All categories

3 years ago

.Please explain to me how to find the momentum

Physics

1 answer:

nexus9112 [7]3 years ago

8 0

Answer:

momentum = mass x velocity

Explanation:

So ultimately, find the mass of the object (be careful since this is not the same as weight. Weight is your gravitational pull to the earth. This means that your weight different amounts on different planets. But mass stays the same on all planets, it's what most people call their weight.) and then find the velocity (2 meters/second north, etc.)

Now you can apply this to your problem

You might be interested in

Global Precipitation Measurement (GPM) is a tool scientists use to forecast weather. Which statements describe GPM? Select three

lyudmila [28]

Answer:

B.It is a satellite that collects data about rain and snow

C.Its orbit covers 90 percent of Earth’s surface

F.The sensors measure microwaves

5 0

3 years ago

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0

3 years ago

A train 471 m long is moving on a straight track with a speed of 75.1 km/h. The engineer applies the brakes at a crossing, and l

hram777 [196]

Answer:

$t = 37.6 s$

Explanation:

As we know that train is initially moving with the speed

$v_i = 75.1 km/h$

now we know that

$v_i = 20.86 m/s$

now the final speed of the train when it crossed the crossing

$v_f = 15 km/h$

$v_f = 4.17 m/s$

now we can use kinematics here

$v_f^2 - v_i^2 = 2 a d$

$4.17^2 - 20.86^2 = 2 a(471)$

$a = -0.44 m/s^2$

Now the time to cross that junction is given as

$v_f - v_i = at$

$4.17 - 20.86 = a(-0.44)$

$t = 37.6 s$

4 0

2 years ago

Which of the following is a unit of acceleration?

olasank [31]

${\tt{\red{\underline{\underline{\huge{Answer:}}}}}}$

$\longrightarrow$ The rate of change of velocity per unit time is called acceleration.

$\longrightarrow$ Its SI unit is m/s².

$\huge\boxed{\fcolorbox{blue}{red}{Thank you}}$

7 0

2 years ago

Three beads are placed on the vertices of an equilateral triangle of side d = 3.4cm. The first bead of mass m1=140gis placed on

Vlad [161]

Answer:

Xcm = 1.95 cm and Ycm = 1.76 cm

Explanation:

The very useful concept of mass center is

R cm = 1/M ∑ $m_{i}$ $r_{i}$

Where ri, mi are the mass positions of the bodies from some reference point by selecting and M is the total mass of the body.

Let's look for the total mass

M = m₁ + m₂ + m₃

M = 140 + 45 + 85

M = 270 g

Let's look for the position of each point

Point 1. top vertex, if the triangle has as side d

R₁ = d / 2 i ^ + d j ^

R₁ = (1.7 cm i ^ + 3.4 j ^) cm

Point 2. left vertex. What is the origin of the system?

R₂ = 0

Point 3. Right vertex

R₃ = d i ^

R₃ = 3.4 i ^ cm

a) The x component of the massage center

Xcm = 1 / M (m₁ x₁ + m₂ x₂ + m₃ x₃)

Xcm = 1 / M (m₁ d / 2 + 0 + m₃ d)

Xcm = d / M (m₁ / 2 + m₃)

b) Let's write the mass center component x

Xcm = 1/270 (1.7 140 + 0 + 3.4 85)

Xcm = 238/270

Xcm = 1.95 cm

c) let's find the component and center of mass

Ycm = 1 / M (m₁ y₁ + m₂ y₂ + m₃ y₃)

Ycm = 1 / M (m₁ d + 0 + 0)

Ycm = m₁ / M d

d) let's calculate

Y cm = 1/270 (140 3.4 + 0 + 0)

Ycm = 1.76 cm

8 0

3 years ago