All categories

3 years ago

Hello, I need help. Due right now

Physics

2 answers:

ryzh [129]3 years ago

7 0

Answer: See below

Explanation:

$\\ \mathrm{Given:} \\\mathrm{Mass \ of \ first \ car}$\left(m_{1}\right)=1383 \mathrm{~kg}$ \\ Velocity $\left(\overrightarrow{V_{1}}\right)=-11.2\ {\math} \mathrm{m} / \mathrm{s}$\\ Mass of second car -$\left(m_{2}\right)=1732 \mathrm{~kg}$\\ Velocity $\left(\vec{v}_{2}\right)=31.3 {\math} \mathrm{m} / \mathrm{s}$$

$m_{1} \vec{v}_{1}+m_{2} \vec{v}_{2}=\left(m_{1}+m_{2}\right) \vec{v} \\1383(-11.2 {\math})+1732(31.3 {\math}) \\=(1383+1732) \vec{v} \\-15489.6 {\math}+54211.6 {\math}=3115 \vec{v} \\ \vec{v}=(17.4 {\math}-4.972 {\math}) \ \mathrm{m/s}$

$\text { So magnitude } \vec{|v|} =\sqrt{(17.4)^{2}+(-4.972)^{2}} \\ =\sqrt{327.605} \\ =18.099 \mathrm \ {m/s} \\$

$\text { Direction } \\\theta=\tan ^{-1}\left(\frac{-4.972}{17.4}\right) \\\theta=-15.945^{\circ}\end{gathered}$

Therefore, both cars move with a velocity of 18.099 m/s in the direction of 15.945° downward from the x-axis (east)

olga_2 [115]3 years ago

3 0

Answer:

Explanation:

Apply conservation of momentum in N-S direction:

Initial momentum: 1383*(-11.2) + 1732*(0) = -15490

Final momentum: (1383 + 1732) * Final N-S velocity = -15490

Final velocity = 1549/(1383+1732) = -4.97 m/s (-ve means South)

Repeat the same for E-W direction:

Initial momentum: 1383*(0) + 1732*(31.3) = 54211.6

Final momentum: (1383 + 1732) * Final E-W velocity = 54211.6

Final velocity = 54211.6/(1383+1732) = 17.4 m/s (+ve means East)

Combining, the final velocity = sqrt (-4.97^2 + 17.4^2) = 18.10 m/s

at direction of arctan(17.4/4.97) = 74 degree South of East.

You might be interested in

How can Magnets cause objects to have kinetic energy?

lianna [129]

Answer:

If there is a system of magnets being held in place, there is potential energy. When you let go, the potential energy converts to kinetic energy and the magnets move. Putting the system of magnets close together which creates an opposing force.

Explanation:

4 0

3 years ago

Read 2 more answers

Which characteristic should a good scientific question have? A) It should lead to a hypothesis that is not testable. B) it shoul

yarga [219]

Answer:

B. IT should have a very broad focus with many variables.

4 0

3 years ago

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is:

$F_{B_y}=qvB_y$

where:

$q=+4.9\cdot 10^{-6}C$ is the charge

$v=345 m/s$ is the velocity

$B_y = +1.9 T$ is the magnetic field

Substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

For the field $B_x$ , the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

$F_{B_x}=qvB_x$

And by substituting,

$F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field $B_y$ , the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

$F_{B_y}=qvB_y$

And by substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0

4 years ago

How far does a 26" bicycle wheel travel at 8 miles per hour?

Luda [366]

Well if it was traveling for an hour then the answer is 8 miles.

8 0

4 years ago

Read 2 more answers

The density of a hippo is approximately 1030kg/m^3,so it sinks to the bottom of the freshwater lakes and rivers. A 1500kg hippo

hram777 [196]

Answer:

14,700 N

Explanation:

The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:

$W-N=0$ (1)

the weight of the hyppo is

$W=mg=(1500 kg)(9.8 m/s^2)=14,700 N$

where m is the hyppo's mass and g is the gravitational acceleration; therefore, solving eq.(1) for N, we find

$N=W=14,700 N$

8 0

3 years ago