difference of 1.8kV

$KE_b = PE_b$

Generelly

$KE_b = \frac{1}{2} m v^2$

And $PE_b = q V_a$

Equating this two formulas

$\frac{1}{2} mv^2 = q V_a$

making v the subject

$v = \sqrt{\frac{q V_a}{2 m} }$

Substituting value

$v = \sqrt{\frac{ 2* 1.602 *10^{-19} * 1.8 *10^{3}}{2 * 6.64 *10^{-27}} }$

$v = 41.65*10^4 m/s$

Generally the electric field between the plates is mathematically represented as

$E = \frac{V_b}{d}$

Substituting value

$E = \frac{120}{8*10^{-3}}$

$E = 15 *10^3 NC^{-1}$

the magnetic field is mathematically evaluate

$B = \frac{E}{v}$

$B = \frac{15 *10^{3}}{41.65 *10^4}$

$B = 0.0036T$

6 0

3 years ago

Substances A and B, initially at different temperatures, come in contact with each other and reach thermal equilibrium. The mass

Gnesinka [82]

Answer: .2) The final temperature will be exactly midway between the initial temperatures of substances A and B.

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_A\times c_A\times (T_{final}-T_A)=-[m_B\times c_B\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_A$ = mass of A = 2x

$m_B$ = mass of B = x

$T_{final}$ = final temperature = z

$T_A$ = temperature of A

$T_2$ = temperature of B

$c_A$ = specific heat capacity of A = y

$c_B$ = specific heat capacity of B = 2y

Now put all the given values in equation (1), we get

$2x\times y\times (z-T_A)=-[x\times 2y\times (z-T_B)]$

$2z=T_B+T_A$

$z=\frac{T_B+T_A}{2}$

Therefore, the final temperature of the mixture will be exactly midway between the initial temperatures of substances A and B.

3 0

3 years ago

Two forces P and Q act on an object of mass 11.0 kg with Q being the larger of the two forces. When both forces are directed to

alexgriva [62]

Answer:

The magnitude of force on P is 2.75 N

The magnitude of force on Q is 7.15 N

Solution:

As per the question:

Mass of the object, M = 11.0 kg

Acceleration of the object when the forces are directed leftwards, a = $0.900 m/s^{2}$

Acceleration when the forces are in opposite direction, a' = $0.400 m/s^{2}$

Now,

The net force on the object in first case is given by:

$F_{net} = |\vec{F_{P}}| + |\vec{F_{Q}}| = Ma$ (1)

The net force on the object in second case is given by:

$F_{net} = |\vec{F_{P}}| - |\vec{F_{Q}}| = Ma'$ (2)

Adding both eqn (1) and (2):

$2|\vec{F_{Q}}| = M(a + a')$

$|\vec{F_{Q}}| = \frac{11.0(0.900 + 0.400)}{2} = 7.15 N$

Putting the above value in eqn (1):

$|\vec{F_{P}}| = 11\times 0.900 - |\vec{F_{Q}}|$

$|\vec{F_{P}}| = 9.900 - 7.15 = 2.75$

6 0

4 years ago