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BaLLatris [955]
3 years ago
13

The volleyball team practiced their serves, rotations, and ________. Which choice completes the sentence with correct parallel s

tructure?
a. spikes.
b. spiking.
c. spikes the ball.
d. spikes over the net.
Physics
2 answers:
Phoenix [80]3 years ago
5 0

Answer

Correct Answer is a.(spikes)

Explanation:

option a is correct because here serves, rotation and spikes all the three words are using as object.

The volleyball team practiced their serves, rotations, and spikes.

option b cannot be correct because here spiking is used as a verb and sentence(The volleyball team practiced their serves, ) is in past indefinite form and we cant use word spiking here because this form of verb is used in continues form of sentences only.

Option c cannot be correct as because in this option ""spikes the ball" spikes is used as a verb  which can be used in present indefinite form of sentence only but here sentence is in past indefinite form.

Again option d cannot be correct because  in this option ""spikes over the net" spikes is used as a verb  which can be used in present indefinite form of sentence only but here sentence is in past indefinite form.

pychu [463]3 years ago
3 0

Answer:

Option (A)

Explanation:

Parallel structure in a sentence refers to the usage of the same type of words that describes multiple purposes comprised of equal significance. These parallel structures are usually connected in a sentence by using coordinating conjunctions. For example, and, or.

In the given question, the correct answer is "spikes" that completes the above sentence with the correct parallel structure.

A volleyball spike is defined as a game-plan that is usually set by the team players, where the ball is to pass through over the net in such a way that nobody in the opposite team can receive the ball, thereby adding points to the team that throws the ball. In order to do this, a good practice is required by the players.

Thus, the correct answer is option (A).

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Atoms of the same element that have the same number of protons but different numbers of neutrons are called_____
saveliy_v [14]
The answer is:  "isotopes" .
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7 0
3 years ago
In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it
bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

      $=\sqrt{2 \times 9.8 \times 0.35}$      (h = 35 cm = 0.35 m)

      = 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}

      $=\frac{0.175}{0.525}+0$

     = 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

 

3 0
3 years ago
which planet should punch travel to if his goal is to weigh in at 118 lb? refer to the table of planetary masses and radii given
Harrizon [31]

The planet that Punch should travel to in order to weigh 118 lb is Pentune.

<h3 /><h3 /><h3>The given parameters:</h3>
  • Weight of Punch on Earth = 236 lb
  • Desired weight = 118 lb

The mass of Punch will be constant in every planet;

W = mg\\\\m = \frac{W}{g}\\\\m = \frac{236}{g}

The acceleration due to gravity of each planet with respect to Earth is calculated by using the following relationship;

F = mg = \frac{GmM}{R^2} \\\\g = \frac{GM}{R^2}

where;

  • M is the mass of Earth = 5.972 x 10²⁴ kg
  • R is the Radius of Earth = 6,371 km

For Planet Tehar;

g_T =\frac{G \times 2.1M}{(0.8R)^2} \\\\g_T = 3.28(\frac{GM}{R^2} )\\\\g_T = 3.28 g

For planet Loput:

g_L =\frac{G \times 5.6M}{(1.7R)^2} \\\\g_L = 1.94(\frac{GM}{R^2} )\\\\g_L = 1.94g

For planet Cremury:

g_C =\frac{G \times 0.36M}{(0.3R)^2} \\\\g_C = 4(\frac{GM}{R^2} )\\\\g_C = 4 g

For Planet Suven:

g_s =\frac{G \times 12M}{(2.8R)^2} \\\\g_s = 1.53(\frac{GM}{R^2} )\\\\g_s = 1.53 g

For Planet Pentune;

g_P =\frac{G \times 8.3 }{(4.1R)^2} \\\\g_P = 0.5(\frac{GM}{R^2} )\\\\g_P = 0.5 g

For Planet Rams;

g_R =\frac{G \times 9.3M}{(4R)^2} \\\\g_R = 0.58(\frac{GM}{R^2} )\\\\g_R = 0.58 g

The weight Punch on Each Planet at a constant mass is calculated as follows;

W = mg\\\\W_T = mg_T\\\\W_T = \frac{236}{g} \times 3.28g = 774.08 \ lb\\\\W_L = \frac{236}{g} \times 1.94g =457.84 \ lb\\\\ W_C = \frac{236}{g}\times 4g = 944 \ lb \\\\ W_S = \frac{236}{g} \times 1.53g = 361.08 \ lb\\\\W_P = \frac{236}{g} \times 0.5 g = 118 \ lb\\\\W_R = \frac{236}{g} \times 0.58 g = 136.88 \ lb

Thus, the planet that Punch should travel to in order to weigh 118 lb is Pentune.

<u>The </u><u>complete question</u><u> is below</u>:

Which planet should Punch travel to if his goal is to weigh in at 118 lb? Refer to the table of planetary masses and radii given to determine your answer.

Punch Taut is a down-on-his-luck heavyweight boxer. One day, he steps on the bathroom scale and "weighs in" at 236 lb. Unhappy with his recent bouts, Punch decides to go to a different planet where he would weigh in at 118 lb so that he can compete with the bantamweights who are not allowed to exceed 118 lb. His plan is to travel to Xobing, a newly discovered star with a planetary system. Here is a table listing the planets in that system (<em>find the image attached</em>).

<em>In the table, the mass and the radius of each planet are given in terms of the corresponding properties of the earth. For instance, Tehar has a mass equal to 2.1 earth masses and a radius equal to 0.80 earth radii.</em>

Learn more about effect of gravity on weight here: brainly.com/question/3908593

5 0
2 years ago
Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S
Alja [10]

Answer:

16294 rad/s

Explanation:

Given that

M(ns) = 2M(s), where

M(s) = 1.99*10^30 kg, so that

M(ns) = 3.98*10^30 kg

Again, R(ns) = 10 km

Using the law of gravitation, the force between the Neutron star and the sun is..

F = G.M(ns).M(s) / R²(ns), where

G = 6.67*10^-11, gravitational constant

Again, centripetal force of the neutron star is given as

F = M(ns).v² / R(ns)

Recall that v = wR(ns), so that

F = M(s).w².R(ns)

For a circular motion, it's been established that the centripetal force is equal to the gravitational force, hence

F = F

G.M(ns).M(s) / R²(ns) = M(s).w².R(ns)

Making W subject of formula, we have

w = √[{G.M(ns).M(s) / R²(ns)} / {M(s).R(ns)}]

w = √[{G.M(ns)} / {R³(ns)}]

w = √[(6.67*10^-11 * 3.98*10^30) / 10000³]

w = √[2.655*10^20 / 1*10^12]

w = √(2.655*10^8)

w = 16294 rad/s

7 0
3 years ago
Ax = 4.3 m
Dimas [21]

Answer:

the answer will be option no b plss mark me brainliest

7 0
3 years ago
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