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3 years ago

"the total angular momentum of an object changes when a net external force acts on the system." This statement is:

1 answer:

6 0

Here is the answer of the given question above. "The total angular momentum of an object changes when a net external force acts on the system." This statement is "sometimes true. it depends on the forces point of application". The answer is option D.

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Martine also has an eraser.it has a mass of 3g,and a volume if 1cm3.what is its density

shtirl [24]

Answer:

3g/cm³

Explanation:

Use the formula:

density = mass ÷ volume

Substitute (plug in) the values:

density = 3 ÷ 1 = 3g/cm³

4 0

2 years ago

A certain first-order reaction is 58% complete in 95 s. What are the values of the rate constant and the half-life for this proc

guajiro [1.7K]

Answer:

$0.005734 s^{-1}$ and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

Explanation:

Expression for rate law for first order kinetics is given by:

$a_o=a\times e^{-kt}$

where,

k = rate constant

t = age of sample

$a_o$ = let initial amount of the reactant

a = amount left after decay process

We have :

$a_o=x$

$a=58\%\times x=0.58x$

t = 95 s

$0.58x=x\times e^{-k\times 95 s}$

$\k= 0.005734 s^{-1}$

Half life is given by for first order kinetics::

$t_{1/2}=\frac{0.693}{k}$

$=\frac{0.693}{0.005734 s^{-1}}=120.86 s$

$0.005734 s^{-1}$ and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

3 0

2 years ago

Give two differences between Mass and weight

nydimaria [60]

Answer:

my opi

Explanation:

Mass is like the inside of the weight. weight is just home much it weighs

8 0

3 years ago

A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is broug

natali 33 [55]

Answer:

Approximately $1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}$ , assuming friction between the vehicle and the ground is negligible.

Explanation:

Let $m$ denote the mass of the vehicle. Let $v$ denote the initial velocity of the vehicle. Let $k$ denote the spring constant (needs to be found.) Let $x$ denote the maximum displacement of the spring.

Convert velocity of the vehicle to standard units (meters per second):

$\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}$ .

Initial kinetic energy ( ${\rm KE}$ ) of the vehicle:

$\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}$ .

When the vehicle is brought to a rest, the elastic potential energy ( $\text{EPE}$ ) stored in the spring would be:

$\displaystyle \frac{1}{2}\, k\, x^{2}$ .

By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial $\text{KE}$ of the vehicle should be equal to the ${\rm EPE}$ of the vehicle. In other words:

$\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}$ .

Rearrange this equation to find an expression for $k$ , the spring constant:

$\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}$ .

Substitute in the given values $m = 1508\; {\rm kg}$ , $v \approx 1.908\; {\rm m\cdot s^{-1}}$ , and $x = 6.87\; {\rm m}$ :

$\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}$

8 0

1 year ago

Which would best help a student determine the net force acting on a rollercoaster car as it moves from one point on its track to

Lelu [443]

Answer:

you can show them a vid

Explanation:

6 0

2 years ago