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3 years ago

A line _____ on a typical Speed vs. Time graph means an object is experiencing a constant acceleration.

Physics

2 answers:

Savatey [412]3 years ago

7 0

Answer:

The answer should be C. slanted upward to the right.

Hope this helps. :-)

Mila [183]3 years ago

4 0

C should be your answer

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While on vacation, Carine rides in a hot-air balloon. To make the balloon rise, the pilot turns on a flame to warm the air insid

andre [41]

The air inside becomes warmer, and that makes it LESS DENSE.

Less-dense air FLOATS in denser air. Since it's inside the balloon, it picks the balloon up when it floats.

3 0

3 years ago

A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc

Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at $v_c = 23m/s$ speed is

$s_c = \Delta t v_c = 23\Delta t$

The distance traveled by the motorcycle after Δt (seconds) at $m_m = 23 m/s$ speed and acceleration of a = 8 m/s2 is

$s_m = \Delta t v_m + a\Delta t^2/2$

$s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2$

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

$s_m = s_c + 58$

$23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58$

$4 \Delta t^2 = 58$

$\Delta t^2 = 14.5$

$\Delta t = \sqrt{14.5} = 3.807s$

(b)

$s_m = 23 \Delta t + 4 \Delta t^2$

$s_m = 23*3.807 + 58 = 145.581 m$

5 0

2 years ago

A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn

Lelechka [254]

Answer:

a. 16 s b. -1.866 kJ

Explanation:

a. Since the initial rotational speed ω₀= 3313 rev/min = 3313/60 × 2π rad/s = 346.94 rad/s. Its rotational speed becomes ω₁ = 0.75ω₀ in time t = 4 s.

We find it rotational acceleration using α = (ω₁ - ω₀)/t = (0.75ω₀ - ω₀)/t = ω₀(0.75 - 1)/t = -0.25ω₀/t = (-0.25 × 346.94 rad/s)/4 s = -21.68 rad/s².

Since the turntable stops at ω = 0, the time it takes to stop is gotten from

ω = ω₀ + αt and t = (ω - ω₀)/α = (0 - 346.94 rad/s)/-21.68 rad/s² = (-346.94/-21.68) s = 16 s.

So it takes the turntable 16 s to stop.

b. The workdone by the turntable to stop W equals its rotational kinetic energy change.

So, W = 1/2Iω² - 1/2Iω₀² = 1/2 × 0.031 kgm² × 0² - 1/2 × 0.031 kgm² × (346.94 rad/s)² = 0 - 1865.7 J = -1865.7 J = -1.8657 kJ ≅ -1.866 kJ

3 0

3 years ago

Thallium consists of 29.5%TI-203 and 70.5%TI-205. What is the relative atomic mass of thallium?

Kazeer [188]

Since you are referring to the TI-203 and TI-205, you need to know the actual masses of these two isotopes. TI-203 has 202.9723 amu and TI-205 has 204.9744 amu. Since you are concluding that this Thallium have 29.5% (Ti-203) and 70.5% (Ti-205), you need to multiply the percentage to the actual masses of the isotopes. With that, you should be able to get 204.3833 amu

7 0

2 years ago

Consider an e. coli to be a cylinder with a diameter of 1 micrometer (um) and with a length of 2 micrometer (um).

julia-pushkina [17]

Answer:

$A=6.28\ \mu m^2$