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shutvik [7]
3 years ago
6

Can someone help me :)

Physics
2 answers:
zhenek [66]3 years ago
6 0

Answer:

undergone a chemical change

Explanation:

aivan3 [116]3 years ago
6 0

Answer:

undergone a chemical change

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A rubber ball and a lump of clay have equal mass. They are thrown with equal speed against a wall. The ball bounces back with ne
gregori [183]

Answer:

The ball experiences the greater momentum change

Explanation:

The momentum change of each object is given by:

\Delta p = m \Delta v= m (v-u)

where

m is the mass of the object

v is the final velocity

u is the initial velocity

Both objects have same mass m and same initial velocity u. So we have:

- For the ball, the final velocity is

v=-u

Since it bounces back (so, opposite direction --> negative sign) with same speed (so, the magnitude of the final velocity is still u). So the change in momentum is

\Delta p=m(v-u)=m((-u)-u)=-2mu

- For the clay, the final velocity is

v=0

since it sticks to the wall. So, the change in momentum is

\Delta p = m(v-u)=m(0-u)=-mu

So we see that the greater momentum change (in magnitude) is experienced by the ball.

3 0
3 years ago
The height of the Empire State Building is 318.00 meters. If a stone is dropped from the top of the building, what is the stone'
Mars2501 [29]
I belive its like 1200 mile per hour ive done he math for it 

3 0
3 years ago
Which of the following is true about inertia
Tamiku [17]

Answer:

The correct answers are It is the resistance of an object to changes in its motion, and It is a force

3 0
3 years ago
A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police
Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. x=x_{0}+vt

2. x=x_{0}+v_{0}t+\frac{at^{2}}{2}

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

vt = v_{0}t+\frac{at^2}{2}\\\\   16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}

We simplify the fraction present and rearrange for our formula so that it equals 0:

0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0

In the very last step we factored a common factor t. There is two possible solutions to the equation at t=0 and:

0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\  0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at t=0 s (when the SUV passed the police car) and t=15.56s(when the police car catches up to the SUV)

8 0
3 years ago
Please indicate how long each bar is in centimeters and millimeters.
Maru [420]

Answer:

1) 74 cm or 740 mm

2) 33 cm or 330 mm

3) 76 cm or 760 mm

4) 17 cm or 170 mm

brainliest !?!?

6 0
2 years ago
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