describe the forces acting on a car as it movies along a level highway in still air at a constant speed

Where is the natural light display called aurora borealis located?

Xelga [282]

The natural light display called aurora borealis is located in the northern

hemisphere.

There are two types of aurora which are called aurora borealis and aurora

australis. The aurora borealis is located in the Northern hemisphere while

the aurora australis is located in the Southern hemisphere.

They receive their energy through the interaction of charged particles

on the Sun and Earth to produce the light display. An example

of the interaction involves solar wind with atoms of the upper atmosphere.

Read more on brainly.com/question/20191244

5 0

2 years ago

A 1 170.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 800.0 kg t

kykrilka [37]

Explanation:

It is given that,

Mas of the car, $m_1=1170\ kg$

Initial speed of the car, $u_1=25\ m/s$

Mass of the truck, $m_2=9800\ kg$

Initial speed of the car, $u_2=20\ m/s$

Final speed of the car, $v_1=18\ m/s$

(a) It is a case of elastic collision. Let $v_2$ is the final velocity of the truck right after the collision. Using the conservation of linear momentum to find it :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\dfrac{1170\times 25+9800\times 20-1170\times 18}{9800}$

$v_2=20.83\ m/s$

(b) Initial kinetic energy is given by :

$k_i=\dfrac{1}{2}(m_1u_1^2+m_2u_2^2)$

$k_i=\dfrac{1}{2}\times (1170\times (25)^2+9800\times (20)^2)$

$k_i=2325625\ J$

Final kinetic energy is given by :

$k_f=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)$

$k_f=\dfrac{1}{2}\times (1170\times (18)^2+9800\times (20.83)^2)$

$k_f=2315595.61\ J$

The change in mechanical energy of the car truck system in the collision:

$\Delta K=k_f-k_i$

$\Delta K=2315595.61-2325625$

$\Delta k=-10029.39\ J$

The loss in kinetic energy is 10029.39 Joules.

(c) The change in mechanical energy gets changed energy gets changed in the form of heat and light.

Hence, this is the required solution.

6 0

2 years ago

The temperature at which all molecular motion stops is

34kurt

Answer:

Zero Kelvin

Explanation:

The average kinetic energy of the particles in a gas is related to the absolute temperature of the gas by (for an ideal monoatomic gas):

$E_K = \frac{3}{2}kT$

where

k is the Boltzmann constant

T is the absolute temperature

The average kinetic energy is the energy possessed by the particles due to their motion; we see that this energy becomes zero when T = 0, which means when the substance reaches a temperature of zero Kelvin. Therefore, this means that at this temperature all the particles stop moving.

7 0

2 years ago

A 12.0-kg package in a mail-sorting room slides 2.00 m down a chute that is inclined at 53.0° below the horizontal. The coeffici

zvonat [6]

Answer

given,

mass of the package = 12 kg

slides down distance = 2 m

angle of inclination = 53.0°

coefficient of kinetic friction = 0.4

a) work done on the package by friction is

W_f = -μk R d

= -μk (mg cos 53°)(2.0)

=-(0.4)(8.0 )(9.8)(cos 53°)(2.0)

= -37.75 J

b)

work done on the package by gravity is

W_g = m (g sin 53°) d

= (8.0 )(9.8 )(sin 53°)(2.0 )

=125.23 J

c)

the work done on the package by the normal force is

W_n = 0

d)

the net work done on the package is

W = -37.75 + 125.23 + 0

W = 87.84 J

7 0

3 years ago

By applying a force of one Newton, one can hold a body of mass _

SVEN [57.7K]

Answer:

By applying a force of one Newton, one can hold a body of mass of 102 gram.

Explanation:

Force is the pull or push of an object. It can be mathematically measured as, F= m* g.

where, F= force in newton

m= mass in kg

g= acceleration due to gravity ( $meter/second^{2}$ )

For 1 newton force,

F= m* 9.8

or, m= $\frac{1}{9.8}$ = 0.102 kg

or, m= 102 gram.

Hence, 102 gram mass can be hold by one Newton force.

3 0

3 years ago