Ask question

Ask question

All categories

3 years ago

10

Q:-What is speed????

1 answer:

Margarita [4]3 years ago

5 0

Explanation:

In everyday use and in kinematics, the speed of an object is the magnitude of the rate of change of its position with time or the magnitude of the change of its position per unit of time; it is thus a scalar quantity.

SI unit: m/s, m s−1

s=d/t

You might be interested in

Changing subscripts is one of the steps of balancing a chemical equation

Sergeeva-Olga [200]

<u>Answer:</u>

No, Changing subscripts is not the step of balancing a chemical equation

<u>Explanation:</u>

In balancing a chemical equation you have to keep one thing in mind that the no of atoms and elements on both sides such as reactant (before reaction) and product( after a reaction) must be equal. Coefficients are added to equal the no of atoms and elements, but we will not change the subscript.

7 0

3 years ago

The part of the shore that represents the water's edge and migrates back and forth with the tide is known as the ________.

Dennis_Churaev [7]

Answer :It is know as the shore line

Explanation:

7 0

3 years ago

A 0.030 kg lead bullet hits a steel plate, both initially at 20?C. The bullet melts and splatters on impact. Assume that 80% of

Dmitry_Shevchenko [17]

Answer:

(a). The required is 1871.2 J.

(b). The speed the lead bullet is 395 m/s.

(c). The 20% of energy must have gone into collision between steel plate and bullet.

Explanation:

Given that,

Mass of bullet = 0.030 kg

Temperature = 20°C

(a). We need to heat required to increase the temperature of the lead bullet and melt it

Using formula of heat

$Q=mS\Delta T+mL$

Where, m = mass of lead bullet

S = specific heat

L = latent heat

T = Temperature

Put the value into the formula

$Q=0.030\times130\times(327.5-20)+0.030\times22400$

$Q=1871.2\ J$

The required is 1871.2 J.

(b). Assume that 80% of the bullet’s kinetic energy goes into increasing its temperature and then melting it

We need to calculate the energy

$Q=\dfrac{1871.2}{0.8}$

$Q=2339 J$

We need to calculate the speed the lead bullet

Using formula of speed

$K.E=Q$

$\dfrac{1}{2}mv^2=2339$

$v^2=\dfrac{2339\times2}{0.030}$

$v=\sqrt{\dfrac{2339\times2}{0.030}}$

$v=394.8 = 395\ m/s$

The speed the lead bullet is 395 m/s.

(c). The 20% of energy must have gone into collision between steel plate and bullet.

Hence, This is the required solution.

6 0

3 years ago

Calculate the magnitude of the electric field at one corner of a square 2.42 m on a side if the other three corners are occupied

miv72 [106K]

5.70 x 10 -6 c - 2.42

7 0

3 years ago

A 1000 kg train car traveling at 12 m/s on a level track has 400 kg of sand dumped, vertically, into it. What is the final speed

crimeas [40]

Answer:

The final speed of the train car is 8.57 m/s.

Explanation:

Given that,

Mass of train car = 1000 kg

Mass of sand = 400 kg

Speed of train car = 12 m/s

We need to calculate the final speed of the train car after the sand is dumped

Using conservation of momentum

$m_{1}u_{1}=(m_{1}+m_{2})v$ ...(I)

Put the value in the equation (I)

$1000\times12=(1000+400)v$

$v=\dfrac{1000\times12}{1400}$

$v=8.57\ m/s$

Hence, The final speed of the train car is 8.57 m/s.

5 0

3 years ago