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Lostsunrise [7]
3 years ago
14

Jupiter’s gravity might have kept a planet from forming. What planet?

Physics
2 answers:
kherson [118]3 years ago
4 0

Answer:

The gravity of Jupiter affects every planet to one degree or another. It is strong enough to tear asteroids apart and capture 64 moons at least. Some scientist think that Jupiter destroyed many celestial objects in the ancient past as well as prevented other planets from forming.Jun 17, 2008

liraira [26]3 years ago
3 0

Answer:

Saturn

Explanation:

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Scientists use models to
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D Is Thw Correct answer
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4 years ago
A projectile is launched from the ground with an initial velocity of 12ms at an angle of 30° above the horizontal. The projectil
netineya [11]

vi^{2}sin2thita/g =12^{2}sin2[30]/9.8=12.7Answer:

Explanation:

range is given as

6 0
3 years ago
Read 2 more answers
An arrow of 43 g moving at 84 m/s to the right, strikes an apple at rest. The arrow sticks to the apple and both travel at 16.8
Aloiza [94]

Answer:

<em>The mass of the apple is 0.172 kg (172 g)</em>

Explanation:

<u>The Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of two bodies, then the total momentum is the sum of both momentums:

P=m_1v_1+m_2v_2

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

We are given the mass of an arrow m1=43 g = 0.043 kg traveling at v1=84 m/s to the right (positive direction). It strikes an apple of unknown mass m2 originally at rest (v2=0). The common speed after they collide is v'=16.8 m/s.

We need to solve the last equation for m2:

m_2v_2-m_2v'=m_1v'-m_1v_1

Factoring m2 and m1:

m_2(v_2-v')=m_1(v'-v_1)

Solving:

\displaystyle m_2=\frac{m_1(v'-v_1)}{v_2-v'}

Substituting:

\displaystyle m_2=\frac{0.043(16.8-84)}{0-16.8}

\displaystyle m_2=\frac{-2.8896}{-16.8}

\displaystyle m_2=0.172\ kg

The mass of the apple is 0.172 kg (172 g)

3 0
3 years ago
A(n) 69.8 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 60.4 m away from the shuttle
alukav5142 [94]

Answer:

The time taken is 6.7  min

Explanation:

Using the linear momentum conservation theorem, we have:

m_1*v_{o1}+m_2*v_{o2}=m_1*v_{f1}+m_2*v_{f2}

when she was 60.4m from the shuttle, she has zero speed, so the initial velocity is zero.

m_1*0+m_2*0=m_1*v_{f1}+m_2*v_{f2}\\m_1*_{f1}=-m_2*v_{f2}\\v_{f1}=-\frac{m_2*v_{f2}}{m_1}\\\\v_{f1}=-\frac{0.886kg*12m/s}{69.8kg}\\\\V_{f1}=-0.15m/s

That is 0.15m/s in the opposite direction of the camera.

the time taken to get to the shuttle is given by:

t=\frac{d}{v_{f1}}\\\\t=\frac{60.4m}{0.15m/s}\\\\t=403s\\t_{min}=403s*\frac{1min}{60s}=6.7min

6 0
4 years ago
A long, rigid conductor, lying along an x axis, carries a current of 4.99 A in the negative x direction. A magnetic field is pre
Tcecarenko [31]

Answer with Explanation:

We are given that

Current in conductor=I=4.99 A  (-x direction)

Magnetic field=B=3.72\hat{i}+8.72x^2\hat{j}mT=(3.72i+8.72x^2j)\times 10^{-3}

(1mT=10^{-3} T)

x(in m) and B (in mT)

Length of conductor is given in negative x- direction

\vec{L}=-x\hat{i}

dL=-dx\hat{i}

Force on current carrying conductor is given by

F=I(L\times B)

dF=I(dL\times B)

Integrating on both sides then we get

\vec{F}=\int_{1.41}^{2.77}(4.99)(-dx\hat{i}\times (3.72\hat{i}+8.72x^2\hat{j}))\times 10^{-3}

\vec{F}=-\int_{1.41}^{2.77}(4.99\times 10^{-3})\cdot 8.72(x^2\hat{k})dx  (i\times i=0, i\times j=k

\vec{F}=-(4.99\times 10^{-3}\times 8.72)[\frac{x^3\hat{k}}{3}]^{2.77}_{1.41}

\vec{F}=-\frac{(4.99\times 10^{-3}\cdot 8.72)}{3}((2.77)^3-(1.41)^3)\hat{k}

\vec{F}=-0.268 \hat{k} N

a. x- component of force=0

b.y- component of force=0

c.z- component of force=-0.268 N

5 0
3 years ago
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