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3 years ago

a large truck has a mass of 20000kg. it is traveling at 28m/s along a staright road . calculate the kinetic energy

Physics

1 answer:

Semmy [17]3 years ago

3 0

Answer:

The answer is

Explanation:

The kinetic energy KE of an object given it's mass and velocity can be found by using the formula

$KE = \frac{1}{2} m {v}^{2} \\$

where

m is the mass

v is the velocity

From the question

m = 20000kg

v = 28 m/s

It's kinetic energy is

$KE = \frac{1}{2} \times 20000 \times {28}^{2} \\ = 10000 \times 784$

We have the final answer as

Hope this helps you

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3 years ago

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3 years ago

A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

Ivahew [28]

Answer:

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of the force is

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$

$F_{31 +32} = 2.16 *10^{-5} (4i + 3j) - 12*10^{-5} j$

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of $F_{31 +32}$ is mathematically evaluated as

$|F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}$

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction is obtained as

$tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}$

$\theta = tan ^{-1} [-0.63889]$

$\theta = - 32.57 ^o$

$\theta = 360 - 32.57$

$\theta =327.43 ^o$

5 0

3 years ago

What mass of steam at 100°C must be mixed with 119 g of ice at its melting point, in a thermally insulated container, to produce

N76 [4]

Answer: $M=27.92\ gm$

Explanation:

Given

mass of ice $m=119\ gm$

Final temperature of liquid $T_f=57^{\circ}C$

Specific heat of water $c=4186\ J/kg-K$

Latent heat of fusion $L=333\ kJ/kg$

Latent heat of vaporization $L_v=2256\ kJ/kg$

Suppose M is the mass of steam at $100^{\circ} C$

Heat required to melt ice and convert it to water at $57^{\circ}C$

$Q_1=mL+mc(T_f-0)$

Heat released by steam

$Q_2=ML_v+Mc(100-T_f)$

$Q_1$ and $Q_2$ must be equal as the heat gained by ice is equal to Heat released by steam

$Q_1=Q_2$

$\Rightarrow mL+mc(T_f-0)= ML_v+Mc(100-T_f)$

$\Rightarrow M=\dfrac{m[L+c\times T_f]}{L_v+c(100-T_f)}$

$\Rightarrow M=\dfrac{119[333\times 10^3+4186\times 57]}{2256\times 10^3+4186\times (100-57)}$

$\Rightarrow M=119\times 0.2346$

$M=27.92\ gm$

7 0

3 years ago

The rounded atomic mass of K is _______<br> K stands for _____ and has _____ valence electron(s).

netineya [11]

Answer:

Potassium

k = Potassium

Potassium Has 19 Electrons

the atomic mass is 39.0983

4 0

3 years ago

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a large truck has a mass of 20000kg. it is traveling at 28m/s along a staright road . calculate the kinetic energy ​

a large truck has a mass of 20000kg. it is traveling at 28m/s along a staright road . calculate the kinetic energy