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Alla [95]
3 years ago
12

Which of these does not have volume? select one:

Physics
2 answers:
lozanna [386]3 years ago
6 0
I think the answer is b, energy.
Temka [501]3 years ago
4 0

The answer is B. Energy does not have any volume.

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What does the object on the screen below model?
dybincka [34]

Answer:

A, molecules

Explanation:

Molecules were viewable around the time period of the computers shown on the screen

5 0
3 years ago
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In a World Cup soccer match, Juan is running due north toward the goal with a speed of 7.60 m/s relative to the ground. A teamma
spayn [35]

Answer:

Part a)

v_{bj} = 11.03 m/s

Part b)

\theta = 4.57 degree East of South

Explanation:

Part a)

Velocity of Juan is given as

v_1 = 7.60 m/s \hat j

velocity of the ball is given as

v_2 = 12.9(cos31.4 \hat i + sin31.4\hat j)

now we have

v_2 = 11\hat i + 6.72\hat j

Part a)

We need to find velocity of ball with respect to Juan

so it is given as

v_{bJ} = \vec v_b - \vec v_j

v_{bj} = 11\hat i + 6.72 \hat j - 7.6\hat j

v_{bj} = 11\hat i - 0.88\hat j

magnitude of the speed is given as

v_{bj} = \sqrt{11^2 + 0.88^2}

v_{bj} = 11.03 m/s

Part b)

direction of velocity of the ball

tan\theta = \frac{v_y}{v_x}

tan\theta = \frac{-0.88}{11}

\theta = 4.57 degree East of South

3 0
3 years ago
HELP ASAP TIMED TEST
balu736 [363]

Answer:

<em>Correct choice: b 4H</em>

Explanation:

<u>Conservation of the mechanical energy</u>

The mechanical energy is the sum of the gravitational potential energy GPE (U) and the kinetic energy KE (K):

E = U + K

The GPE is calculated as:

U = mgh

And the kinetic energy is:

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

Where:

m = mass of the object

g = gravitational acceleration

h = height of the object

v = speed at which the object moves

When the snowball is dropped from a height H, it has zero speed and therefore zero kinetic energy, thus the mechanical energy is:

U_1 = mgH

When the snowball reaches the ground, the height is zero and the GPE is also zero, thus the mechanical energy is:

\displaystyle U_2=\frac{1}{2}mv^2

Since the energy is conserved, U1=U2

\displaystyle mgH=\frac{1}{2}mv^2    \qquad\qquad [1]

For the speed to be double, we need to drop the snowball from a height H', and:

\displaystyle mgH'=\frac{1}{2}m(2v)^2

Operating:

\displaystyle mgH'=4\frac{1}{2}m(v)^2 \qquad\qquad [2]

Dividing [2] by [1]

\displaystyle \frac{mgH'}{mgH}=\frac{4\frac{1}{2}m(v)^2}{\frac{1}{2}m(v)^2}

Simplifying:

\displaystyle \frac{H'}{H}=4

Thus:

H' = 4H

Correct choice: b 4H

4 0
3 years ago
Review Conceptual Example 6 as background for this problem. A car is traveling to the left, which is the negative direction. The
DiKsa [7]

Answer:

(a) 1.21 m/s² (b) 1.75 m/s²

Explanation:

The initial speed of the car, u = 17.8 m/s

Case 1.

Final speed of the car, v = 23.5 m/s

Time, t = 4.68-s

Acceleration = rate of change of velocity

a=\dfrac{23.5 -17.8 }{4.68}\\\\a=1.21\ m/s^2

Case 2.

Final speed of the car, v = 15.3 m/s

a=\dfrac{23.5 -15.3}{4.68}\\\\a=1.75\ m/s^2

Hence, this is the required solution.

3 0
2 years ago
Which statement did Kepler’s investigations of the movement of the planets explain?
wlad13 [49]

Answer:

It’s the fourth one :)

Explanation:

7 0
2 years ago
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