A moving sidewalk has a velocity of 1.3 m/s north. If a man walks south on

HELP

Sergeu [11.5K]

Answer:

The answer is: To accelerate an object the force applied to the object has to increase.

Explanation:

the acceleration of an object increases with increased force and decreases with increased mass.

3 0

3 years ago

I have a bottle of gas, the bottle can expand and contract. Initially the gas is at 1 kpa of pressure and a volume of 1 Liter, a

drek231 [11]

Answer:

P₂ = 1.22 kPa

Explanation:

This problem can be solved using the equation of state:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

where,

P₁ = initial pressure = 1 KPa

P₂ = final pressure = ?

V₁ = initial Volume = 1 liter

V₂ = final volume = 1.1 liter

T₁ = initial temperature = 290 k

T₂ = final temperature = 390 k

Therefore,

$\frac{(1\ kPa)(1\ liter)}{290\ k} =\frac{(P_2)(1.1\ liter)}{390\ k}\\\\P_2= \frac{(1\ kPa)(1\ liter)(390\ k)}{(290\ k)(1.1\ liter)}$

P₂ = 1.22 kPa

7 0

2 years ago

Helppppppppppppppp.......

kow [346]

Answer:

9013 m/s

Explanation:

hope it helped!!!

8 0

3 years ago

Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e $\frac{dv}{dt} =10 cm^3/s$

Consider r be the radius of the balloon.

The volume of of a sphere is

$v=\frac{4}{3} \pi r^3$

Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

Given that r = 5 cm

$[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}$ =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0

3 years ago

Which statement correctly explains scientific theories?

Elan Coil [88]

Scientific theories are tested and proven over time; they are then considered scientific laws.

Sometimes however, they are proven wrong, and so they do not become laws

hope this helps

4 0

3 years ago

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