Power = work+ time work time time work Fxd Ext Fxd

A rocket moves upward, starting from rest with an acceleration of 29.4 m/s^2 for 3.98 s. it runs out of fuel at the end of 3.98

ValentinkaMS [17]

The distance d₁ it rises from rest while the engine is burning is given by
d₁ = d₀ + v₀t + (1/2)at²
d₁ = 0 + 0 + (1/2)·(29.4 m/s²)·(3.98 s)² = 232.85 m
So it gets to 232.85 m and then runs out of fuel. Its velocity v₁ at this point is given by
v₁ = v₀ + at = (29.4 m/s²)·(3.98 s) = 117 m/s
At this point, gravity begins to slow it down until it reaches its peak where its velocity v₂ is zero.
v₂² = v₁² + 2ad₂
where d₂ is the distance it rises until v=0
Since gravity is decelerating the rocket, a = -g, and we have
0² = (117 m/s)² + 2(-9.8 m/s²)d₂
0 = (117)² - (19.6)·d₂
0 = 13,689 - (19.6)·d₂
d₂ = 13,689/19.6 = 698.42 m
So the total height it rises is given by
d₁ + d₂ = 232.85 m + 698.42 m
= 931.27 m

5 0

3 years ago

2. A jack exerts a vertical force of 4.5 X 103

skad [1K]

Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

work is done by the jack?

$\\ \\$

Given :-

$\star \sf \small force = 4.5 \times {10}^{3} \: newton$

$\star \sf \small distance = 0.25 \: meter$

$\\ \\$

To find:-

$\sf \star \: work = \: ?$

$\\ \\$

Solution:-

we know :-

$\bf \dag \boxed{ \rm work = force \times distance}$

$\\ \\$

So:-

$\dashrightarrow \sf work = force \times distance$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{0 \cancel.5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \cancel \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{4\cancel.5}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{3 - 1} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{1} \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times \cancel{10}}{ \cancel2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times 5}{ 1} \\$

$\\ \\$

$\dashrightarrow \sf work =225 \times 10$

$\\ \\$

$\dashrightarrow \bf work =\red{2250\: joule}$

5 0

3 years ago

a block has a volume of 0.09m3 and a density of 4,000kg/m3. what's the force of gravity acting on the block in water

Lunna [17]

   Density = (mass) / (volume)

      4,000 kg/m³ = (mass) / (0.09 m³)

Multiply each side
by 0.09 m³ :           (4,000 kg/m³) x (0.09 m³) = mass

   mass = 360 kg .

Force of gravity = (mass) x (acceleration of gravity)

   = (360 kg) x (9.8 m/s²)

   = (360 x 9.8) kg-m/s²

   =   3,528 newtons .

That's the force of gravity on this block, and it doesn't matter
what else is around it. It could be in a box on the shelf or at
the bottom of a swimming pool . . . it's weight is 3,528 newtons
(about 793.7 pounds).

Now, it won't seem that heavy when it's in the water, because
there's another force acting on it in the upward direction, against
gravity. That's the buoyant force due to the displaced water.

The block is displacing 0.09 m³ of water. Water has 1,000 kg of
mass in a m³, so the block displaces 90 kg of water. The weight
of that water is (90) x (9.8) = 882 newtons (about 198.4 pounds),
and that force tries to hold the block up, against gravity.

So while it's in the water, the block seems to weigh

   (3,528 - 882) = 2,646 newtons (about 595.2 pounds) .

But again ... it's not correct to call that the "force of gravity acting
on the block in water". The force of gravity doesn't change, but
there's another force, working against gravity, in the water.

5 0

3 years ago

An automobile with a mass of 1232 kg accelerates at a rate of 2 m/s² in the forward direction. What is the net force acting on t

boyakko [2]

Based on Newton's second law:
Force = mass x acceleration

For this problem: we have the mass = 1232 kg and the acceleration = 2 m/s^2
So, just substitute with the givens in the above equation to calculate the force as follows:
Force = 1232 x 2 = 2464 Newtons

7 0

3 years ago

Masses are stacked on top of the block until the top of the block is level with the waterline. This requires 20 g of mass. What

Kobotan [32]

Answer:

Mass of the wooden Block is 20g.

Explanation:

The buoyant force equation will be used here

Buoyant Force= ρ*g*1/2V Here density used is of water

m*g= ρ*g*1/2V

Simplifying the above equation

2m= ρ*V Eq-1

Also we know from the question that

ρ*V = m + 0.020 Eq-2 ( Density = (Mass+20g)/Volume )

Equating Eq-1 & Eq-2 we get

2m = m+0.020

m = 0.020kg

m = 20g

6 0

3 years ago