Ask question

Ask question

All categories

3 years ago

5

An automobile with an initial speed of 5.13 m/s accelerates uniformly at the rate of 3.0 m/s2. Find the final speed of the car a

fter 4.9 s. Answer in units of m/s.

1 answer:

Digiron [165]3 years ago

7 0

Answer:

Final velocity v=19.83 m/sec

Explanation:

We have given initial velocity u =5.13 m/sexc

Acceleration of automobile $a=3m/sec^2$

Time t =4.9 sec

We have to find the final velocity v

According to first law of motion v = u+at ,here v is the final velocity , a is acceleration and t is time

So $v=5.13+3\times 4.9=19.83m/sec$

So the final velocity is 19.83 m/sec

You might be interested in

Which of these is a TRUE statement? A) Sound travels through gases easier than liquids. B) Sound travels through solids easier t

lakkis [162]

It’s B. Sound travels faster through solids than liquids. Have you ever put your head on a desk, and tap the desk? That’s an example of it going faster through solids

6 0

3 years ago

Read 2 more answers

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

3 years ago

In a chemical reaction, molecules of hydrogen gas (H2) react with molecules of oxygen gas (O2) in a sealed reaction chamber to p

11Alexandr11 [23.1K]

Answer:

Option (B) is correct.

Explanation:

Given that the molecules of hydrogen gas ( $H_2$ ) react with molecules of oxygen gas ( $O_2$ ) in a sealed reaction chamber to produce water ( $H_2O$ ).

The governing equation for the reaction is

$2H_2 +O_2 \rightarrow 2H_2O$

From the given, the only fact that can be observed that 2 moles of $H_2$ and 1 mole of $O_2$ reacts to produce 2 moles of $H_2O$ .

As the mass of 1 mole of $H_2 = 2$ grams ... (i)

The mass of 1 mole of $O_2 = 32$ grams ...(ii)

The mass of 1 mole of $H_2O = 18$ grams (iii)

Now, the mass of the reactant = Mass of 2 moles of $H_2$ + mass 1 mole of $O_2$

$= 2 \times 2 + 32$ [ using equations (i) and (ii)]

$=4+32 = 36$ grams.

Mass of the product = Mass of 2 moles of $H_2O$

$=2\times 18$ [ using equations (iii)]

=36 grams

As the mass of reactants = mass of the product.

So, mass is conserved.

Hence, option (B) is correct.

8 0

3 years ago

A 5.50-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical

iris [78.8K]

Answer:

17.71N/m

Explanation:

The period of the spring is expressed according to the expression;

$T = 2 \pi \sqrt{\frac{m}{k} } \\$

m is the mass of the object

k is the force constant

Given

m = 5.50kg

T = 3.50s

Substitute into the formula;

$T = 2 \pi \sqrt{\frac{m}{k} } \\3.5 = 2 (3.14) \sqrt{\frac{5.5}{k} } \\3.5 = 6.28 \sqrt{\frac{5.5}{k} } \\\frac{3.5}{6.28} = \sqrt{\frac{5.5}{k} } \\0.557 = \sqrt{\frac{5.5}{k} } \\square \ both \ sides\\0.557^2 = (\sqrt{\frac{5.5}{k} })^2 \\0.3106 = \frac{5,5}{k}\\k = \frac{5.5}{0.3106}\\k = 17.71N/m$

Hence the force constant of the spring is 17.71N/m

4 0

3 years ago

The periodic wave in the diagram below has a frequency of 80. hertz.

ASHA 777 [7]

Answer:

hi how are you

2) 5/8

4) 5/8

6 0

3 years ago