A bowling ball is pushed with a force of 22.0N and accerlates at 5.5 m/s square. What is the mass of the bowling ball

Calculate the speed of a body covering a distance of 320 km in 4h

Mice21 [21]

We have that the speed of a body covering a distance of 320 km in 4h is mathematically given as

V=22.22m/s is

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Speed</h3>

From the question we are told

calculate the speed of a body covering a distance of 320 km in 4h

Generally the equation for the Speed is mathematically given as

$V=\frac{distance }{time}\\\\Therefore\\\\V=\frac{320*1000}{4*60*60}\\\\$

V=22.22m/s

Hence

The speed of a body covering a distance of 320 km in 4h is

V=22.22m/s

For more information on Speed visit

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4 0

2 years ago

State and prove bessel inequality

maria [59]

Statement :- We assume the orthagonal sequence ${{\{\phi\}}_{1}^{\infty}}$ in Hilbert space, now ${\forall \sf \:v\in \mathbb{V}}$ , the Fourier coefficients are given by:

${\quad \qquad \longrightarrow \sf a_{i}=(v,{\phi}_{i})}$

Then Bessel's inequality give us:

${\boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Proof :- We assume the following equation is true

${\quad \qquad \longrightarrow \displaystyle \sf v_{n}=\sum_{i=1}^{n}a_{i}{\phi}_{i}}$

So that, ${\bf v_n}$ is projection of ${\bf v}$ onto the surface by the first ${\bf n}$ of the ${\bf \phi_{i}}$ . For any event, ${\sf (v-v_{n})\perp v_{n}}$

Now, by Pythagoras theorem:

${:\implies \quad \sf \Vert v\Vert^{2}=\Vert v-v_{n}\Vert^{2}+\Vert v_{n}\Vert^{2}}$

${:\implies \quad \displaystyle \sf ||v||^{2}=\Vert v-v_{n}\Vert^{2}+\sum_{i=1}^{n}\vert a_{i}\vert^{2}}$

Now, we can deduce that from the above equation that;

${:\implies \quad \displaystyle \sf \sum_{i=1}^{n}\vert a_{i} \vert^{2}\leqslant \Vert v\Vert^{2}}$

For ${\sf n\to \infty}$ , we have

${:\implies \quad \boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Hence, Proved

5 0

2 years ago

Read 2 more answers

A diamond is cut such that the angle between its top surface and its bottom surface is α. For α=45∘, find the largest possible v

Reil [10]

Answer:

i = 61 degree

Explanation:

Given,

$\alpha =45^{o}$

Now, by the snell's law

$sin\theta_c=\frac{1}{n}\\sin\theta_c=\frac{1}{2.450}\\\theta_c=24.09^o$

Now,

Sin i / sin r = n 2 / n 1

sin i / sin r (45 - 24.09) = 2.45 / 1

i = 60.97 degree

4 0

4 years ago

Two objects that are not initially in thermal equilibrium are placed in close contact. After a while, the temperature of the cod

Dima020 [189]

Answer:

If the temperature of the colder object rises by the same amount as the temperature of the hotter object drops, then <u>the specific heats of both objects will be equal.</u>

Explanation:

If the temperature of the colder object rises by the same amount as the temperature of the hotter object drops when the two<u> objects of same mass</u> are brought into contact, then their specific heat capacity is equal.

<u>We can prove this by the equation of heat for the two bodies:</u>

<em>According to given condition,</em>

$\Delta T_1=\Delta T_2$

$\frac{Q_1}{m_1.c_1} = \frac{Q_2}{m_2.c_2}$

<em>when there is no heat loss from the system of two bodies then </em> $Q_1=Q_2$

$\frac{1}{m.c_1} =\frac{1}{m.c_2}$

$\Rightarrow c_1=c_2$

Thermal conductivity is ultimately affects the rate of heat transfer, however the bodies will attain their final temperature based upon their mass and their specific heat capacities.

The temperature of the colder object will rise twice as much as the temperature of the hotter object only in two cases:

when the specific heat of the colder object is half the specific heat of the hotter object while mass is equal for both.

OR

the mass of colder object is half the mass of the hotter object while their specific heat is same.

3 0

3 years ago

Why is it good practice for scientists to repeat trails

tensa zangetsu [6.8K]

It makes the data thet they collect more reliable so if they need the data again, they have already tested it a few times so therefor they know that it is right.

8 0

3 years ago