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Effectus [21]
3 years ago
10

Two objects of different masses accelerate at the same rate. According to Newton’s second law of motion, the more massive obje

ct requires________.
A.less speed to accelerate
B.less force to accelerate
C.more force to accelerate
D.more distance to accelerate
Physics
1 answer:
babymother [125]3 years ago
7 0

Newton's Second law of motion is

F = m * a   ( force = mass * acceleration).

So if a is constant   then a = F 1 / m1  and  a = F2 / m2  then

F1 /m1 = F2 /  m2 and if  m1 > m2 then

F1 must be greater than F2.

Answer is C.

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At what point in its swing is potential energy a maximum?<br> E<br> B<br> D<br> A<br> C
cestrela7 [59]
I think the answer is A.

Hope this helps :)
3 0
3 years ago
Why does the moon turn red during a lunar eclipse?.
SOVA2 [1]

Answer:

During a lunar eclipse, the Moon turns red because the only sunlight reaching the Moon passes through Earth's atmosphere. The more dust or clouds in Earth's atmosphere during the eclipse, the redder the Moon will appear.

4 0
2 years ago
A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a
katen-ka-za [31]

Answer:b

Explanation:

Given

mass of first cart m_1=6 kg

mass of second cart m_2=3 kg

velocity of first cart v_1=3 m/s

conserving momentum

m_1v_1+m_2v_2=(m_1+m_2)v

6\times 3+3\times 0=(9)\cdot v

v=2 m/s

Initial kinetic Energy K.E._1=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2

K.E._1=\frac{1}{2}\cdot 6\cdot 3^2+0

K.E._1=27 J

Final Kinetic Energy

K.E._2=\frac{1}{2}(m_1+m_2)v^2

K.E._2=\frac{1}{2}(6+3)\cdot 2^2=18 J

Ratio of initial Kinetic Energy to the Final Kinetic Energy

=\frac{27}{18}=1.5

6 0
3 years ago
In a standard tensile test, a steel rod of 7 8-in. diameter is subjected to a tension force of 17 kips. Knowing that ν = 0.30 an
natali 33 [55]

Answer:

(a) Elongation of the rod==5.61×10⁻⁹m

(b) Change in diameter=1.640×10⁻⁸m

Explanation:

Given data

Diameter d=78 in=1.9812 m

Cross Area is:

A=(\pi /4)d^{2} \\A=(\pi /4)(1.9812m)^{2}\\A=3.08m^{2}

Applied Load P=17 KN=17×10³N

E=29 × 106 psi=1.99947961×10¹¹Pa

Stress and Strain in x direction

Stress in x direction

σ=P/A

=\frac{17*10^{3}N }{3.08m^{2} }\\ =5517.25Pa

σ=5517.25 Pa

Strain in x direction

ε=σ/E

=\frac{5517.25}{1.99947961*10^{11} } \\=2.76*10^{-8}

ε=2.76×10⁻⁸

Part (a)

Elongation of the rod=Lε

=(0.2032)(2.76×10⁻⁸)

Elongation of the rod==5.61×10⁻⁹m

Part(b) Change in diameter

Strain in y direction

ε₁= -vε

ε₁= -(0.30)(2.76×10⁻⁸)

ε₁=-8.28×10⁻⁹

Change in diameter=d×ε₁

Change in diameter=(1.9812m)×(-8.28×10⁻⁹)

Change in diameter=1.640×10⁻⁸m

4 0
3 years ago
In Rutherford's experiment, which of the following proved that the
vampirchik [111]

Answer:

b) the alpha particles were found to be attracted to the nucleus

Explanation:

5 0
3 years ago
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