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2 years ago

An object has kinetic energy of 324 J. If it’s speed is 9m/s, what is it’s mass?

Physics

2 answers:

harkovskaia [24]2 years ago

6 0

It is 8 kilograms

I had to type more so here you go

Illusion [34]2 years ago

3 0

Answer:

8 kg.

Explanation:

Remember that the formula to calculate the kinetick energy is half of the mass mutiplied by the squared velocity:

$Ke= .5mv^2$

So in order to solve this problem we will have to solve the equation for mass:

$M=\frac{324J}{.5*81} \\m=\frac{324}{40.5}\\ m=8 kg$

So by clearing the equation you now know that the answer is that there are 8 kg in an object that is moving at 9m/s and has a kinetic energy of 324J.

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A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. The coefficie

Rudiy27

Answer:

a) xf = 5.1 m

b) u = 0.304

c) x = 10.3 m

Explanation:

we will use the following formula:

u = 0.1 + A*x

Si x = 12.5 m, u = 0.6

Clearing A:

A = 0.5/12.5 = 0.04 m^-1

a) we have to:

W = Ekf - Eki

where Ekf = final kinetic energy

Eki = initial kinetic energy

9.8*(0.1xf + ((0.04*xf^2)/(2))) = (4.5^2)/(2)

Clearing xf, we have:

xf = 5.1 m

b) Replacing values for u:

u = 0.1 + (0.04*5.1) = 0.304

c) Wf = Ekf - Eki

-u*m*x*g = 0 - (m*v^2)/2

Clearing x:

x = v^2/(2*u*g) = (4.5^2)/(2*0.1*9.8) = 10.3 m

4 0

2 years ago

Read 2 more answers

. A little car has a maximum acceleration of 2.57 m/s2. What is the new maximum acceleration of the little car if it tows anothe

valkas [14]

Answer:

$a'=1.285\ m/s^2$

Explanation:

Let m be the mass of a little car and m' be the mass of another car.

We know that,

Force = mass × acceleration

ATQ,

m × a = 2m × a'

a = 2 × a'

$a'=\dfrac{a}{2}\\\\a'=\dfrac{2.57}{2}\\\\a'=1.285\ m/s^2$

So, the acceleration of another little car is equal to $1.285\ m/s^2$ .

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2 years ago

A student placed an ice cube on a table and observed it for five minutes. He noticed that the ice cube seemed to get smaller and

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Answer:

the ice cube melted due to the absorption of heat from the surrounding of the ice, the has a heat capacity of zero so it turns it from its solid state to its liquid state which leads to the formation of the puddle

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2 years ago

Two blocks, 1 and 2, are connected by a rope R1 of negligible mass. A second rope R2, also of negligible mass, is tied to block

alekssr [168]

Answer:

Explanation:

Given

Two block are connected by rope $R_1$

$R_2$ rope is attached to block 2

suppose $F_2$ is a force applied to Rope $R_2$

Applied force $F_2$ =Tension in Rope 2

$F_2=(m_1+m_2)a---1$

where a=acceleration of system

Tension in rope $R_1$ is denoted by $F_1$

$F_1=m_1a---2$

divide 1 and 2 we get

$\frac{F_2}{F_1}=\frac{(m_1+m_2)a}{m_1a}$

also $m_1=2.11\cdot m_2$

$\frac{F_2}{F_1}=\frac{2.11m_2+m_2}{2.11m_2}$

$\frac{F_2}{F_1}=\frac{3.11}{2.11}$

$\frac{F_1}{F_2}=\frac{2.11}{3.11}$

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2 years ago

A slender rod is 80.0 cm long and has mass 0.370 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05

nataly862011 [7]

Answer:

1.10 m/s

Explanation:

Linear speed is given by

$v=r\omega$

Kinetic energy is given by

$KE=0.5I\omega^{2}$

Potential energy

PE= mgh

From the law of conservation of energy, KE=PE hence

$0.5I\omega^{2}=mgh$ where m is mass, I is moment of inertia, $\omega$ is angular velocity, g is acceleration due to gravity and h is height

Substituting m2-m1 for m and 0.5l for h, $\frac {2v}{L}$ for $\omega$ we obtain

$0.5I(\frac {2v}{L})^{2}=0.5Lg(m2-m1)$

$(\frac {2v}{L})^{2}=\frac {gl(m2-m1)}{I}$ and making v the subject

$v^{2}=\frac {gl^{3}(m2-m1)}{4I}$

$v=\sqrt {\frac {gl^{3}(m2-m1)}{4I}}$

For the rod, moment of inertia $I=\frac {ML^{2}}{12}$ and for sphere $I=MR^{2}$ hence substituting 0.5L for R then $I=M(0.5L)^{2}$

For the sphere on the left hand side, moment of inertia I

$I=m1(0.5L)^{2}$ while for the sphere on right hand side, $I=m2(0.5L)^{2}$

The total moment of inertia is therefore given by adding

$I=\frac {ML^{2}}{12}+ m1(0.5L)^{2}+ m2(0.5L)^{2}=\frac {L^{2}(M+3m1+3m2)}{12}$

Substituting $\frac {L^{2}(M+3m1+3m2)}{12}$ for I in the equation $v=\sqrt {\frac {gL^{3}(m2-m1)}{4I}}$

Then we obtain

$v=\sqrt {\frac {gL^{3}(m2-m1)}{4(\frac {L^{2}(M+3m1+3m2)}{12})}}=\sqrt {\frac {3gL^{3}(m2-m1)}{L^{2}(M+3m1+3m2)}}$

This is the expression of linear speed. Substituting values given we get

$v=\sqrt {\frac {3*9.81*0.8^{3}(0.05-0.02)}{0.8^{2}(0.39+3(0.02)+3(0.05))}} \approx 1.08 m/s$

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2 years ago