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Art [367]
3 years ago
11

I'm giving out a lot of points for whoever helps me and answers correctly! 3 questions to answer! Please answer them ALL!!! :D

Physics
2 answers:
lions [1.4K]3 years ago
5 0

1-a 2-b 3-b    

srry had to write more, i go to this school














                                                                                                                 

Lemur [1.5K]3 years ago
3 0

1. A) A ball sits motionless on the ground

Newton's First Law of inertia states that:

"When the forces acting on an object are balanced, the object stays at rest (if it was at rest) or continues its motion with constant velocity (if it was in motion)"

Among the options given, we see that the option

A) A ball sits motionless on the ground

is an example of this law: in fact, the ball sits at rest on the ground, the forces acting on it are balanced, so the ball stays at rest.

2. B) The earth pulls more on the shuttle

The gravitational attraction exerted by a planet on the shuttle is given by

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the shuttle

r is the distance between the shuttle and the planet

In this problem, we are comparing the gravitational pulls exerted by the Earth and the Moon on the shuttle. The distance, r, is the same (because the shuttle is half-way), the mass of the shuttle m is the same, and G is the same. However, the mass of the Earth is greater than the mass of the Moon, so in the formula M is greater for the Earth: therefore, the Earth pulls more on the shuttle.

3. B) The velocity of the car reduced from 50 km/h to 35 km/h over one minute

The acceleration of the car is defined as the ratio between the change in velocity and the time taken:

a=\frac{v_f-v_i}{\Delta t}

where vf is the final velocity, vi is the initial velocity, and \Delta t is the time taken.

In order to have a negative acceleration, the numerator must be negative: this means that the initial velocity, vi, must be greater than the final velocity, vf, and this situation occurs in option B), since the initial velocity is 50 km/h while the final velocity is 35 km/h.

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When energy is added to a wave, how can the wave change?.
nadya68 [22]

Answer:

When waves overlap in-phase (crest meets crest or trough meets trough) the waves energy is additive and the amplitude increases.

Explanation:

When waves overlap out-of-phase (crest meets trough) the waves cancel and the amplitude (energy) decreases. When two interfering waves cancel each other out.

6 0
2 years ago
Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an
ddd [48]

Answer:

Al's mass is 102.92  kg  

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

F_{net} = 0

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

F_{net} = \frac{dp_{horizontal}}{dt} = 0

p_{horizontal_i }= p_{horizontal_f}

where the suffix i and f means initial and final respectively.

The initial momentum will be:

p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}

But, as they are at rest, initially

p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0

p_{horizontal_i} = 0

So, this means:

p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0

We know that the have an combined mass of 195 kg:

m_{total} = m_{Al} + m_{Jo} = 195 \ kg.

so:

m_{Jo} = 195 \ kg - m_{Al}.

m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0

m_{Al} \  v_{Al_f} - m_{Al} \  v_{Jo_f}= - 195 \ kg \  v_{Jo_f}

m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}

m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } {  v_{Al_f} - v_{Jo_f} }

m_{Al} = \frac{195 \ kg  \ v_{Jo_f} } {    v_{Jo_f} - v_{Al_f} }

Now, we can use the values:

v_{Al_f}= 10.2 \frac{m}{s}

v_{Jo_f}= - 11.4 \frac{m}{s}

where the minus sign appears as they are moving at opposite directions

m_{Al} = \frac{195 \ kg  ( - 11.4 \frac{m}{s} ) } {   (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }

m_{Al} = 102.92 \ kg

and this is the Al's mass.

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