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2 years ago

Physics Momentum topic question. I will mark brainliest please help.

Physics

1 answer:

yawa3891 [41]2 years ago

7 0

Answer:

10 m/s

Explanation:

Given

mass of ball 1, m1 = 1 kg
mass of ball 2, m2 = 0.25 kg
initial velocity of ball 1, u1 = 10 m/s
initial velocity of ball 2, u2 = 0 m/s [stationary]
final velocity of ball 1, v1 = 7.5 m/s
momentum of ball 2 after collision = 2.5 kg m/s

Solving :

We do not require all of this information
Only require points 2 and 6
Momentum = mass x velocity
⇒ velocity = 2.5 / 0.25
⇒ velocity = 10 m/s

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A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.

Luba_88 [7]

Answer:

The final equilibrium temperature of the system is $T = 12.48^oC$

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice $M_{i}$ = 129 g = 0.129 kg

Mass of the steam $M_s$ = 19 g = 0.019 kg

Initial temperature is $T_i$ = 0°C

Temperature of steam $T_s$ = 100°C

Following the change of state of water in the question

The energy required by ice to change to water is mathematically given as

$Q_A = M_iL_f$

Where $L_f$ is a constant known as heat of fusion and the value is $334*10^3 J/kg$

$Q_A = 0.129 *334 *10^3 = 43086 J$

The energy been released when the steam changes to water is mathematically given as

$Q_B = M_s * L_v$

Where $L_v$ is a constant known as heat of vaporization and the value is $2256*10^3J/kg$

$Q_B = 0.019 * 2256*10^3 = 42864J$

The energy released when the temperature of water decrease from 100°C to 0°C is

$Q_C = M_s *C_water ($ 100°C)

Where $C_{water}$ is the specific heat of water which has a value $4186J/kg \cdot K$

$Q_C = 0.019 *4186*100 = 7953.4$

Looking at the values we obtained we noticed that ]

$Q_B + Q_C > Q_A$

What this means is that the ice will melt

bearing in mind the conservation of energy

looking the way at which water at different temperature were mixed according to the question

Heat lossed by the vapor = heat gained by ice

$Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T$

$T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}$

$T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}$

$T = 12.48^oC$

3 0

4 years ago

Four toy racecars are racing along a circular race track. The cars start at the 3-o'clock position and travel CCW along the trac

xeze [42]

Answer:

in t seconds, Car A sweep out t radian { i.e θ = t radian }

Explanation:

Given the data in the question;

4 toy racecars are racing along a circular race track.

They all start at 3 o'clock position and moved CCW

Car A is constantly 2 feet from the center of the race track and moves at a constant speed

so maximum distance from the center = 2 ft

The angle Car A sweeps out increases at a constant rate of 1 radian per second.

Rate of change of angle = dθ/dt = 1

Now,

since dθ/dt = 1

Hence θ = t + C

where C is the constant of integration

so at t = 0, θ = 0, the value of C will be 0.

Hence, θ = t radian

Therefore, in t seconds, Car A sweep out t radian { i.e θ = t radian }

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3 years ago

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Kay [80]

Answer:

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3 years ago

002 10.0 points note: report your answer in joules here. a proton and an antiproton, both at rest with respect to one another, m

Ulleksa [173]

The mass of the proton is:
$m_p = 1.67 \cdot 10^{-27} kg$
and the mass of the antiproton is exactly the same, so the total mass of the two particles is $2m_p$ .

In the annihilation, all the mass of the two particles is converted into energy, and the amount of this energy is given by Einstein's equivalence between mass and energy:
$E=Mc^2$
where M is the mass converted into energy and c is the speed of light. In this example, $M=2m_p$ , therefore the energy released is
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3 years ago

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Troyanec [42]

This indicates that carob dioxide is a compound

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3 years ago

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