Answer:
Average speed of Elain = 60 km/h
Explanation:
Total Distance covered by Jack = 360km
Average Speed of Jack = 80 km/h
Time taken by Jack to complete his journey = Distance / Average speed = 360 km / 80 km/h
Time taken by Jack to complete his journey = 4.5 hours
As it is given the both Jack and Elain travelled the same amount of distance:
Total distance travelled by Elain = 360 km
It is given that Elain took 1.5 hourse more than Jack to cover the distance, so Time taken by Elain to cover the distance is = 4.5 hours + 1.5 hours = 6 hours
Average speed of Elain = Distance/ time = 360 km / 6 hours
Average speed of Elain = 60 km/h
Answer:
1050 kg
Explanation:
The formula for kinetic energy is:
KE (kinetic energy) = 1/2 × m × v² where <em>m</em> is the <em>mass in kg </em>and <em>v</em> is the velocity or <em>speed</em> of the object <em>in m/s</em>.
We can now substitute the values we know into this equation.
KE = 472 500 J and v = 30 m/s:
472 500 = 1/2 × m × 30²
Next, we can rearrange the equation to make m the subject and solve for m:
m = 472 500 ÷ (1/2 × 30²)
m = 472 500 ÷ 450
m = 1050 kg
Hope this helps!
Answer:
The final velocity of the thrower is
and the final velocity of the catcher is
.
Explanation:
Given:
The mass of the thrower,
.
The mass of the catcher,
.
The mass of the ball,
.
Initial velocity of the thrower, 
Final velocity of the ball, 
Initial velocity of the catcher, 
Consider that the final velocity of the thrower is
. From the conservation of momentum,

Consider that the final velocity of the catcher is
. From the conservation of momentum,

Thus, the final velocity of thrower is
and that for the catcher is
.
First,

where
is density,
is mass, and
is volume. We can compute the volume of the roll:


When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness
. Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).
So we have

where
is the given area, so


If we're taking significant digits into account, the volume we found would have been
, in turn making the thickness
.
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