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Oliga [24]
3 years ago
15

Which of the following states of matter has a define shape

Physics
1 answer:
Ulleksa [173]3 years ago
3 0

Answer:

Solids have a definite shape and volume. Liquids have a definite volume, but take the shape of the container. Gases have no definite shape or volume.

Explanation:

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You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is
Arlecino [84]

Answer:

a) v=0.5405\ m.s^{-1}

b) v_p'=0.1143\ m.s^{-1}

Explanation:

Given:

mass of ball, m_b=4\ kg

initial speed of the ball, v_b=10\ m.s^{-1}

mass of the person, m_p=70\ kg

a)

Using the conservation of linear momentum:

When the person catches the ball, assuming that the person catches it with an impact without absorbing the shock.

m_b.v_b=(m_b+m_p)v

4\times 10=(4+70)\times  v

v=0.5405\ m.s^{-1}

b)

When the ball hits the person and bounces off with the velocity of v_b'=8\ m.s^{-1}.

Using the conservation of linear momentum:

m_b.v_b+m_p.v_p=m_b.v_b'+m_p.v_p'

where:

v_b'= final speed of the ball after collision

v_p'= final speed of the person after collision

v_p= initial velocity of the person = 0

putting the respective values in the above eq.

4\times 10+0=4\times 8+70\times v_p'

v_p'=0.1143\ m.s^{-1}

8 0
3 years ago
Read 2 more answers
What is an analogy that describes the process of electron carriers?
Vladimir79 [104]
<span>An analogy is a comparison between one thing and another, typically for the purpose of explanation or clarification.</span><span>

An analogy of an electron carrier molecule and high energy electrons can be just like moving a potato. </span>A laundry basket filled with warm laundry can also be compared to an electron carrier. In this analogy, the laundry basket represents the electron carrier and the warm laundry represents the high energy electrons. There is another analogy that describes the process of electron carriers is a laundry basket filled with warm laundry can also be compared to an electron carrier. In this analogy, the laundry basket represents the electron carrier and the warm laundry represents the high energy electrons.  
4 0
3 years ago
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spectroscope or a spectrometer is an instrument which is used for separating the components of light, which have different wavel
ohaa [14]

Answer:

The answer is B

Explanation:

The line spectrum shows distinct light different energies and wavelengths

7 0
3 years ago
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Oil having a density of 922kg/m^3 floats on water. A rectangular block of wood 3.97 cm high and with a density of 963 kg/m^3 flo
Blizzard [7]

Explanation:

For the equilibrium:

\rho_{wood}gh-\rho_{oil}g(h-x)-\rho_{water}gx=0ρ

wood

gh−ρ

oil

g(h−x)−ρ

water

gx=0

\rho_{wood}h-\rho_{oil}(h-x)-\rho_{water}x=0ρ

wood

h−ρ

oil

(h−x)−ρ

water

x=0

(974)(3.97)-928(3.97-x)-1000x=0(974)(3.97)−928(3.97−x)−1000x=0

x=2.54\ cmx=2.54 cm

3 0
2 years ago
The earth's radius is 6.37×106m; it rotates once every 24 hours. What is the earth's angular speed? What is the speed of a point
Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

c) v1 = 440.433 m/s

Explanation:

Given:-

- The radius of the earth,  R = 6.37 * 10 ^6 m

- The time period for 1 revolution T = 24 hrs

Find:

What is the earth's angular speed?

What is the speed of a point on the equator?

What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

Solution:

- The angular speed w of the earth can be related with the Time period T of the earth revolution by:

                                  w = 2π / T

                                  w = 2π / 24*3600

                                  w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

                                 v1 = R*w

                                 v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

                                 v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

                                 π/2  ........... s

                                 x     ............ 1/5 s

                                 x = π/2*5 = 18°    

- The radius of the earth R' at point where θ = 18° from the equator is:

                                R' = R*cos(18)

                                R' = (6.37 * 10 ^6)*cos(18)

                                R' = 6058230.0088 m

- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

                              v2 = R'*w

                              v2 = (6058230.0088)*(7.27 * 10^-5)

                              v2 = 440.433 m/s

5 0
3 years ago
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