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3 years ago

In a collision, a 25.0 kg mass moving at 3.0 m/s transfers all of its momentum to a 5.0 kg mass.

Physics

1 answer:

nadezda [96]3 years ago

8 0

Answer:

Explanation:

The momentum of the 25 kg mass is

$p=mv$

$p=25kg*3m/s= 75kg*m/s$

If this whole momentum of the object is transferred to the 5.0 kg object then according to the law of conservation of momentum, the momentum of the 25.0 kg object must be transferred to the 5.0 kg object:

$75kg*m/s = 5.0kg*v$

$v=\dfrac{75}{5}$

$\boxed{v=15m/s}$

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When hydrogen is heated or subjected to an electric discharge, it emits light with specific visible wavelengths. A diffraction g

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$d sin\theta = m\lambda$

Here,

d = Distance between Slits

m = Order of the fringes

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$\text{Spacing between adjacent lines}$ = $d = 4926nm$

$\text{Wavelength of light} = \lambda = 656nm$

Rearranging to find the angle,

$sin\theta = \frac{m\lambda}{d}$

$\theta = sin^{-1}(\frac{m\lambda}{d})$

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Therefore the angle that the fourth order bright fringe occur for this specific wavelenth of light occur is 32.19°

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3 years ago

In terms of newton first law of motion why is it important to wear a seat belt

lilavasa [31]

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Seatbelts stop you

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Any passengers in the car will also be decelerated to rest if they are strapped to the car by seat belts.

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4 years ago

The basic barometer can be used as an altitude-measuring device in airplanes. The ground control reports a barometric reading of

kipiarov [429]

Answer:

Δh_air=714m

Explanation:

Given data

$P_{1}=753mmHg\\P_{2}=690mmHg\\ p_{air}=1.2kg/m^{3}\\ g=9.8m/s^{2}$

Solution

ΔP=P₁-P₂

=(ΔhHg)×pHg×g

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3 years ago

WILL GIVE BRAINLIEST TO CORRECT ANSWER PLEASE HELP ME

koban [17]

Answer:

The total distance is 381.5 [m]

Explanation:

In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the motorcyclist starts from rest, i.e. its initial speed is zero.

$v_{f} =v_{o} +(a*t)$

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0

a = acceleration = 2 [m/s²]

t = time = 7 [s]

Vf = 0 + (2*7)

Vf = 14 [m/s]

With this velocity, we can calculate the displacement using the following expression.

$v_{f} ^{2} =v_{o} ^{2} +2*a*x$

where

x = distance traveled [m]

14² = 0 + (2*7*x)

x = 196/(14)

x = 14 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.

We must calculate the final velocity.

$v_{f}= v_{i} +(a*t)$

Vf = final velocity [m/s]

Vi = initial velocity = 14 [m/s]

a = desacceleration = 4 [m/s²]

t = time = 8 [s]

Vf = 24 + (4*8)

Vf = 56 [m/s]

With this velocity, we can calculate the displacement using the following expression.

$v_{f} ^{2} =v_{o} ^{2} +2*a*x$

where

x = distance traveled [m]

56² = 14² + (2*4*x)

x = 2940/(8)

x = 367.5 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

Therefore the total distance is Xt = 14 + 367.5 = 381.5 [m].

4 0

3 years ago