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3 years ago

Was the object accelerating during the time interval labeled “B”? A) Yes B) No

Physics

2 answers:

Oxana [17]3 years ago

7 0

Yes it was accelerating

Alla [95]3 years ago

5 0

There's not enough information in the graph to determine whether it was or it wasn't. It could have been, but we can't tell.

Before I explain this answer, it's VERY important to know exactly what "accelerating" means:

-- "Accelerating" does NOT always mean "speeding up".

-- "Accelerating" means ANY change in velocity.

-- That can be ==> speeding up, or ==> slowing down, or ==> turning.

This graph is labeled "Velocity vs Time", but it isn't really. "Velocity" means speed AND direction. The graph only shows us the object's speed, but nothing about its direction.

During time-interval-B, the speed isn't changing, but the object could still be changing direction, like turning a corner, or on a curved path. We don't know. The graph doesn't show us that information.

The object could be accelerating during time-interval-B. It's true that the speed wasn't changing. But maybe the direction was. If the direction was changing, then we have to say the object was 'accelerating' during the interval labeled B .

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A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru

r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

= 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

$s = u t +\dfrac{1}{2}at^2$

$-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2$

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

$t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}$

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0

3 years ago

A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi

Lubov Fominskaja [6]

Answer:

The moment of inertia is $I=\frac{M}{12} a^{2}$

Explanation:

The moment of inertia is equal:

$I=\int\limits^a_b {r^{2} } \, dm$

If r is $-\frac{a}{2}$

and $dm=\frac{M}{a} dr$

$I=\int\limits^a_b {r^{2}\frac{M}{a} } \, dr\\a=\frac{a}{2} \\b=-\frac{a}{2}$

$I=\frac{M}{a} \int\limits^a_b {r^{2} } \, dr\\\\I=\frac{M}{a} (\frac{M}{3} )_{b}^{a}\\ I=\frac{M}{3a} (\frac{a^{3} }{8} +\frac{a^{3} }{8} )\\I=\frac{M}{12} a^{2}$

7 0

3 years ago

Electrical power companies sell electrical energy

mixer [17]

Heat used by electric heater :

Q = m • c • ∆T

Q = (75 kg)(4200 J/kg°C)(43°C - 15°C)

Q = 8.82 × 10⁶ J

Cost of electrical energy :

Cost = (8.82 × 10⁶ J)/(3.6 × 10⁶ J) • ($ 0.15)

Cost = $ 0.3675

4 0

2 years ago

A car speed off around a bend at a constant 10m/s explain why it's velocity is not constant

VARVARA [1.3K]

It's velocity is not constant as direction is changing.

We know, velocity is speed with direction, so if direction is changing, velocity can't be constant, doesn't matter that speed is constant.

Hope this helps!

8 0

3 years ago

The period of the earth around the sun is 1 year and its distance is 150 million km from the sun. An asteroid in a circular orbi

SOVA2 [1]

Answer:

5.024 years

Explanation:

T1 = 1 year

r1 = 150 million km

r2 = 440 million km

let the period of asteroid orbit is T2.

Use Kepler's third law

T² ∝ r³

So,

$\left ( \frac{T_{2}}{T_{1}} \right )^2=\left ( \frac{r_{2}}{r_{1}} \right )^3$

$\left ( \frac{T_{2}}{1} \right )^2=\left ( \frac{440}{150} \right )^3$

T2 = 5.024 years

Thus, the period of the asteroid's orbit is 5.024 years.

4 0

3 years ago