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2 years ago

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A 1,100 kg car comes uniformly to a stop. If the vehicle is accelerating at -1.2 m/s2, which force is closest to the net force a

cting on the vehicle?
Group of answer choices

1 answer:

natulia [17]2 years ago

3 0

65234. i say this because the net force is stopping the jig

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A flywheel with radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of 0.400 rad/s2.For a po

oksian1 [2.3K]

Answer:

$0.12\ m/s^2$

Explanation:

Given that,

The radius of a flywheel, r = 0.3 m

Angular acceleration of a flywheel, $\alpha =0.4\ rad/s^2$

We need to find the magnitude of the tangential acceleration after 2.00 s of acceleration.

The relation between the tangential and angular acceleration is given by :

$a_t=r\alpha \\\\a_t=0.3\times 0.4\\\\a_t=0.12\ m/s^2$

So, the required magnitude of tangential acceleration is $0.12\ m/s^2$ .

6 0

2 years ago

A distant galaxy has a redshift z = 5.82 and a recessional velocity vr = 287,000 km/s (about 96% of the speed of light.) Notice

vlabodo [156]

Answer: 4100 Mpc

Explanation:

Since H o = 70 km/s/Mpc

Redshift z = 5.82

Recessional velocity vr = 287,000 km/s

Then, the distance to the galaxy in light years will be:

= Recessional velocity / H o

= 287000 / 70

= 4100 Mpc

7 0

2 years ago

The kinetic energy of a rotating body is generally written as K=12Iω2, where I is the moment of inertia. Find the moment of iner

stira [4]

Answer:

See explanation

Explanation:

We have a mass $m$ revolving around an axis with an angular speed $\omega$ , the distance from the axis is $r$ . We are given:

$\omega = 10 [rad/s]\\r=0.5 [m]\\m=13[Kg]$

and also the formula which states that the kinetic rotational energy of a body is:

$K =\frac{1}{2}I\omega^2$ .

Now we use the kinetic energy formula

$K =\frac{1}{2}mv^2$

where $v$ is the tangential velocity of the particle. Tangential velocity is related to angular velocity by:

$v=\omega r$

After replacing in the previous equation we get:

$K =\frac{1}{2}m(\omega r)^2$

now we have the following:

$K =\frac{1}{2}m(\omega r)^2 =\frac{1}{2}Iw^2$

therefore:

$mr^2=I$

then the moment of inertia will be:

$I = 13*(0.5)^2=3.25 [Kg*m^2]$

3 0

2 years ago

What is the ground-state electron configuration of a neutral atom of titanium?

Georgia [21]

Answer:

$\bold {1s^22s^22p^63s^23p^64s^23d^2}$

Explanation:

electronic configuration of an atom is the spatial arrangement of the electrons around the nucleus in the energy orbits

We know that the atomic number of Titanium is 22

The electronic configuration of Titanium is

$1s^22s^22p^63s^23p^64s^23d^2$

3 0

2 years ago

An iron ball and an aluminum ball of mass 100 g each are heated to the same temperature and then cooled to a temperature of 20°

MrRa [10]

Answer:

The specific heat of aluminum is greater.

Explanation:

It lost the most heat.

7 0

2 years ago

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