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solong [7]
2 years ago
12

Students can take the aspire test in ninth and grade

Physics
2 answers:
Elodia [21]2 years ago
6 0

Answer:

The correct answer is option B, tenth

Explanation:

ASPIRE test is basically given by students of ninth and tenth grade as pre-test to the ACT test. ACT ASPIRE test can be given from third grade till tenth grade. The scoring obtained in this test helps the student, teachers and parents to monitor the preparation of their kids towards completing a successful ACT test.  

Apart from ninth and tenth grade , students belonging to lower grades (i.e from 3rd grade to 8th grade) can opt for online test.  However, higher grades can opt for both online and paper based test.

Simora [160]2 years ago
3 0

<em>The answer is </em>Ninth <em>and </em>Tenth <em>grade so the answer would be</em> B

<em>I hope this helps you </em>


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Romashka [77]

Answer:

the first one is energy level

Explanation:

5 0
3 years ago
When an atom that has no charge looses two electrons it becomes a
alex41 [277]

Answer: it becomes a positive ion

Explanation:

So, when an atom loses 2 electrons there will be no change in the number of neutrons. Therefore, an isotope will not form. Thus, it is concluded that when an atom with no charge loses two electrons, it becomes a positive ion.

8 0
3 years ago
Read 2 more answers
A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el
Irina-Kira [14]

Answer:

71.19 C

Explanation:

25C = 25 + 273 = 298 K

Applying the ideal gas equation we have

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:

T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C

4 0
3 years ago
A capacitor of capacitance C has charge Q and stored energy is U if the charge is increased to 2Q then energy will be A)4U B)2U
Lubov Fominskaja [6]

Answer:

2u

Explanation:

2u

W = Vq

q = CV

W = V.CV

W = CV²

W/C = V²

V = √(W/C)

√(W1/C1) = √(W2/C2)

√(u/c) = √(x/2c)

x = 2u

8 0
3 years ago
A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsul
jeka94

The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

<h3>Given:</h3>

Mass of the astronaut, m_a = 126 kg

Speed he acquires, v_{a}  = 2.70 m/s

Mass of the space capsule, m_{c} = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

P_f = m_av_a + m_cv_c

P_I = 0

m_av_a + m_cv_c = 0

v_c =\frac{- m_a v_a}{m_c}}\\\\

   = \frac{126* 2.70}{1800}

   = - 0.189 m/s

Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

    = 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

  = 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = \frac{1}{2} m v^2

     = \frac{1}{2} × 126 × (2.70) ^2

     = 459.27 J

The Kinetic Energy of the capsule;

K.E = \frac{1}{2} m v^2

     = \frac{1}{2}×1800×(0.189) ^2

     = 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Learn more about kinetic energy here:

brainly.com/question/26520543

#SPJ1

3 0
1 year ago
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