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Kobotan [32]
2 years ago
7

He shoots the tree stump, which has a mass of (M), with a bullet of mass (m) traveling at some velocity (vbullet), and the bulle

t lodges itself inside the tree stump before the system of objects slides along the ice. What is the speed of the tree stump-bullet system immediately after the collision?
Physics
1 answer:
Ymorist [56]2 years ago
4 0

Answer:

Explanation:

Given

mass of tree stump is M

mass bullet is m

velocity of bullet is v

Conserving momentum for bullet and tree stump

Initial Momentum P_i=mv

Suppose v_0 is the velocity of the system

Final Momentum P_f=(M+m)v_0

Initial momentum =Final Momentum

mv=(m+M)v_0

v_0=\frac{mv}{m+M}

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azamat

Answer:

the answer is option d

because in oven which works through electricity ,it will bake the cake (heat energy is produced)

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3 years ago
When droplets of water in the atmosphere act like prisms, the colors in sunlight undergo?
dexar [7]
White light is all the colours of light combined. When the droplets act like prisms, they split the white light into all its colours and also slightly bend the different colours. This is how a rainbow is formed.
4 0
2 years ago
How do i do a Wavelength and frequency problem
KengaRu [80]

Answer:

wavelength = v/f or wavelength equals to velocity over frequency

frequency= v/w or velocity over wavelength

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7 0
3 years ago
A roller skater of 47kg moving with a velocity of 12 m/s to the east picks up a bag of 6.0 kg. What is the final velocity of the
11Alexandr11 [23.1K]

Answer:

v_f = 10.85 m/s

Explanation:

We will apply the law of conservation of momentum here:

m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f}+m_{2}v_{2f}\\

where,

m₁ = mass of roller skater = 47 kg

m₂ = mass of bag = 6 kg

v_1i = initial speed of roller skater = 12 m/s

v_2i = initial speed of the bag = 0 m/s

v_1f = final speed of the roller skater = ?

v_2f = final speed of the bag = ?

Both the bag and the skater will have same speed at the end because kater is carrying the bag:

v_1f = v_2f = v_f

Therefore, the equation will become:

(47\ kg)(12\ m/s)+(6\ kg)(0\ m/s)=(47\ kg)(v_{f})+(5\ kg)(v_{f})\\564\ N.s = (47\ kg+5\ kg)(v_{f})\\v_{f} = \frac{564\ N.s}{52\ kg}\\

<u>v_f = 10.85 m/s</u>

4 0
2 years ago
Ch 27 HW Exercise 27.12 10 of 20 Constants A horizontal rectangular surface has dimensions 2.80 cm by 3.15 cm and is in a unifor
deff fn [24]

Answer:

Magnetic field, B = 0.88 T

Explanation:

It is given that,

The dimension of rectangular surface is 2.80 cm by 3.15 cm. The area of rectangular surface is, A=8.82\ cm^2=0.000882\ m^2

Angle between the uniform magnetic field and the horizontal, \theta=31

Magnetic flux, \phi=4\times 10^{-4}\ Wb

Let B is the magnitude of magnetic field in which the rectangular surface is placed. It is given by :

\phi=BA\ cos\theta

\theta is the angle between magnetic field and the area

Here, \theta=90-31=59^{\circ}

B=\dfrac{\phi}{A\ cos\theta}

B=\dfrac{4\times 10^{-4}}{0.000882\times cos(59)}

B = 0.88 T

So, the magnitude of magnetic field is 0.88 T. Hence, this is the required solution.

7 0
3 years ago
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