All categories

2 years ago

A car of mass 600 Kg is moving at 15m/s. Calculate its momentum.

Physics

1 answer:

mr_godi [17]2 years ago

3 0

Mass=600kg
Velocity =15m/s
Momentum(p)=?
Now,
P=mass x velocity
=600x15
=9000kgm/s

You might be interested in

Water is discharged from a pipeline at a velocity v (in ft/sec) given by v = 1306p(1/2), where p is the pressure (in psi). If th

shepuryov [24]

Answer:

$a=38.5 ft/sec^{2}$

Explanation:

Note that acceleration is the rate change of velocity i.e

$acceleration=\frac{change in velocity}{change in time}\\a=\frac{dv}{dt} \\$ .

Since the velocity is giving as a variable dependent on the pressure, we have to differentiate implicitly both side with respect to time,i.e

$\frac{dv}{dt}=1306*(1/2)p^{-1/2}\frac{dp}{dt} \\$

if we substitute value for the pressure as giving in the question and also since the rate change of pressure is 0.354psi/sec, we have

$a=653*0.1667*0.354\\$

$a=\frac{dv}{dt}=653(36)^{-1/2}*0.354\\ a=38.5 ft/sec^{2}$

3 0

2 years ago

Write your answer to the question below.

ElenaW [278]

Answer:

Hi... Potential energy is converted to kinetic energy and kinetic energy is converted to potential energy

8 0

2 years ago

A 0.0780 kg lemming runs off a

kotegsom [21]

Answer:

5.01 J

Explanation:

Info given:

mass (m) = 0.0780kg

height (h) = 5.36m

velocity (v) = 4.84 m/s

gravity (g) = 9.81m/s^2

1. First, solve for Kinetic energy (KE)

KE = 1/2mv^2

1/2(0.0780kg)(4.84m/s)^2 = 0.91 J

so KE = 0.91 J

2. Next, solve for Potential energy (PE)

PE = mgh

(0.0780kg)(9.81m/s^2)(5.36m) = 4.10 J

so PE = 4.10 J

3. Mechanical Energy , E = KE + PE

Plug in values for KE and PE

KE + PE = 0.91J + 4.10 J = 5.01 J

4 0

2 years ago

PLEASE HELP ME WHATS THE ASNWRR

stealth61 [152]

Answer:

think its b i dunno

Explanation:

7 0

1 year ago

Read 2 more answers

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

2 years ago