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zhannawk [14.2K]
3 years ago
15

Situations during a soccer game in which the ball is determined in or out of play

Physics
1 answer:
NISA [10]3 years ago
7 0
In play because I just did an assignment with this same question
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What is the weight of a 225kg space probe on the moon and the acceleration of gravity on the moon is 1.62
olga nikolaevna [1]
<span>364N should be your answer.. hope this helps

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7 0
3 years ago
Read 2 more answers
20% Part (a) Use an "E Field Sensor" and move it along either equipotential. What can you say about the E field along an equipot
Alex

Answer:

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

Explanation:

As we know that the relation between electric field and electric potential is given as

\Delta E = -\frac{dV}{dr}

here if we say that potential is constant because electric field sensor is moving along equi-potential line.

Then we will say

V = constant

so we have

\Delta E = 0

so electric field will remain constant always in magnitude and always remains perpendicular to the surface

so we have

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

6 0
3 years ago
(please help i gotta turn this in a few minutes 10 points!)
pogonyaev

Answer:

3a, 2b,4c,1d

Explanation:

what do I need to explain just something you know

7 0
3 years ago
A 110 kg quarterback is running the ball downfield at 4.5 m/s in the positive direction when he is tackled head-on by a 150 kg l
melisa1 [442]

Answer:

v_f=-0.29\frac{m}{s}

Explanation:

The principle of conservation of momentum, states that if the sum of the forces acting on a system is null, the initial total momentum of the system before a collision equals the final total momentum of the system after the collision. The collision is completely inelastic, which means that the players remain stick to each other after the collision:

p_i=p_f\\m_1v_1+m_2v_2=(m_1+m_2)v_f\\v_f=\frac{m_1v_1+m_2v_2}{(m_1+m_2)}\\\\v_f=\frac{(110kg)4.5\frac{m}{s}+150kg(-3.8\frac{m}{s})}{(110kg+150kg)}\\v_f=-0.29\frac{m}{s}

5 0
3 years ago
You are the juror of a case involving a drunken driver whose 1041 kg sports car ran into a stationary 1928 kg station wagon stop
mina [271]
1. Find the force of friction between the sports car and the station wagon stuck together and the road. The total mass m = 1928kg + 1041kg = 2969kg. The only force in the x-direction is friction: F = μ*N = μ * m * g 
2. Find the acceleration due to friction: 
F = m*a =  μ * m * g => a = μ * g = 0.6 * 9.81
3. Find the time it took the two cars stuck together to slide 12m:
x = 0.5*a*t² 
t = sqrt(2*x / a) = sqrt(2 * x / (μ * g) )
4. Find the initial velocity of the two cars:
v = a*t = μ * g * sqrt(2 * x / (μ * g) ) = sqrt( 2 * x * μ * g)
5. Use the initial velocity of the two cars combined to find the velocity of the sports car. Momentum must be conserved:

m₁ mass of sports car
v₁ velocity of sports car before the crash
m₂ mass of station wagon
v₂ velocity of station wagon before the crash = 0
v velocity after the crash

m₁*v₁ + m₂*v₂ = (m₁+m₂) * v = m₁*v₁ 
v₁ = (m₁+m₂) * v / m₁ = (m₁+m₂) * sqrt( 2 * x * μ * g) / m₁
v₁ = 33.9 m/s


7 0
3 years ago
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