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3 years ago

A toy car of mass 0.15kg accelerates from a speed of 10 cm/s to a speed of 15 cm/s. What is the impulse acting on the car?

Physics

2 answers:

OLga [1]3 years ago

7 0

v=v2-v1=15-10=5 cm/s

p=mv=0.15•5=0.75 kg•cm/s

marusya05 [52]3 years ago

6 0

Answer : The correct option is, (A) 7.5 mN.s

Explanation :

Impulse : It is defined as the force applied to an object for a certain amount of time.

Formula used :

$I=F\times \Delta t$

As we know that:

$F=m\times \Delta a\\\\\Delta a=\frac{\Delta v}{\Delta t}$

Thus, formula of impulse will be:

$I=m\times \Delta v$

where,

I = impulse = ?

m = mass = 0.15 kg

$\Delta v$ = change in speed of velocity = $v_{final}-v_{initial}=15cm/s-10cm/s=5cm/s=0.05m/s$ (1 m = 100 cm)

Now put all the given values in the above formula, we get:

$I=(0.15kg)\times (0.05m/s)$

$I=0.0075kg.m/s=0.0075N.s=7.5mN.s$ (1 N.s = 1000 mN.s)

Therefore, the impulse acting on the car is, 7.5 mN.s

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Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=

drek231 [11]

Answer:

Their final relative velocity is 0.190 m/s

Explanation:

The relative velocity of the satellites, v = 0.190 m/s

The mass of the first satellite, m₁ = 4.00 × 10³ kg

The mass of the second satellite, m₂ = 7.50 × 10³ kg

Given that the satellites have elastic collision, we have;

$v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2$

$v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2$

Given that the initial velocities are equal in magnitude, we have;

u₁ = u₂ = v/2

u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s

v₁ and v₂ = The final velocities of the satellites

We get;

$v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

$v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

The final relative velocity of the satellite, $v_f$ = v₁ + v₂

∴ $v_f$ = 0.095 + 0.095 = 0.190

The final relative velocity of the satellite, $v_f$ = 0.190 m/s

4 0

2 years ago

A student of weight 678 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude

baherus [9]

Answer:

(a) Magnitude of seat force at lowest point = 678 + 92 = 770

(b) Force exerted by the seat (highest point) = 310 N

(c) Force exerted by seat (lowest point) = 1046 N

Explanation:

At the highest point the magnitude of force by the seat = 586 N

The weight of the student = 678 N

Thus, at the highest point, the difference in this force is due to the centrifugal force acting on the boy. This can be calculated as follows:

Centrifugal Force = 678 - 586 = 92 N

(a) The magnitude of force on the student by the seat at the lowest point will have to appose both the student's weight and centrifugal force acting towards the ground. This would mean:

Magnitude of seat force at lowest point = 678 + 92 = 770 N

(b) If the wheel's speed is doubled, the centrifugal force will change accordingly. The equation of centrifugal force is given below:

$F = m * v^2 / r$

We can see from this that the force is directly proportional to the square of the velocity. So if the velocity is doubled, the centrifugal force increases four times.

So at the highest point the centrifugal force will decrease the force of weight acting on the seat. The seat force would then be:

Force exerted by the seat = Weight - Centrifugal force

Force exerted by the seat = 678 - (4 * 92)

Force exerted by the seat (highest point) = 310 N

(c) The force exerted by the seat at the lowest point will be the centrifugal force plus the weight.

This is:

Force exerted by seat = Centrifugal force + Weight

Force exerted by seat = (4 * 92) + 678

Force exerted by seat (lowest point) = 1046 N

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3 years ago