All categories

3 years ago

A stationary shell is exploded in to three fragments A, B, C of masses in the ratio 1:2:3. A travels

Physics

1 answer:

spin [16.1K]3 years ago

7 0

Answer:

20 m/s

Explanation:

If the mass of fragment A is m, then the mass of fragment B is 2m, and the mass of fragment C is 3m.

The velocity of A is 60 m/s at angle 0°.

The velocity of B is 30 m/s at angle 120°.

The velocity of C is v at angle θ.

In the x direction:

Momentum before = momentum after

(m + 2m + 3m) (0) = m (60 cos 0°) + 2m (30 cos 120°) + 3m (v cos θ)

0 = 60m − 30m + 3m v cos θ

0 = 30m + 3m v cos θ

-30m = 3m v cos θ

-10 = v cos θ

In the y direction:

Momentum before = momentum after

(m + 2m + 3m) (0) = m (60 sin 0°) + 2m (30 sin 120°) + 3m (v sin θ)

0 = 0 + 30√3 m + 3m v sin θ

-30√3m = 3m v sin θ

-10√3 = v sin θ

Square the two equations and add together:

(-10)² + (-10√3)² = (v cos θ)² + (v sin θ)²

100 + 300 = v² cos² θ + v² sin² θ

400 = v² (cos² θ + sin² θ)

400 = v²

v = 20

The speed of fragment C is 20 m/s.

You might be interested in

A wooden dowel (cylinder) has a diameter of 2.20 cm. it floats in water with 0.60 cm of its diameter above water. determine the

Gre4nikov [31]

In the image the situation is illustrated, also provided is a simple proof for the circle segment area. The problem can be solved without actually knowing the lenght of the dowel.
We will only care for the circular seection of the dowel which is partially submerged. Recall that for a floating body the weight of displaced water is equal to the body's weight.
For the dowel's weight we have:
$D_w=m.g$
where g is the gravitational constant and m the dowel's mass.

Now for the displaced water weight:
$W_w=V_{displaced}.\rho_{water}.g$
where $\rho_{water}$ is the water density, which happens to be $1000kg/m^3$ .
we now have the following:
$V_{displaced}.\rho_{water}.g=m.g$
$\implies V_{displaced}.\rho_{water}=m$
but:
$m=\rho_{dowel}.V$
being V the volume of the dowel, putting the above toghether gives us:
$V_{displaced}.\rho_{water}=V.\rho_{dowel}$
Now $V_{displaced}$ is the volume of the dowel that is submerged.The fraction of the dowel submerged will be according to the image:
$\frac{\pi r^2-\frac{1}{2}r^2(\theta-sin\theta)}{\pi r^2}=\frac{(\pi-\frac{1}{2}\theta+sin\theta)}{\pi}$
So the dowel's volume that is submerged is:
$V_{displaced}=V_{submerged}=\frac{(\pi-\frac{1}{2}\theta+sin\theta)}{\pi}.V$

Plug this into the previous expression to get:
$\frac{(\pi-\frac{1}{2}\theta+sin\theta)}{\pi}.V.\rho_{water}=V.\rho_{dowel}\\
\implies \rho_{dowel}=\frac{(\pi-\frac{1}{2}\theta+sin\theta)}{\pi}.\rho_{water}$

There's only one thing missing, the $\theta$ angle. Refer to the second image to get the following expression:
$\theta=2Acos(\frac{1}{2r})$
where r is the dowel's radius.
Plugging the above in the expression for the dowel's density we get:
$\frac{\left(\pi-Acos(\frac{1}{2r})+sin\left[ Acos\left( \frac{1}{2r}\right )\right]\right)}{\pi}.\rho_{water}\\
$
where:

$sin\left[ Acos\left( \frac{1}{2r}\right)\right]=\sqrt{1-\left( \frac{1}{2r}\right)^2}$
We finally get:
$\rho_{dowel}=\frac{\left(\pi-Acos(\frac{1}{2r})+\sqrt{1-\left( \frac{1}{2r}\right)^2}\right)}{\pi}.\rho_{water}$
This result is in $kg/m^3$ .

6 0

3 years ago

A battery with an emf of 24.0 V is connected to a resistive load. If the terminal voltage of the battery is 16.1 V and the curre

lions [1.4K]

Answer:

2.03 Ω

Explanation:

EMF: This can be defined as the potential difference of a cell when it is not delivering any current. The S.I unit of Emf is Volt.

The formula of emf is given as,

E = I(R+r)............................ Equation 1

Where E = Emf, I = current, R = External resistance, r = internal resistance.

Make r the subject of the equation

r = (E-IR)/I........................ Equation 2

Note: From ohm's law, V = IR.

r = (E-V)/I........................ Equation 3

Where V = Terminal voltage

Given: E = 24 V, I = 3.9 A, V = 16.1 V.

Substitute into equation 3

r = (24-16.1)/3.9

r = 7.9/3.9

r = 2.03 Ω

6 0

3 years ago

A .5 kg toy train car moving forward at 3 m/s collides with and sticks to a .8 kg toy car that is traveling at 2 m/s what is the

Viktor [21]

Here we have perfectly inelastic collision. Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:
$m_{1} * v_{1} + m_{2} * v_{2} = m_{1} * v'_{1}+ m_{2} * v'_{2}$

In case of perfectly inelastic collision v'1 and v'2 are same.

We are given information:
m₁=0.5kg
m₂=0.8kg
v₁=3m/s
v₂=2m/s
v'₁=v'₂=x

0.5*3 + 0.8*2 = 0.5*x + 0.8*x
1.5 + 1.6 = 1.3x
3.1 = 1.3x
x = 2.4 m/s

4 0

3 years ago

Based on the wall of conservation of energy how can we reasonably improve a machines ability to do work?

Margaret [11]

we can reduce the friction between its moving parts.

8 0

3 years ago

A wire carries current in the plane of this screen toward the top of the screen. The wire experiences a magnetic force toward th

GenaCL600 [577]

Answer:

The direction of the magnetic field causing this force is

In the plane of the screen and towards the bottom of the egde

Explanation:

This is by applying Fleming s right hand rule which explains that

When a conductor such as a wire attached to a circuit moves through a magnetic field, an electric current is induced in the wire due to Faraday's law of induction. The current in the wire can have two possible directions. Fleming's right-hand rule gives which direction the current flows.

The right hand is held with the thumb, index finger and middle finger mutually perpendicular to each other (at right angles), as shown in the diagram.[1]

The thumb is pointed in the direction of the motion of the conductor relative to the magnetic field.

The first finger is pointed in the direction of the magnetic field. (north to south)

Then the second finger represents the direction of the induced or generated current within the conductor (from the terminal with lower electric potential to the terminal with higher electric potential, as in a voltage source)

7 0

3 years ago