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3 years ago

Which of the following is an example of matter?

Physics

2 answers:

krek1111 [17]3 years ago

8 0

Answer:

(D) all of the above

Explanation:

All of the choices describe matter.

Gemiola [76]3 years ago

5 0

D cause I learned this in the 5th grade and some how remember this (I think because it’s all up in space)

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A 0.75 kg model car is moving west at a speed of 9.0 m/s when it collides head-on with a 2.00 kg model truck that is traveling e

Lyrx [107]

Answer:

2.5 m/s east

Explanation:

Let east be the positive direction for velocity.

The change in momentum of the 0.75 kg model car is ...

m1·v2 -m1·v1 = (0.75 kg)(11 m/s) -(0.75 kg)(-9 m/s)

= (0.75 kg)(20 m/s) = 15 kg·m/s

The change in momentum of the 2.0 kg model car is the opposite of this, so the total change in momentum is zero.

m2·v2 -m2·v1 = (2 kg)(v2 m/s) -(2 kg)(10 m/s) = 2(v2 -10) kg·m/s

The required relation is ...

15 kg·m/s = -2(v2 -10) kg·m/s

-7.5 = v2 -10 . . . . divide by -2

2.5 = v2 . . . . . . . add 10

The velocity of the model truck after the collision is 2.5 m/s east.

3 0

3 years ago

Osteoporosis is a condition that makes bones more

erastova [34]

the answer for this question is c

7 0

3 years ago

Suppose that the coefficient of kinetic friction between Zak's feet and the floor, while wearing socks, is 0.250. Knowing this,

insens350 [35]

Answer:

The distance travel before stopping is 1.84 m

Explanation:

Given :

coefficient of kinetic friction $\mu_{k} = 0.250$

Zak's speed $v = 3 \frac{m}{s}$

Gravitational acceleration $g = 9.8$ $\frac{m}{s^{2} }$

Work done by frictional force is given by,

$W = \Delta K$

$\mu _{k} mg d = \frac{1}{2} m v^{2}$

$d = \frac{v^{2} }{2 g \mu _{k} }$

$d = \frac{9}{2 \times 9.8 \times 0.250}$

$d = 1.84$ m

Therefore, the distance travel before stopping is 1.84 m

3 0

3 years ago

At an instant when the displacement is equal to a/2, what fraction of the total energy of the system is potential?

madreJ [45]

<span>At an instant when the displacement is equal to a/2, Potential energy U = 1/2ka(square) where a is displacement. when a= a/2 U = 1/4ka(square) U = E/4 Potential Energy = 1/4 Total energy</span>

4 0

3 years ago

Two tiny particles having charges 20.0 μC and 8.00 μC are separated by a distance of 20.0 cm What are the magnitude and directio

Alecsey [184]

Answer:

The magnitude and direction of electric field midway between these two charges is $10.8\times10^{5}\ N/C$ along AB.

Explanation:

Given that,

First charge $q_{1}= 20\mu C$

second charge $q_{2}= 8\mu C$

Distance = 20 cm

We need to calculate the electric field

For first charge,

Using formula of electric field

$E_{1}= \dfrac{kq_{1}}{r^2}$

Put the valueinto the formula

$E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}$

$E_{1}=18\times10^{5}\ N/C$

Direction of electric field along AB

We need to calculate the electric field

For second charge,

Using formula of electric field

$E_{2}= \dfrac{kq_{2}}{r^2}$

Put the valueinto the formula

$E_{2}=\dfrac{9\times10^{9}\times8\times10^{-6}}{10\times10^{-2}}$

$E_{2}=7.2\times10^{5}\ N/C$

Direction of electric field along AO

We need to calculate the net electric field at midpoint

$E_{net}=E_{1}-E_{2}$

$E_{net}=(18-7.2)\times10^{5}\ N/C$

$E_{net}=10.8\times10^{5}\ N/C$

Direction of net electric field along AB

Hence, The magnitude and direction of electric field midway between these two charges is $10.8\times10^{5}\ N/C$ along AB.

8 0

3 years ago